Function local name binding from an outer scope - python

I need a way to "inject" names into a function from an outer code block, so they are accessible locally and they don't need to be specifically handled by the function's code (defined as function parameters, loaded from *args etc.)
The simplified scenario: providing a framework within which the users are able to define (with as little syntax as possible) custom functions to manipulate other objects of the framework (which are not necessarily global).
Ideally, the user defines
def user_func():
Mouse.eat(Cheese)
if Cat.find(Mouse):
Cat.happy += 1
Here Cat, Mouse and Cheese are framework objects that, for good reasons, cannot be bounded to the global namespace.
I want to write a wrapper for this function to behave like this:
def framework_wrap(user_func):
# this is a framework internal and has name bindings to Cat, Mouse and Cheese
def f():
inject(user_func, {'Cat': Cat, 'Mouse': Mouse, 'Cheese': Cheese})
user_func()
return f
Then this wrapper could be applied to all user-defined functions (as a decorator, by the user himself or automatically, although I plan to use a metaclass).
#framework_wrap
def user_func():
I am aware of the Python 3's nonlocal keyword, but I still consider ugly (from the framework's user perspective) to add an additional line:
nonlocal Cat, Mouse, Cheese
and to worry about adding every object he needs to this line.
Any suggestion is greatly appreciated.

The more I mess around with the stack, the more I wish I hadn't. Don't hack globals to do what you want. Hack bytecode instead. There's two ways that I can think of to do this.
1) Add cells wrapping the references that you want into f.func_closure. You have to reassemble the bytecode of the function to use LOAD_DEREF instead of LOAD_GLOBAL and generate a cell for each value. You then pass a tuple of the cells and the new code object to types.FunctionType and get a function with the appropriate bindings. Different copies of the function can have different local bindings so it should be as thread safe as you want to make it.
2) Add arguments for your new locals at the end of the functions argument list. Replace appropriate occurrences of LOAD_GLOBAL with LOAD_FAST. Then construct a new function by using types.FunctionType and passing in the new code object and a tuple of the bindings that you want as the default option. This is limited in the sense that python limits function arguments to 255 and it can't be used on functions that use variable arguments. None the less it struck me as the more challenging of the two so that's the one that I implemented (plus there's other stuff that can be done with this one). Again, you can either make different copies of the function with different bindings or call the function with the bindings that you want from each call location. So it too can be as thread safe as you want to make it.
import types
import opcode
# Opcode constants used for comparison and replacecment
LOAD_FAST = opcode.opmap['LOAD_FAST']
LOAD_GLOBAL = opcode.opmap['LOAD_GLOBAL']
STORE_FAST = opcode.opmap['STORE_FAST']
DEBUGGING = True
def append_arguments(code_obj, new_locals):
co_varnames = code_obj.co_varnames # Old locals
co_names = code_obj.co_names # Old globals
co_argcount = code_obj.co_argcount # Argument count
co_code = code_obj.co_code # The actual bytecode as a string
# Make one pass over the bytecode to identify names that should be
# left in code_obj.co_names.
not_removed = set(opcode.hasname) - set([LOAD_GLOBAL])
saved_names = set()
for inst in instructions(co_code):
if inst[0] in not_removed:
saved_names.add(co_names[inst[1]])
# Build co_names for the new code object. This should consist of
# globals that were only accessed via LOAD_GLOBAL
names = tuple(name for name in co_names
if name not in set(new_locals) - saved_names)
# Build a dictionary that maps the indices of the entries in co_names
# to their entry in the new co_names
name_translations = dict((co_names.index(name), i)
for i, name in enumerate(names))
# Build co_varnames for the new code object. This should consist of
# the entirety of co_varnames with new_locals spliced in after the
# arguments
new_locals_len = len(new_locals)
varnames = (co_varnames[:co_argcount] + new_locals +
co_varnames[co_argcount:])
# Build the dictionary that maps indices of entries in the old co_varnames
# to their indices in the new co_varnames
range1, range2 = xrange(co_argcount), xrange(co_argcount, len(co_varnames))
varname_translations = dict((i, i) for i in range1)
varname_translations.update((i, i + new_locals_len) for i in range2)
# Build the dictionary that maps indices of deleted entries of co_names
# to their indices in the new co_varnames
names_to_varnames = dict((co_names.index(name), varnames.index(name))
for name in new_locals)
if DEBUGGING:
print "injecting: {0}".format(new_locals)
print "names: {0} -> {1}".format(co_names, names)
print "varnames: {0} -> {1}".format(co_varnames, varnames)
print "names_to_varnames: {0}".format(names_to_varnames)
print "varname_translations: {0}".format(varname_translations)
print "name_translations: {0}".format(name_translations)
# Now we modify the actual bytecode
modified = []
for inst in instructions(code_obj.co_code):
# If the instruction is a LOAD_GLOBAL, we have to check to see if
# it's one of the globals that we are replacing. Either way,
# update its arg using the appropriate dict.
if inst[0] == LOAD_GLOBAL:
print "LOAD_GLOBAL: {0}".format(inst[1])
if inst[1] in names_to_varnames:
print "replacing with {0}: ".format(names_to_varnames[inst[1]])
inst[0] = LOAD_FAST
inst[1] = names_to_varnames[inst[1]]
elif inst[1] in name_translations:
inst[1] = name_translations[inst[1]]
else:
raise ValueError("a name was lost in translation")
# If it accesses co_varnames or co_names then update its argument.
elif inst[0] in opcode.haslocal:
inst[1] = varname_translations[inst[1]]
elif inst[0] in opcode.hasname:
inst[1] = name_translations[inst[1]]
modified.extend(write_instruction(inst))
code = ''.join(modified)
# Done modifying codestring - make the code object
return types.CodeType(co_argcount + new_locals_len,
code_obj.co_nlocals + new_locals_len,
code_obj.co_stacksize,
code_obj.co_flags,
code,
code_obj.co_consts,
names,
varnames,
code_obj.co_filename,
code_obj.co_name,
code_obj.co_firstlineno,
code_obj.co_lnotab)
def instructions(code):
code = map(ord, code)
i, L = 0, len(code)
extended_arg = 0
while i < L:
op = code[i]
i+= 1
if op < opcode.HAVE_ARGUMENT:
yield [op, None]
continue
oparg = code[i] + (code[i+1] << 8) + extended_arg
extended_arg = 0
i += 2
if op == opcode.EXTENDED_ARG:
extended_arg = oparg << 16
continue
yield [op, oparg]
def write_instruction(inst):
op, oparg = inst
if oparg is None:
return [chr(op)]
elif oparg <= 65536L:
return [chr(op), chr(oparg & 255), chr((oparg >> 8) & 255)]
elif oparg <= 4294967296L:
return [chr(opcode.EXTENDED_ARG),
chr((oparg >> 16) & 255),
chr((oparg >> 24) & 255),
chr(op),
chr(oparg & 255),
chr((oparg >> 8) & 255)]
else:
raise ValueError("Invalid oparg: {0} is too large".format(oparg))
if __name__=='__main__':
import dis
class Foo(object):
y = 1
z = 1
def test(x):
foo = Foo()
foo.y = 1
foo = x + y + z + foo.y
print foo
code_obj = append_arguments(test.func_code, ('y',))
f = types.FunctionType(code_obj, test.func_globals, argdefs=(1,))
if DEBUGGING:
dis.dis(test)
print '-'*20
dis.dis(f)
f(1)
Note that a whole branch of this code (that relating to EXTENDED_ARG) is untested but that for common cases, it seems to be pretty solid. I'll be hacking on it and am currently writing some code to validate the output. Then (when I get around to it) I'll run it against the whole standard library and fix any bugs.
I'll also probably be implementing the first option as well.

