I'm creating a wrapper for a function with functools.wraps. My wrapper has the effect of overriding a default parameter (and it doesn't do anything else):
def add(*, a=1, b=2):
"Add numbers"
return a + b
#functools.wraps(add)
def my_add(**kwargs):
kwargs.setdefault('b', 3)
return add(**kwargs)
This my_add definition behaves the same as
#functools.wraps(add)
def my_add(*, a=1, b=3):
return add(a=a, b=b)
except that I didn't have to manually type out the parameter list.
However, when I run help(my_add), I see the help string for add, which has the wrong function name and the wrong default argument for the parameter b:
add(*, a=1, b=2)
Add numbers
How can I override the function name and the default argument in this help() output?
(Or, is there a different way to define my_add, using for example some magic function my_add = magic(add, func_name='my_add', kwarg_defaults={'b': 3}) that will do what I want?)
Let me try and explain what happens.
When you call the help functions, this is going to request information about your function using the inspect module. Therefore you have to change the function signature, in order to change the default argument.
Now this is not something that is advised, or often preferred, but who cares about that right? The provided solution is considered hacky and probably won't work for all versions of Python. Therefore you might want to reconsider how important the help function is... Any way let's start with some explanation on how it was done, followed by the code and test case.
Copying functions
Now the first thing we will do is copy the entire function, this is because I only want to change the signature of the new function and not the original function. This decouples the new my_add signature (and default values) from the original add function.
See:
How to create a copy of a python function
How can I make a deepcopy of a function in Python?
For ideas of how to do this (I will show my version in a bit).
Copying / updating signature
The next step is to get a copy of the function signature, for that this post was very useful. Except for the part where we have to adjust the signature parameters to match the new keyword default arguments.
For that we have to change the value of a mappingproxy, which we can see when running the debugger on the return value of inspect.signature(g). Now so far this can only be done by changing the private variables (the values with leading underscores _private). Therefore this solution will be considered hacky and is not guaranteed to withstand possible updates. That said, let's see the solution!
Full code
import inspect
import types
import functools
def update_func(f, func_name='', update_kwargs: dict = None):
"""Based on http://stackoverflow.com/a/6528148/190597 (Glenn Maynard)"""
g = types.FunctionType(
code=f.__code__,
globals=f.__globals__.copy(),
name=f.__name__,
argdefs=f.__defaults__,
closure=f.__closure__
)
g = functools.update_wrapper(g, f)
g.__signature__ = inspect.signature(g)
g.__kwdefaults__ = f.__kwdefaults__.copy()
# Adjust your arguments
for key, value in (update_kwargs or {}).items():
g.__kwdefaults__[key] = value
g.__signature__.parameters[key]._default = value
g.__name__ = func_name or g.__name__
return g
def add(*, a=1, b=2):
"Add numbers"
return a + b
my_add = update_func(add, func_name="my_add", update_kwargs=dict(b=3))
Example
if __name__ == '__main__':
a = 2
print("*" * 50, f"\nMy add\n", )
help(my_add)
print("*" * 50, f"\nOriginal add\n", )
help(add)
print("*" * 50, f"\nResults:"
f"\n\tMy add : a = {a}, return = {my_add(a=a)}"
f"\n\tOriginal add: a = {a}, return = {add(a=a)}")
Output
**************************************************
My add
Help on function my_add in module __main__:
my_add(*, a=1, b=3)
Add numbers
**************************************************
Original add
Help on function add in module __main__:
add(*, a=1, b=2)
Add numbers
**************************************************
Results:
My add : a = 2, return = 5
Original add: a = 2, return = 4
Usages
f: is the function that you want to update
func_name: is optionally the new name of the function (if empty, keeps the old name)
update_kwargs: is a dictionary containing the key and value of the default arguments that you want to update.
Notes
The solution is using copy variables to make full copies of dictionaries, such that there is no impact on the original add function.
The _default value is a private variable, and can be changed in future releases of python.
On Codewars.com I encountered the following task:
Create a function add that adds numbers together when called in succession. So add(1) should return 1, add(1)(2) should return 1+2, ...
While I'm familiar with the basics of Python, I've never encountered a function that is able to be called in such succession, i.e. a function f(x) that can be called as f(x)(y)(z).... Thus far, I'm not even sure how to interpret this notation.
As a mathematician, I'd suspect that f(x)(y) is a function that assigns to every x a function g_{x} and then returns g_{x}(y) and likewise for f(x)(y)(z).
Should this interpretation be correct, Python would allow me to dynamically create functions which seems very interesting to me. I've searched the web for the past hour, but wasn't able to find a lead in the right direction. Since I don't know how this programming concept is called, however, this may not be too surprising.
How do you call this concept and where can I read more about it?
I don't know whether this is function chaining as much as it's callable chaining, but, since functions are callables I guess there's no harm done. Either way, there's two ways I can think of doing this:
Sub-classing int and defining __call__:
The first way would be with a custom int subclass that defines __call__ which returns a new instance of itself with the updated value:
class CustomInt(int):
def __call__(self, v):
return CustomInt(self + v)
Function add can now be defined to return a CustomInt instance, which, as a callable that returns an updated value of itself, can be called in succession:
>>> def add(v):
... return CustomInt(v)
>>> add(1)
1
>>> add(1)(2)
3
>>> add(1)(2)(3)(44) # and so on..
50
In addition, as an int subclass, the returned value retains the __repr__ and __str__ behavior of ints. For more complex operations though, you should define other dunders appropriately.
As #Caridorc noted in a comment, add could also be simply written as:
add = CustomInt
Renaming the class to add instead of CustomInt also works similarly.
Define a closure, requires extra call to yield value:
The only other way I can think of involves a nested function that requires an extra empty argument call in order to return the result. I'm not using nonlocal and opt for attaching attributes to the function objects to make it portable between Pythons:
def add(v):