Edited answer -- restores namespace dict after calling user_func()
Tested using Python 2.7.5 and 3.3.2
File framework.py:
# framework objects
class Cat: pass
class Mouse: pass
class Cheese: pass
_namespace = {'Cat':Cat, 'Mouse':Mouse, 'Cheese':Cheese } # names to be injected
# framework decorator
from functools import wraps
def wrap(f):
func_globals = f.func_globals if hasattr(f,'func_globals') else f.__globals__
#wraps(f)
def wrapped(*args, **kwargs):
# determine which names in framework's _namespace collide and don't
preexistent = set(name for name in _namespace if name in func_globals)
nonexistent = set(name for name in _namespace if name not in preexistent)
# save any preexistent name's values
f.globals_save = {name: func_globals[name] for name in preexistent}
# temporarily inject framework's _namespace
func_globals.update(_namespace)
retval = f(*args, **kwargs) # call function and save return value
# clean up function's namespace
for name in nonexistent:
del func_globals[name] # remove those that didn't exist
# restore the values of any names that collided
func_globals.update(f.globals_save)
return retval
return wrapped
Example usage:
from __future__ import print_function
import framework
class Cat: pass # name that collides with framework object
#framework.wrap
def user_func():
print('in user_func():')
print(' Cat:', Cat)
print(' Mouse:', Mouse)
print(' Cheese:', Cheese)
user_func()
print()
print('after user_func():')
for name in framework._namespace:
if name in globals():
print(' {} restored to {}'.format(name, globals()[name]))
else:
print(' {} not restored, does not exist'.format(name))
Output:
in user_func():
Cat: <class 'framework.Cat'>
Mouse: <class 'framework.Mouse'>
Cheese: <class 'framework.Cheese'>
after user_func():
Cheese not restored, does not exist
Mouse not restored, does not exist
Cat restored to <class '__main__.Cat'>

Sounds like you maybe want to be using exec code in dict, where code is the user's function and dict is a dictionary you provide which can
be pre-filled with references to objects that the user code should be able to use
store any functions or variables declared by the user's code for later use by your framework.
Docs for exec: http://docs.python.org/reference/simple_stmts.html#the-exec-statement
However, I'm pretty sure that this would only work if the user's code is being brought in as a string and you need to exec it. If the function is already compiled, it will already have its global bindings set. So doing something like exec "user_func(*args)" in framework_dict won't work, because user_func's globals are already set to the module in which it was defined.
Since func_globals is readonly, I think you'll have to do something like what martineau suggests in order to modify the function globals.
I think it likely (unless you're doing something unprecedentedly awesome, or I'm missing some critical subtlety) that you probably would be better off putting your framework objects into a module, and then have the user code import that module. Module variables can be reassigned to or mutated or accessed quite readily by code that's been defined outside of that module, once the module has been imported.
I think this would be better for code readibility also, because user_func will end up having explicit namespacing for Cat, Dog, etc. rather than readers unfamiliar with your framework having to wonder where they came from. E.G. animal_farm.Mouse.eat(animal_farm.Cheese), or maybe lines like
from animal_farm import Goat
cheese = make_cheese(Goat().milk())
If you are doing something unprecedently awesome, I think you'll need to use the C API to pass arguments to a code object. It looks like the function PyEval_EvalCodeEx is the one you want.

If your application is strictly Python 3, I don't see how using Python 3's nonlocal is any uglier than writing a decorator to manipulate function's local namespace. I say give the nonlocal solution a try or rethink this strategy.

Related

Trouble naming csv file I'm writing a dataframe to in pandas in Python [duplicate]

I already read How to get a function name as a string?.
How can I do the same for a variable? As opposed to functions, Python variables do not have the __name__ attribute.
In other words, if I have a variable such as:
foo = dict()
foo['bar'] = 2
I am looking for a function/attribute, e.g. retrieve_name() in order to create a DataFrame in Pandas from this list, where the column names are given by the names of the actual dictionaries:
# List of dictionaries for my DataFrame
list_of_dicts = [n_jobs, users, queues, priorities]
columns = [retrieve_name(d) for d in list_of_dicts]
With Python 3.8 one can simply use f-string debugging feature:
>>> foo = dict()
>>> f'{foo=}'.split('=')[0]
'foo'
One drawback of this method is that in order to get 'foo' printed you have to add f'{foo=}' yourself. In other words, you already have to know the name of the variable. In other words, the above code snippet is exactly the same as just
>>> 'foo'
Even if variable values don't point back to the name, you have access to the list of every assigned variable and its value, so I'm astounded that only one person suggested looping through there to look for your var name.
Someone mentioned on that answer that you might have to walk the stack and check everyone's locals and globals to find foo, but if foo is assigned in the scope where you're calling this retrieve_name function, you can use inspect's current frame to get you all of those local variables.
My explanation might be a little bit too wordy (maybe I should've used a "foo" less words), but here's how it would look in code (Note that if there is more than one variable assigned to the same value, you will get both of those variable names):
import inspect
x, y, z = 1, 2, 3
def retrieve_name(var):
callers_local_vars = inspect.currentframe().f_back.f_locals.items()
return [var_name for var_name, var_val in callers_local_vars if var_val is var]
print(retrieve_name(y))
If you're calling this function from another function, something like:
def foo(bar):
return retrieve_name(bar)
foo(baz)
And you want the baz instead of bar, you'll just need to go back a scope further. This can be done by adding an extra .f_back in the caller_local_vars initialization.
See an example here: ideone
The only objects in Python that have canonical names are modules, functions, and classes, and of course there is no guarantee that this canonical name has any meaning in any namespace after the function or class has been defined or the module imported. These names can also be modified after the objects are created so they may not always be particularly trustworthy.
What you want to do is not possible without recursively walking the tree of named objects; a name is a one-way reference to an object. A common or garden-variety Python object contains no references to its names. Imagine if every integer, every dict, every list, every Boolean needed to maintain a list of strings that represented names that referred to it! It would be an implementation nightmare, with little benefit to the programmer.
TL;DR
Use the Wrapper helper from python-varname:
from varname.helpers import Wrapper
foo = Wrapper(dict())
# foo.name == 'foo'
# foo.value == {}
foo.value['bar'] = 2
For list comprehension part, you can do:
n_jobs = Wrapper(<original_value>)
users = Wrapper(<original_value>)
queues = Wrapper(<original_value>)
priorities = Wrapper(<original_value>)
list_of_dicts = [n_jobs, users, queues, priorities]
columns = [d.name for d in list_of_dicts]
# ['n_jobs', 'users', 'queues', 'priorities']
# REMEMBER that you have to access the <original_value> by d.value
I am the author of the python-varname package. Please let me know if you have any questions or you can submit issues on Github.
The long answer
Is it even possible?
Yes and No.
We are retrieving the variable names at runtime, so we need a function to be called to enable us to access the previous frames to retrieve the variable names. That's why we need a Wrapper there. In that function, at runtime, we are parsing the source code/AST nodes in the previous frames to get the exact variable name.
However, the source code/AST nodes in the previous frames are not always available, or they could be modified by other environments (e.g: pytest's assert statement). One simple example is that the codes run via exec(). Even though we are still able to retrieve some information from the bytecode, it needs too much effort and it is also error-prone.
How to do it?
First of all, we need to identify which frame the variable is given. It's not always simply the direct previous frame. For example, we may have another wrapper for the function:
from varname import varname
def func():
return varname()
def wrapped():
return func()
x = wrapped()
In the above example, we have to skip the frame inside wrapped to get to the right frame x = wrapped() so that we are able to locate x. The arguments frame and ignore of varname allow us to skip some of these intermediate frames. See more details in the README file and the API docs of the package.
Then we need to parse the AST node to locate where the variable is assigned value (function call) to. It's not always just a simple assignment. Sometimes there could be complex AST nodes, for example, x = [wrapped()]. We need to identify the correct assignment by traversing the AST tree.
How reliable is it?
Once we identify the assignment node, it is reliable.
varname is all depending on executing package to look for the node. The node executing detects is ensured to be the correct one (see also this).
It partially works with environments where other AST magics apply, including pytest, ipython, macropy, birdseye, reticulate with R, etc. Neither executing nor varname is 100% working with those environments.
Do we need a package to do it?
Well, yes and no, again.
If your scenario is simple, the code provided by #juan Isaza or #scohe001 probably is enough for you to work with the case where a variable is defined at the direct previous frame and the AST node is a simple assignment. You just need to go one frame back and retrieve the information there.
However, if the scenario becomes complicated, or we need to adopt different application scenarios, you probably need a package like python-varname, to handle them. These scenarios may include to:
present more friendly messages when the source code is not available or AST nodes are not accessible
skip intermediate frames (allows the function to be wrapped or called in other intermediate frames)
automatically ignores calls from built-in functions or libraries. For example: x = str(func())
retrieve multiple variable names on the left-hand side of the assignment
etc.
How about the f-string?
Like the answer provided by #Aivar Paalberg. It's definitely fast and reliable. However, it's not at runtime, meaning that you have to know it's foo before you print the name out. But with varname, you don't have to know that variable is coming:
from varname import varname
def func():
return varname()
# In external uses
x = func() # 'x'
y = func() # 'y'
Finally
python-varname is not only able to detect the variable name from an assignment, but also:
Retrieve variable names directly, using nameof
Detect next immediate attribute name, using will
Fetch argument names/sources passed to a function using argname
Read more from its documentation.
However, the final word I want to say is that, try to avoid using it whenever you can.
Because you can't make sure that the client code will run in an environment where the source node is available or AST node is accessible. And of course, it costs resources to parse the source code, identify the environment, retrieve the AST nodes and evaluate them when needed.
On python3, this function will get the outer most name in the stack:
import inspect
def retrieve_name(var):
"""
Gets the name of var. Does it from the out most frame inner-wards.
:param var: variable to get name from.
:return: string
"""
for fi in reversed(inspect.stack()):
names = [var_name for var_name, var_val in fi.frame.f_locals.items() if var_val is var]
if len(names) > 0:
return names[0]
It is useful anywhere on the code. Traverses the reversed stack looking for the first match.
I don't believe this is possible. Consider the following example:
>>> a = []
>>> b = a
>>> id(a)
140031712435664
>>> id(b)
140031712435664
The a and b point to the same object, but the object can't know what variables point to it.
def name(**variables):
return [x for x in variables]
It's used like this:
name(variable=variable)
>> my_var = 5
>> my_var_name = [ k for k,v in locals().items() if v == my_var][0]
>> my_var_name
'my_var'
In case you get an error if myvar points to another variable, try this (suggested by #mherzog)-
>> my_var = 5
>> my_var_name = [ k for k,v in locals().items() if v is my_var][0]
>> my_var_name
'my_var'
locals() - Return a dictionary containing the current scope's local variables.
by iterating through this dictionary we can check the key which has a value equal to the defined variable, just extracting the key will give us the text of variable in string format.
from (after a bit changes)
https://www.tutorialspoint.com/How-to-get-a-variable-name-as-a-string-in-Python
I wrote the package sorcery to do this kind of magic robustly. You can write:
from sorcery import dict_of
columns = dict_of(n_jobs, users, queues, priorities)
and pass that to the dataframe constructor. It's equivalent to:
columns = dict(n_jobs=n_jobs, users=users, queues=queues, priorities=priorities)
Here's one approach. I wouldn't recommend this for anything important, because it'll be quite brittle. But it can be done.
Create a function that uses the inspect module to find the source code that called it. Then you can parse the source code to identify the variable names that you want to retrieve. For example, here's a function called autodict that takes a list of variables and returns a dictionary mapping variable names to their values. E.g.:
x = 'foo'
y = 'bar'
d = autodict(x, y)
print d
Would give:
{'x': 'foo', 'y': 'bar'}
Inspecting the source code itself is better than searching through the locals() or globals() because the latter approach doesn't tell you which of the variables are the ones you want.
At any rate, here's the code:
def autodict(*args):
get_rid_of = ['autodict(', ',', ')', '\n']
calling_code = inspect.getouterframes(inspect.currentframe())[1][4][0]
calling_code = calling_code[calling_code.index('autodict'):]
for garbage in get_rid_of:
calling_code = calling_code.replace(garbage, '')
var_names, var_values = calling_code.split(), args
dyn_dict = {var_name: var_value for var_name, var_value in
zip(var_names, var_values)}
return dyn_dict
The action happens in the line with inspect.getouterframes, which returns the string within the code that called autodict.
The obvious downside to this sort of magic is that it makes assumptions about how the source code is structured. And of course, it won't work at all if it's run inside the interpreter.
>>> locals()['foo']
{}
>>> globals()['foo']
{}
If you wanted to write your own function, it could be done such that you could check for a variable defined in locals then check globals. If nothing is found you could compare on id() to see if the variable points to the same location in memory.
If your variable is in a class, you could use className.dict.keys() or vars(self) to see if your variable has been defined.
This function will print variable name with its value:
import inspect
def print_this(var):
callers_local_vars = inspect.currentframe().f_back.f_locals.items()
print(str([k for k, v in callers_local_vars if v is var][0])+': '+str(var))
***Input & Function call:***
my_var = 10
print_this(my_var)
***Output**:*
my_var: 10
I have a method, and while not the most efficient...it works! (and it doesn't involve any fancy modules).
Basically it compares your Variable's ID to globals() Variables' IDs, then returns the match's name.
def getVariableName(variable, globalVariables=globals().copy()):
""" Get Variable Name as String by comparing its ID to globals() Variables' IDs
args:
variable(var): Variable to find name for (Obviously this variable has to exist)
kwargs:
globalVariables(dict): Copy of the globals() dict (Adding to Kwargs allows this function to work properly when imported from another .py)
"""
for globalVariable in globalVariables:
if id(variable) == id(globalVariables[globalVariable]): # If our Variable's ID matches this Global Variable's ID...
return globalVariable # Return its name from the Globals() dict
In Python, the def and class keywords will bind a specific name to the object they define (function or class). Similarly, modules are given a name by virtue of being called something specific in the filesystem. In all three cases, there's an obvious way to assign a "canonical" name to the object in question.
However, for other kinds of objects, such a canonical name may simply not exist. For example, consider the elements of a list. The elements in the list are not individually named, and it is entirely possible that the only way to refer to them in a program is by using list indices on the containing list. If such a list of objects was passed into your function, you could not possibly assign meaningful identifiers to the values.
Python doesn't save the name on the left hand side of an assignment into the assigned object because:
It would require figuring out which name was "canonical" among multiple conflicting objects,
It would make no sense for objects which are never assigned to an explicit variable name,
It would be extremely inefficient,
Literally no other language in existence does that.
So, for example, functions defined using lambda will always have the "name" <lambda>, rather than a specific function name.
The best approach would be simply to ask the caller to pass in an (optional) list of names. If typing the '...','...' is too cumbersome, you could accept e.g. a single string containing a comma-separated list of names (like namedtuple does).
I think it's so difficult to do this in Python because of the simple fact that you never will not know the name of the variable you're using. So, in his example, you could do:
Instead of:
list_of_dicts = [n_jobs, users, queues, priorities]
dict_of_dicts = {"n_jobs" : n_jobs, "users" : users, "queues" : queues, "priorities" : priorities}
Many of the answers return just one variable name. But that won't work well if more than one variable have the same value. Here's a variation of Amr Sharaki's answer which returns multiple results if more variables have the same value.
def getVariableNames(variable):
results = []
globalVariables=globals().copy()
for globalVariable in globalVariables:
if id(variable) == id(globalVariables[globalVariable]):
results.append(globalVariable)
return results
a = 1
b = 1
getVariableNames(a)
# ['a', 'b']
just another way to do this based on the content of input variable:
(it returns the name of the first variable that matches to the input variable, otherwise None. One can modify it to get all variable names which are having the same content as input variable)
def retrieve_name(x, Vars=vars()):
for k in Vars:
if isinstance(x, type(Vars[k])):
if x is Vars[k]:
return k
return None
If the goal is to help you keep track of your variables, you can write a simple function that labels the variable and returns its value and type. For example, suppose i_f=3.01 and you round it to an integer called i_n to use in a code, and then need a string i_s that will go into a report.
def whatis(string, x):
print(string+' value=',repr(x),type(x))
return string+' value='+repr(x)+repr(type(x))
i_f=3.01
i_n=int(i_f)
i_s=str(i_n)
i_l=[i_f, i_n, i_s]
i_u=(i_f, i_n, i_s)
## make report that identifies all types
report='\n'+20*'#'+'\nThis is the report:\n'
report+= whatis('i_f ',i_f)+'\n'
report+=whatis('i_n ',i_n)+'\n'
report+=whatis('i_s ',i_s)+'\n'
report+=whatis('i_l ',i_l)+'\n'
report+=whatis('i_u ',i_u)+'\n'
print(report)
This prints to the window at each call for debugging purposes and also yields a string for the written report. The only downside is that you have to type the variable twice each time you call the function.
I am a Python newbie and found this very useful way to log my efforts as I program and try to cope with all the objects in Python. One flaw is that whatis() fails if it calls a function described outside the procedure where it is used. For example, int(i_f) was a valid function call only because the int function is known to Python. You could call whatis() using int(i_f**2), but if for some strange reason you choose to define a function called int_squared it must be declared inside the procedure where whatis() is used.
Maybe this could be useful:
def Retriever(bar):
return (list(globals().keys()))[list(map(lambda x: id(x), list(globals().values()))).index(id(bar))]
The function goes through the list of IDs of values from the global scope (the namespace could be edited), finds the index of the wanted/required var or function based on its ID, and then returns the name from the list of global names based on the acquired index.
Whenever I have to do it, mostly while communicating json schema and constants with the frontend I define a class as follows
class Param:
def __init__(self, name, value):
self.name = name
self.value = value
Then define the variable with name and value.
frame_folder_count = Param({'name':'frame_folder_count', 'value':10})
Now you can access the name and value using the object.
>>> frame_folder_count.name
'frame_folder_count'
>>> def varname(v, scope=None):
d = globals() if not scope else vars(scope); return [k for k in d if d[k] == v]
...
>>> d1 = {'a': 'ape'}; d2 = {'b': 'bear'}; d3 = {'c': 'cat'}
>>> ld = [d1, d2, d3]
>>> [varname(d) for d in ld]
[['d1'], ['d2'], ['d3']]
>>> d5 = d3
>>> [varname(d) for d in ld]
[['d1'], ['d2'], ['d3', 'd5']]
>>> def varname(v, scope=None):
d = globals() if not scope else vars(scope); return [k for k in d if d[k] is v]
...
>>> [varname(d) for d in ld]
[['d1'], ['d2'], ['d3', 'd5']]
As you see and is noted here, there can be multiple variables with the same value or even address, so using a wrapper to keep the names with the data is best.
Following method will not return the name of variable but using this method you can create data frame easily if variable is available in global scope.
class CustomDict(dict):
def __add__(self, other):
return CustomDict({**self, **other})
class GlobalBase(type):
def __getattr__(cls, key):
return CustomDict({key: globals()[key]})
def __getitem__(cls, keys):
return CustomDict({key: globals()[key] for key in keys})
class G(metaclass=GlobalBase):
pass
x, y, z = 0, 1, 2
print('method 1:', G['x', 'y', 'z']) # Outcome: method 1: {'x': 0, 'y': 1, 'z': 2}
print('method 2:', G.x + G.y + G.z) # Outcome: method 2: {'x': 0, 'y': 1, 'z': 2}
A = [0, 1]
B = [1, 2]
pd.DataFrame(G.A + G.B) # It will return a data frame with A and B columns
Some of the previous cases would fail if there are two variables with the same value. So it is convenient to alert it:
Defining function:
# Variable to string of variable name
def var_name(variable,i=0):
results = []
for name in globals():
if eval(name) == variable:
results.append(name)
if len(results) > 1:
print('Warning:' )
print(' var_name() has found',len(results), 'possible outcomes.')
print(' Please choose the suitable parameter "i". Where "i" is the index')
print(' that matches your choice from the list below.')
print(' ',results) ; print('')
return results[i]
Use:
var_1 = 10
var_name(var_1) # Output will be "var_1"
If you have 2 variables with the same value like var_1 = 8 and var_2 = 8, then a warning will appear.
var_1 = 8
var_2 = 8
var_name(var_2) # Output will be "var_1" too but Warning will appear
You can get your variable as kwargs and return it as string:
var=2
def getVarName(**kwargs):
return list(kwargs.keys())[0]
print (getVarName(var = var))
Note: variable name must be equal to itself.
I try to get name from inspect locals, but it cann't process var likes a[1], b.val.
After it, I got a new idea --- get var name from the code, and I try it succ!
code like below:
#direct get from called function code
def retrieve_name_ex(var):
stacks = inspect.stack()
try:
func = stacks[0].function
code = stacks[1].code_context[0]
s = code.index(func)
s = code.index("(", s + len(func)) + 1
e = code.index(")", s)
return code[s:e].strip()
except:
return ""
You can try the following to retrieve the name of a function you defined (does not work for built-in functions though):
import re
def retrieve_name(func):
return re.match("<function\s+(\w+)\s+at.*", str(func)).group(1)
def foo(x):
return x**2
print(retrieve_name(foo))
# foo
When finding the name of a variable from its value,
you may have several variables equal to the same value,
for example var1 = 'hello' and var2 = 'hello'.
My solution:
def find_var_name(val):
dict_list = []
global_dict = dict(globals())
for k, v in global_dict.items():
dict_list.append([k, v])
return [item[0] for item in dict_list if item[1] == val]
var1 = 'hello'
var2 = 'hello'
find_var_name('hello')
Outputs
['var1', 'var2']
Compressed version of iDilip's answer:
import inspect
def varname(x):
return [k for k,v in inspect.currentframe().f_back.f_locals.items() if v is x][0]
hi = 123
print(varname(hi))
It's totally possible to get the name of an instance variable, so long as it is the property of a class.
I got this from Effective Python by Brett Slatkin. Hope it helps someone:
The class must implement the get, set, and set_name dunder methods, which are part of the "Descriptor Protocol"
This worked when I ran it:
class FieldThatKnowsItsName():
def __init__(self):
self.name = None
self._value= None
self.owner = None
def __set_name__(self, owner, name):
self.name = name
self.owner = owner
self.owner.fields[self.name] = self
def __get__(self, instance, instance_type):
return self
def __set__(self, instance, value):
self = value
class SuperTable:
fields = {}
field_1=FieldThatKnowsItsName()
field_2=FieldThatKnowsItsName()
table = SuperTable()
print(table.field_1.name)
print(table.field_2.name)
You can then add methods and or extend your datatype as you like.
As a bonus, the set_name(self, owner, name) dunder also passes the parent instance, so the Field class instance can register itself with the parent.
I got this from Effective Python by Brett Slatkin. It took a while to figure out how to implement.
How can I do the same for a variable? As opposed to functions, Python variables do not have the __name__ attribute.
The problem comes up because you are confused about terminology, semantics or both.
"variables" don't belong in the same category as "functions". A "variable" is not a thing that takes up space in memory while the code is running. It is just a name that exists in your source code - so that when you're writing the code, you can explain which thing you're talking about. Python uses names in the source code to refer to (i.e., give a name to) values. (In many languages, a variable is more like a name for a specific location in memory where the value will be stored. But Python's names actually name the thing in question.)
In Python, a function is a value. (In some languages, this is not the case; although there are bytes of memory used to represent the actual executable code, it isn't a discrete chunk of memory that your program logic gets to interact with directly.) In Python, every value is an object, meaning that you can assign names to it freely, pass it as an argument, return it from a function, etc. (In many languages, this is not the case.) Objects in Python have attributes, which are the things you access using the . syntax. Functions in Python have a __name__ attribute, which is assigned when the function is created. Specifically, when a def statement is executed (in most languages, creation of a function works quite differently), the name that appears after def is used as a value for the __name__ attribute, and also, independently, as a variable name that will get the function object assigned to it.
But most objects don't have an attribute like that.
In other words, if I have a variable such as:
That's the thing: you don't "have" the variable in the sense that you're thinking of. You have the object that is named by that variable. Anything else depends on the information incidentally being stored in some other object - such as the locals() of the enclosing function. But it would be better to store the information yourself. Instead of relying on a variable name to carry information for you, explicitly build the mapping between the string name you want to use for the object, and the object itself.

Python - How to obtain a dictionary(or even a list) of variables in a specific scope; more specific than locals/globals()

So, the title pretty much says it all.
for instance let's look at the below example code:
## How can I obtain a dict/list (like locals()) of all the variables in second and/or third layer scopes via a command
# coming from the first layer?
## Or another example would be how could I obtain the variables "locals() style" from the thirdlayer via a
# command from the second layer?
# essentially can a parent function/class access a list/dict of a child function
# or class??
def firstLayer():
a = 4.7
q = locals()
print(q)
# local vars of 1st layer
def secondlayer():
b = 7
r = locals()
print(r)
# local vars of 2nd layer
def thirdlayer():
c = False
s = locals()
i = globals()
print('c:\n', c, "\nglobals from 3rd layer:\n\t", i)
# local vars of 3rd layer
thirdlayer()
secondlayer()
firstLayer()
sample_var = globals()
print(sample_var)
# returns the list of global variables
to reiterate what I said in the comments in the code, essentially is their any way I can get a list of all the variables local to a 'child' scope? I know functions are shut off, but if their is no way to do this is their any more complicated code that could achieve this and I could integrate it into a function or class if necessary.
EDIT:
To elaborate further; here's the situation i'm in.
def varsfunc():
font1 = "Harlow Solid"
grey = '#454545'
font2 = 'Mistral'
font3 = 'Italic 35px Times New Roman'
pnk = 'pink'
grn = 'green'
return locals()
Essentially, I am creating a module and the user must create some type of function that they list all of they variables they would like to declare to be used to modify a css file. Essentially, I would like to allow the user to not have to type "return locals()". I want to achieve it by having the end-users wrap the above example function in a decorator that will do the equivalent of returning locals() of the exact scope I want. The decorator does not work for me because it is in an outer scope.
TO BE EVEN MORE CLEAR:
I need a decorator/function that wraps another function(i.e. a decorator), that can access and create a list of a child element.
def module_decorator_func_thing():
r = command_that_acts_like_locals()_but_for_child_scopes
def user_var_list():
font1 = 'green'
font2 = 'pink'
# back in "module_decorator_func_thing"'s scope
print(r) # this variable should contain only a dict/list containing the
# the following:
# r = {'font1': 'green', 'font2': 'pink')
currently users need to do this:
def vars_func_container():
font1 = 'green'
font2 = 'pink'
return locals() # <---- I want the user to not have to type this and for
# a function decorator to take care of it instead possibly.
Info for #aguy and others wishing for more info.
The dictionary/list that I am obtaining via your guys' tips will be sent to this function to do the real job of the program.
(If I were to start using lists, i'd need to convert to a dictionary but that's no problem for me to solve.)
The dict of variables is used with this function to "compile/compyle"(Pun on the word 'Python' + 'compile) and is insert in the "variables" parameter. e.g. you execute the function like this.
compyle("My sample title", return_stylesheet_from_func(*insert .css filename),
return_variables_from_function(*insert function containing variables*), "**True/False to turn on compilation**",
"**True/False to turn on annotations/suggestions**")
def compyle(title, style_sheet, variables, boolean=False, boolean2=True):
"""
:param title: The name you wish your .css file to be named.
:param style_sheet: The name of the multi-line string that will compose your .css file
:param variables: The name of the dictionary containing your .pcss variables
:param boolean: A.K.A the "Compiler Parameter" - Turns the compiler on or off
:param boolean2: A.K.A the "Annotation Parameter" - Turns annotations on or off
:return: returns compiled .pcss text as normal .css style text to be utilized with .html
"""
# -----------------------------------
file_name = title + ".css"
replace_num = len(variables.keys())
counter = replace_num
content = style_sheet
# -----------------------------------
# add theme support with namedtuple's formatted to mimic structs in C/C++
# this will be a major feature update as well as a nice way to allow the future prospect of integrating C/C++ into
# the compiler. Info: https://stackoverflow.com/questions/35988/c-like-structures-in-python
for k, v in variables.items():
counter -= 1
content = content.replace(k, v, replace_num)
if counter == 0:
break
else:
pass
looped_content = str(content)
id_content = looped_content.replace("hash_", "#")
output = id_content.replace("dot_", ".")
if boolean is True:
if boolean2 is True:
output = " /* --- Pyle Sheet --- */\n" + output
with open(file_name, 'w') as writ:
writ.write(output)
writ.close()
print('compiled successfully; The file was saved as ' + "\"" + file_name + "\".")
elif boolean2 is False:
pass
else:
logging.warning("An Error Occurred - see module, documentation, or online Q&A for assistance.")
elif boolean is False:
if boolean2 is True:
print('compiled successfully; The file ' + "\"" + file_name + "\"" + "was not saved/created.")
elif boolean2 is False:
pass
else:
logging.warning("An Error Occurred - see module, documentation, or online Q&A for assistance.")
else:
logging.warning('An Error Occurred with the Compile Parameter (See: boolean in pyle_sheets source file) - \ '
'see module, documentation, or online Q&A for assistance.')
I can't see any way to do this without getting pretty deep; what follows is the simplest solution I've come up with.
how it works
Using the ast module, we go through the code of the given function and find all the assignments. These are evaluated in a given namespace and this namespace is returned.
the code
import ast
import functools
import inspect
def returnAssignments(f):
#functools.wraps(f)
def returner():
assignments = dict()
for node in ast.walk(ast.parse(inspect.getsource(f))):
if isinstance(node, ast.Assign):
exec(compile(ast.Module([node]), '<ast>', 'exec'),
globals(),
assignments)
return assignments
return returner
usage
from ra import returnAssignments
#returnAssignments
def foo():
this = 'something'
that = 37
the_other = object()
print(foo())
output
rat#pandion:~/tmp$ python test.py
{'this': 'something', 'that': 37, 'the_other': <object object at 0x10205b130>}
I wonder if such a crude solution as I provide here might be useful to you. Note that I haven't tested it on all cases, so it might be a bit rough. Also, it returns everything as a string, a behavior which you might want to further change.
Define the function:
def get_local_vars_from_function(f):
import inspect
s = inspect.getsourcelines(f)[0]
d = {}
for l in s:
if '=' in l:
var, val = l.split('=')
var = var.strip()
val = val.strip()
d[var] = val
return d
Then to use it:
In[91]: get_local_vars_from_function(user_var_list)
Out[91]: {'font1': "'green'", 'font2': "'pink'"}

How to create variables local to with statement?

I would like to create temporary variables visible in a limited scope.
It seems likely to me that you can do this with a "with" statement, and I would think there is a construct that makes it easy to do, but I cannot seem to find it.
I would like something like the following (but it does not work this way of course):
pronunciation = "E_0 g z #_1 m p l"
# ...
with pronunciation.split() as phonemes:
if len(phonemes) > 2 or phonemes[0].startswith('E'):
condition = 1
elif len(phonemes) < 3 and phonemes[-1] == '9r':
condition = 2
So is there a simple way to make this work, using built-ins?
Thanks!
Python creates local variables with function scope (once a name is used it stays alive until the end of the function).
If you really want to limit scope then "del <var>" when you want it explicitly discarded, or create separate function to act as a container for a more limited scope.
You can create a method
def process_pronunciation(pronunciation):
phonemes = pronunciation.split()
if len(phonemes) > 2 or phonemes[0].startswith('E'):
condition = 1
elif len(phonemes) < 3 and phonemes[-1] == '9r':
condition = 2
return condition
When you call the method, the local variable phonemes won't be available in the global namespace.
pronunciation = "E_0 g z #_1 m p l"
condition = process_phonemes(pronunciation)
You could do it with with, but I don't think it's worth the trouble. Basically (in a python function) you have two scopes - global or local, that's it. If you want a symbol to have a lifespan shorter than the function you'll have to delete it afterwards using del. You could define your own context manager to make this happen:
class TempVar:
def __init__(self, loc, name, val):
self.loc = loc
self.name = name
self.val
def __enter__(self):
if self.name in self.loc:
self.old = self.loc[self.name]
self.loc[self.name] = self.val
def __exit__(self, *exc):
if hasattr(self, "old"):
self.loc[self.name] = self.old
else:
del self.loc[self.name]
then you can use it to get a temporary variable:
with TempVar(locals(), "tempVar", 42):
print(tempVar)
The working is that it modifies the dict containing local variables (which is supplied to the constructor via locals()) on entry and restoring it when leaving. Please note that this relies on that modifying the result returned by locals() actually modifies the local namespace - the specification does NOT guarantee this behaviour.
Another (and safer) alternative that was pointed out is that you could define a separate function which would have it's own scope. Remember it's perfectly legal to nest functions. For example:
def outer():
def inner(tempVar):
# here tempVar is in scope
print(tempVar)
inner(tempVar = 42)
# here tempVar is out of scope
with statement does not have its own scope , it uses the surrounding scope (like if the with statement is directly inside the script , and not within any function, it uses global namespace , if the with statement is used inside a function, it uses the function's namespace(scope)).
If you want the statements inside a with block to run in its own local scope, one possible way would be to move the logic to a function , that way the logic would be running in its own scope (and not the surrounding scope of with.
Example -
def function_for_with(f):
#Do something.
with pronunciation.split() as phonemes:
function_for_with(phonemes)
Please note, the above will not stop phonemes from being defined in the surrounding scope.
If you want that as well (move the phonemes into its own scope), you can move the complete with statement inside a function. Example -
def function_with(pronunciation):
with pronunciation.split() as phonemes:
#do stuff
pronunciation = "E_0 g z #_1 m p l"
function_with(pronunciation)
Expanding on #skyking's answer, here's an even more magical implementation of the same idea that reads almost exactly like you wrote. Introducing: the with var statement!1
class var:
def __init__(self, value):
import inspect
self.scope = inspect.currentframe().f_back.f_locals
self.old_vars = set(self.scope.keys())
self.value = value
def __enter__(self):
return self.value
def __exit__(self, type, value, traceback):
for name in set(self.scope.keys()) - self.old_vars:
del self.scope[name]
### Usage:
line = 'a b c'
with var (line.split()) as words:
# Prints "['a', 'b', 'c']"
print(words)
# Causes a NameError
print(words)
It does all the nasty extracting of local variables and names for you! How swell. If you space it quirkily like I did and hide the definition in a from boring_stuff import * statement, you can even pretend var is a keyword to all of your confused co-workers.
[1] If you actually use this, the ghost of a dead parrot will probably haunt you forever. The other answers provide much saner solutions; this one is more of a joke.

Getting the name of a variable as a string

I already read How to get a function name as a string?.
How can I do the same for a variable? As opposed to functions, Python variables do not have the __name__ attribute.
In other words, if I have a variable such as:
foo = dict()
foo['bar'] = 2
I am looking for a function/attribute, e.g. retrieve_name() in order to create a DataFrame in Pandas from this list, where the column names are given by the names of the actual dictionaries:
# List of dictionaries for my DataFrame
list_of_dicts = [n_jobs, users, queues, priorities]
columns = [retrieve_name(d) for d in list_of_dicts]
With Python 3.8 one can simply use f-string debugging feature:
>>> foo = dict()
>>> f'{foo=}'.split('=')[0]
'foo'
One drawback of this method is that in order to get 'foo' printed you have to add f'{foo=}' yourself. In other words, you already have to know the name of the variable. In other words, the above code snippet is exactly the same as just
>>> 'foo'
Even if variable values don't point back to the name, you have access to the list of every assigned variable and its value, so I'm astounded that only one person suggested looping through there to look for your var name.
Someone mentioned on that answer that you might have to walk the stack and check everyone's locals and globals to find foo, but if foo is assigned in the scope where you're calling this retrieve_name function, you can use inspect's current frame to get you all of those local variables.
My explanation might be a little bit too wordy (maybe I should've used a "foo" less words), but here's how it would look in code (Note that if there is more than one variable assigned to the same value, you will get both of those variable names):
import inspect
x, y, z = 1, 2, 3
def retrieve_name(var):
callers_local_vars = inspect.currentframe().f_back.f_locals.items()
return [var_name for var_name, var_val in callers_local_vars if var_val is var]
print(retrieve_name(y))
If you're calling this function from another function, something like:
def foo(bar):
return retrieve_name(bar)
foo(baz)
And you want the baz instead of bar, you'll just need to go back a scope further. This can be done by adding an extra .f_back in the caller_local_vars initialization.
See an example here: ideone
The only objects in Python that have canonical names are modules, functions, and classes, and of course there is no guarantee that this canonical name has any meaning in any namespace after the function or class has been defined or the module imported. These names can also be modified after the objects are created so they may not always be particularly trustworthy.
What you want to do is not possible without recursively walking the tree of named objects; a name is a one-way reference to an object. A common or garden-variety Python object contains no references to its names. Imagine if every integer, every dict, every list, every Boolean needed to maintain a list of strings that represented names that referred to it! It would be an implementation nightmare, with little benefit to the programmer.
TL;DR
Use the Wrapper helper from python-varname:
from varname.helpers import Wrapper
foo = Wrapper(dict())
# foo.name == 'foo'
# foo.value == {}
foo.value['bar'] = 2
For list comprehension part, you can do:
n_jobs = Wrapper(<original_value>)
users = Wrapper(<original_value>)
queues = Wrapper(<original_value>)
priorities = Wrapper(<original_value>)
list_of_dicts = [n_jobs, users, queues, priorities]
columns = [d.name for d in list_of_dicts]
# ['n_jobs', 'users', 'queues', 'priorities']
# REMEMBER that you have to access the <original_value> by d.value
I am the author of the python-varname package. Please let me know if you have any questions or you can submit issues on Github.
The long answer
Is it even possible?
Yes and No.
We are retrieving the variable names at runtime, so we need a function to be called to enable us to access the previous frames to retrieve the variable names. That's why we need a Wrapper there. In that function, at runtime, we are parsing the source code/AST nodes in the previous frames to get the exact variable name.
However, the source code/AST nodes in the previous frames are not always available, or they could be modified by other environments (e.g: pytest's assert statement). One simple example is that the codes run via exec(). Even though we are still able to retrieve some information from the bytecode, it needs too much effort and it is also error-prone.
How to do it?
First of all, we need to identify which frame the variable is given. It's not always simply the direct previous frame. For example, we may have another wrapper for the function:
from varname import varname
def func():
return varname()
def wrapped():
return func()
x = wrapped()
In the above example, we have to skip the frame inside wrapped to get to the right frame x = wrapped() so that we are able to locate x. The arguments frame and ignore of varname allow us to skip some of these intermediate frames. See more details in the README file and the API docs of the package.
Then we need to parse the AST node to locate where the variable is assigned value (function call) to. It's not always just a simple assignment. Sometimes there could be complex AST nodes, for example, x = [wrapped()]. We need to identify the correct assignment by traversing the AST tree.
How reliable is it?
Once we identify the assignment node, it is reliable.
varname is all depending on executing package to look for the node. The node executing detects is ensured to be the correct one (see also this).
It partially works with environments where other AST magics apply, including pytest, ipython, macropy, birdseye, reticulate with R, etc. Neither executing nor varname is 100% working with those environments.
Do we need a package to do it?
Well, yes and no, again.
If your scenario is simple, the code provided by #juan Isaza or #scohe001 probably is enough for you to work with the case where a variable is defined at the direct previous frame and the AST node is a simple assignment. You just need to go one frame back and retrieve the information there.
However, if the scenario becomes complicated, or we need to adopt different application scenarios, you probably need a package like python-varname, to handle them. These scenarios may include to:
present more friendly messages when the source code is not available or AST nodes are not accessible
skip intermediate frames (allows the function to be wrapped or called in other intermediate frames)
automatically ignores calls from built-in functions or libraries. For example: x = str(func())
retrieve multiple variable names on the left-hand side of the assignment
etc.
How about the f-string?
Like the answer provided by #Aivar Paalberg. It's definitely fast and reliable. However, it's not at runtime, meaning that you have to know it's foo before you print the name out. But with varname, you don't have to know that variable is coming:
from varname import varname
def func():
return varname()
# In external uses
x = func() # 'x'
y = func() # 'y'
Finally
python-varname is not only able to detect the variable name from an assignment, but also:
Retrieve variable names directly, using nameof
Detect next immediate attribute name, using will
Fetch argument names/sources passed to a function using argname
Read more from its documentation.
However, the final word I want to say is that, try to avoid using it whenever you can.
Because you can't make sure that the client code will run in an environment where the source node is available or AST node is accessible. And of course, it costs resources to parse the source code, identify the environment, retrieve the AST nodes and evaluate them when needed.
On python3, this function will get the outer most name in the stack:
import inspect
def retrieve_name(var):
"""
Gets the name of var. Does it from the out most frame inner-wards.
:param var: variable to get name from.
:return: string
"""
for fi in reversed(inspect.stack()):
names = [var_name for var_name, var_val in fi.frame.f_locals.items() if var_val is var]
if len(names) > 0:
return names[0]
It is useful anywhere on the code. Traverses the reversed stack looking for the first match.
I don't believe this is possible. Consider the following example:
>>> a = []
>>> b = a
>>> id(a)
140031712435664
>>> id(b)
140031712435664
The a and b point to the same object, but the object can't know what variables point to it.
def name(**variables):
return [x for x in variables]
It's used like this:
name(variable=variable)
>> my_var = 5
>> my_var_name = [ k for k,v in locals().items() if v == my_var][0]
>> my_var_name
'my_var'
In case you get an error if myvar points to another variable, try this (suggested by #mherzog)-
>> my_var = 5
>> my_var_name = [ k for k,v in locals().items() if v is my_var][0]
>> my_var_name
'my_var'
locals() - Return a dictionary containing the current scope's local variables.
by iterating through this dictionary we can check the key which has a value equal to the defined variable, just extracting the key will give us the text of variable in string format.
from (after a bit changes)
https://www.tutorialspoint.com/How-to-get-a-variable-name-as-a-string-in-Python
I wrote the package sorcery to do this kind of magic robustly. You can write:
from sorcery import dict_of
columns = dict_of(n_jobs, users, queues, priorities)
and pass that to the dataframe constructor. It's equivalent to:
columns = dict(n_jobs=n_jobs, users=users, queues=queues, priorities=priorities)
Here's one approach. I wouldn't recommend this for anything important, because it'll be quite brittle. But it can be done.
Create a function that uses the inspect module to find the source code that called it. Then you can parse the source code to identify the variable names that you want to retrieve. For example, here's a function called autodict that takes a list of variables and returns a dictionary mapping variable names to their values. E.g.:
x = 'foo'
y = 'bar'
d = autodict(x, y)
print d
Would give:
{'x': 'foo', 'y': 'bar'}
Inspecting the source code itself is better than searching through the locals() or globals() because the latter approach doesn't tell you which of the variables are the ones you want.
At any rate, here's the code:
def autodict(*args):
get_rid_of = ['autodict(', ',', ')', '\n']
calling_code = inspect.getouterframes(inspect.currentframe())[1][4][0]
calling_code = calling_code[calling_code.index('autodict'):]
for garbage in get_rid_of:
calling_code = calling_code.replace(garbage, '')
var_names, var_values = calling_code.split(), args
dyn_dict = {var_name: var_value for var_name, var_value in
zip(var_names, var_values)}
return dyn_dict
The action happens in the line with inspect.getouterframes, which returns the string within the code that called autodict.
The obvious downside to this sort of magic is that it makes assumptions about how the source code is structured. And of course, it won't work at all if it's run inside the interpreter.
This function will print variable name with its value:
import inspect
def print_this(var):
callers_local_vars = inspect.currentframe().f_back.f_locals.items()
print(str([k for k, v in callers_local_vars if v is var][0])+': '+str(var))
***Input & Function call:***
my_var = 10
print_this(my_var)
***Output**:*
my_var: 10
>>> locals()['foo']
{}
>>> globals()['foo']
{}
If you wanted to write your own function, it could be done such that you could check for a variable defined in locals then check globals. If nothing is found you could compare on id() to see if the variable points to the same location in memory.
If your variable is in a class, you could use className.dict.keys() or vars(self) to see if your variable has been defined.
I have a method, and while not the most efficient...it works! (and it doesn't involve any fancy modules).
Basically it compares your Variable's ID to globals() Variables' IDs, then returns the match's name.
def getVariableName(variable, globalVariables=globals().copy()):
""" Get Variable Name as String by comparing its ID to globals() Variables' IDs
args:
variable(var): Variable to find name for (Obviously this variable has to exist)
kwargs:
globalVariables(dict): Copy of the globals() dict (Adding to Kwargs allows this function to work properly when imported from another .py)
"""
for globalVariable in globalVariables:
if id(variable) == id(globalVariables[globalVariable]): # If our Variable's ID matches this Global Variable's ID...
return globalVariable # Return its name from the Globals() dict
In Python, the def and class keywords will bind a specific name to the object they define (function or class). Similarly, modules are given a name by virtue of being called something specific in the filesystem. In all three cases, there's an obvious way to assign a "canonical" name to the object in question.
However, for other kinds of objects, such a canonical name may simply not exist. For example, consider the elements of a list. The elements in the list are not individually named, and it is entirely possible that the only way to refer to them in a program is by using list indices on the containing list. If such a list of objects was passed into your function, you could not possibly assign meaningful identifiers to the values.
Python doesn't save the name on the left hand side of an assignment into the assigned object because:
It would require figuring out which name was "canonical" among multiple conflicting objects,
It would make no sense for objects which are never assigned to an explicit variable name,
It would be extremely inefficient,
Literally no other language in existence does that.
So, for example, functions defined using lambda will always have the "name" <lambda>, rather than a specific function name.
The best approach would be simply to ask the caller to pass in an (optional) list of names. If typing the '...','...' is too cumbersome, you could accept e.g. a single string containing a comma-separated list of names (like namedtuple does).
I think it's so difficult to do this in Python because of the simple fact that you never will not know the name of the variable you're using. So, in his example, you could do:
Instead of:
list_of_dicts = [n_jobs, users, queues, priorities]
dict_of_dicts = {"n_jobs" : n_jobs, "users" : users, "queues" : queues, "priorities" : priorities}
Many of the answers return just one variable name. But that won't work well if more than one variable have the same value. Here's a variation of Amr Sharaki's answer which returns multiple results if more variables have the same value.
def getVariableNames(variable):
results = []
globalVariables=globals().copy()
for globalVariable in globalVariables:
if id(variable) == id(globalVariables[globalVariable]):
results.append(globalVariable)
return results
a = 1
b = 1
getVariableNames(a)
# ['a', 'b']
just another way to do this based on the content of input variable:
(it returns the name of the first variable that matches to the input variable, otherwise None. One can modify it to get all variable names which are having the same content as input variable)
def retrieve_name(x, Vars=vars()):
for k in Vars:
if isinstance(x, type(Vars[k])):
if x is Vars[k]:
return k
return None
If the goal is to help you keep track of your variables, you can write a simple function that labels the variable and returns its value and type. For example, suppose i_f=3.01 and you round it to an integer called i_n to use in a code, and then need a string i_s that will go into a report.
def whatis(string, x):
print(string+' value=',repr(x),type(x))
return string+' value='+repr(x)+repr(type(x))
i_f=3.01
i_n=int(i_f)
i_s=str(i_n)
i_l=[i_f, i_n, i_s]
i_u=(i_f, i_n, i_s)
## make report that identifies all types
report='\n'+20*'#'+'\nThis is the report:\n'
report+= whatis('i_f ',i_f)+'\n'
report+=whatis('i_n ',i_n)+'\n'
report+=whatis('i_s ',i_s)+'\n'
report+=whatis('i_l ',i_l)+'\n'
report+=whatis('i_u ',i_u)+'\n'
print(report)
This prints to the window at each call for debugging purposes and also yields a string for the written report. The only downside is that you have to type the variable twice each time you call the function.
I am a Python newbie and found this very useful way to log my efforts as I program and try to cope with all the objects in Python. One flaw is that whatis() fails if it calls a function described outside the procedure where it is used. For example, int(i_f) was a valid function call only because the int function is known to Python. You could call whatis() using int(i_f**2), but if for some strange reason you choose to define a function called int_squared it must be declared inside the procedure where whatis() is used.
Maybe this could be useful:
def Retriever(bar):
return (list(globals().keys()))[list(map(lambda x: id(x), list(globals().values()))).index(id(bar))]
The function goes through the list of IDs of values from the global scope (the namespace could be edited), finds the index of the wanted/required var or function based on its ID, and then returns the name from the list of global names based on the acquired index.
Whenever I have to do it, mostly while communicating json schema and constants with the frontend I define a class as follows
class Param:
def __init__(self, name, value):
self.name = name
self.value = value
Then define the variable with name and value.
frame_folder_count = Param({'name':'frame_folder_count', 'value':10})
Now you can access the name and value using the object.
>>> frame_folder_count.name
'frame_folder_count'
>>> def varname(v, scope=None):
d = globals() if not scope else vars(scope); return [k for k in d if d[k] == v]
...
>>> d1 = {'a': 'ape'}; d2 = {'b': 'bear'}; d3 = {'c': 'cat'}
>>> ld = [d1, d2, d3]
>>> [varname(d) for d in ld]
[['d1'], ['d2'], ['d3']]
>>> d5 = d3
>>> [varname(d) for d in ld]
[['d1'], ['d2'], ['d3', 'd5']]
>>> def varname(v, scope=None):
d = globals() if not scope else vars(scope); return [k for k in d if d[k] is v]
...
>>> [varname(d) for d in ld]
[['d1'], ['d2'], ['d3', 'd5']]
As you see and is noted here, there can be multiple variables with the same value or even address, so using a wrapper to keep the names with the data is best.
Following method will not return the name of variable but using this method you can create data frame easily if variable is available in global scope.
class CustomDict(dict):
def __add__(self, other):
return CustomDict({**self, **other})
class GlobalBase(type):
def __getattr__(cls, key):
return CustomDict({key: globals()[key]})
def __getitem__(cls, keys):
return CustomDict({key: globals()[key] for key in keys})
class G(metaclass=GlobalBase):
pass
x, y, z = 0, 1, 2
print('method 1:', G['x', 'y', 'z']) # Outcome: method 1: {'x': 0, 'y': 1, 'z': 2}
print('method 2:', G.x + G.y + G.z) # Outcome: method 2: {'x': 0, 'y': 1, 'z': 2}
A = [0, 1]
B = [1, 2]
pd.DataFrame(G.A + G.B) # It will return a data frame with A and B columns
Some of the previous cases would fail if there are two variables with the same value. So it is convenient to alert it:
Defining function:
# Variable to string of variable name
def var_name(variable,i=0):
results = []
for name in globals():
if eval(name) == variable:
results.append(name)
if len(results) > 1:
print('Warning:' )
print(' var_name() has found',len(results), 'possible outcomes.')
print(' Please choose the suitable parameter "i". Where "i" is the index')
print(' that matches your choice from the list below.')
print(' ',results) ; print('')
return results[i]
Use:
var_1 = 10
var_name(var_1) # Output will be "var_1"
If you have 2 variables with the same value like var_1 = 8 and var_2 = 8, then a warning will appear.
var_1 = 8
var_2 = 8
var_name(var_2) # Output will be "var_1" too but Warning will appear
You can get your variable as kwargs and return it as string:
var=2
def getVarName(**kwargs):
return list(kwargs.keys())[0]
print (getVarName(var = var))
Note: variable name must be equal to itself.
I try to get name from inspect locals, but it cann't process var likes a[1], b.val.
After it, I got a new idea --- get var name from the code, and I try it succ!
code like below:
#direct get from called function code
def retrieve_name_ex(var):
stacks = inspect.stack()
try:
func = stacks[0].function
code = stacks[1].code_context[0]
s = code.index(func)
s = code.index("(", s + len(func)) + 1
e = code.index(")", s)
return code[s:e].strip()
except:
return ""
You can try the following to retrieve the name of a function you defined (does not work for built-in functions though):
import re
def retrieve_name(func):
return re.match("<function\s+(\w+)\s+at.*", str(func)).group(1)
def foo(x):
return x**2
print(retrieve_name(foo))
# foo
When finding the name of a variable from its value,
you may have several variables equal to the same value,
for example var1 = 'hello' and var2 = 'hello'.
My solution:
def find_var_name(val):
dict_list = []
global_dict = dict(globals())
for k, v in global_dict.items():
dict_list.append([k, v])
return [item[0] for item in dict_list if item[1] == val]
var1 = 'hello'
var2 = 'hello'
find_var_name('hello')
Outputs
['var1', 'var2']
Compressed version of iDilip's answer:
import inspect
def varname(x):
return [k for k,v in inspect.currentframe().f_back.f_locals.items() if v is x][0]
hi = 123
print(varname(hi))
It's totally possible to get the name of an instance variable, so long as it is the property of a class.
I got this from Effective Python by Brett Slatkin. Hope it helps someone:
The class must implement the get, set, and set_name dunder methods, which are part of the "Descriptor Protocol"
This worked when I ran it:
class FieldThatKnowsItsName():
def __init__(self):
self.name = None
self._value= None
self.owner = None
def __set_name__(self, owner, name):
self.name = name
self.owner = owner
self.owner.fields[self.name] = self
def __get__(self, instance, instance_type):
return self
def __set__(self, instance, value):
self = value
class SuperTable:
fields = {}
field_1=FieldThatKnowsItsName()
field_2=FieldThatKnowsItsName()
table = SuperTable()
print(table.field_1.name)
print(table.field_2.name)
You can then add methods and or extend your datatype as you like.
As a bonus, the set_name(self, owner, name) dunder also passes the parent instance, so the Field class instance can register itself with the parent.
I got this from Effective Python by Brett Slatkin. It took a while to figure out how to implement.
How can I do the same for a variable? As opposed to functions, Python variables do not have the __name__ attribute.
The problem comes up because you are confused about terminology, semantics or both.
"variables" don't belong in the same category as "functions". A "variable" is not a thing that takes up space in memory while the code is running. It is just a name that exists in your source code - so that when you're writing the code, you can explain which thing you're talking about. Python uses names in the source code to refer to (i.e., give a name to) values. (In many languages, a variable is more like a name for a specific location in memory where the value will be stored. But Python's names actually name the thing in question.)
In Python, a function is a value. (In some languages, this is not the case; although there are bytes of memory used to represent the actual executable code, it isn't a discrete chunk of memory that your program logic gets to interact with directly.) In Python, every value is an object, meaning that you can assign names to it freely, pass it as an argument, return it from a function, etc. (In many languages, this is not the case.) Objects in Python have attributes, which are the things you access using the . syntax. Functions in Python have a __name__ attribute, which is assigned when the function is created. Specifically, when a def statement is executed (in most languages, creation of a function works quite differently), the name that appears after def is used as a value for the __name__ attribute, and also, independently, as a variable name that will get the function object assigned to it.
But most objects don't have an attribute like that.
In other words, if I have a variable such as:
That's the thing: you don't "have" the variable in the sense that you're thinking of. You have the object that is named by that variable. Anything else depends on the information incidentally being stored in some other object - such as the locals() of the enclosing function. But it would be better to store the information yourself. Instead of relying on a variable name to carry information for you, explicitly build the mapping between the string name you want to use for the object, and the object itself.

Any method to denote object assignment?

I've been studying magic methods in Python, and have been wondering if there's a way to outline the specific action of:
a = MyClass(*params).method()
versus:
MyClass(*params).method()
In the sense that, perhaps, I may want to return a list that has been split on the '\n' character, versus dumping the raw list into the variable a that keeps the '\n' intact.
Is there a way to ask Python if its next action is about to return a value to a variable, and change action, if that's the case? I was thinking:
class MyClass(object):
def __init__(params):
self.end = self.method(*params)
def __asgn__(self):
return self.method(*params).split('\n')
def __str__(self):
"""this is the fallback if __asgn__ is not called"""
return self.method(*params)
No. You cannot change what happens when you assign to a bare name.
You can change what happens if the assignment target on the left hand side is an attribute or item of an object. You can override a[blah] = ... with __setitem__ and a.blah = ... with __setattr__ (although you can only hook into these on a, not on the object being assigned). But you can't override or in any way influence a = ....
Note that having the right-hand side change based on what is "going to happen" would be even stranger, and very bad. That would mean that
someFunc(MyClass().method())
could be different than
a = MyClass().method()
someFunc(a)
In Python names are just labels attached to objects. Objects don't get to know what labels are attached to them, and that's a good thing. You might assign the result a computation to an intermediate variable just to make subsequent lines more readable, and you don't want that assignment to change the result of that computation.
There should be no difference between calling MyClass(*params).method() directly and assigning it to a variable. What you may be seeing here is your interpreter automatically printing return results, which is why it appears to be split while the variable value contains EOL markers.
There is no way to override default assignment to a variable. However, by using an object, you can easily provide your own hooks:
class Assigner(object):
def __init__(self, assignment_callback):
self.assignment = assignment_callback
def __setattr__(self, key, value):
if hasattr(self, 'assignment'):
value = self.assignment(value)
super(Assigner, self).__setattr__( key, value )
def uppercase(value):
# example function to perform on each attribute assignment
return value.upper()
Then in your code, rather than assigning to a variable directly you assign to attributes on your object:
>>> my = Assigner(uppercase)
>>> my.a = 'foo'
>>> print my.a
FOO
Yes.* Python allows inspecting its own stack, which can be used to peek ahead at the next instruction.
#!/usr/bin/env python3
import dis
import inspect
from itertools import dropwhile
class MyClass(object):
def method(self):
# inspect the stack to get calling line of code
frame = inspect.stack()[1].frame
# disassemble stack frame
ins = dis.get_instructions(frame.f_code)
# move to last instruction
ins = dropwhile(lambda x: x.offset < frame.f_lasti, ins)
# the last call would have been to this method/function
current_instruction = ins.__next__()
assert current_instruction.opname.startswith('CALL_')
# peek ahead at the next instruction
next_instruction = ins.__next__()
# vary behaviour depending on the next instruction
if next_instruction.opname.startswith('STORE_'):
return "returning to assignment"
elif next_instruction.opname.startswith('CALL_'):
return "returning to function/method call"
elif next_instruction.opname == 'POP_TOP':
print("return value thrown away")
return "return ignored"
elif next_instruction.opname == 'PRINT_EXPR':
return "return to interactive console"
else:
return "return to {}".format(next_instruction.opname)
This will result in the following behaviour:
a = MyClass().method()
print(a)
# returning to assignment
def someFunc(x):
return x.split()
b = someFunc(MyClass().method())
print(b)
# ['returning', 'to', 'function/method', 'call']
MyClass().method()
# return value thrown away (if called as program)
# return to interactive console (if run interactively)
* Though as the accepted answer points out, doing so is "very bad". It's also fragile, as it can be affected by bytecode optimisation. See also: Nested dictionary that acts as defaultdict when setting items but not when getting items

Categories

Resources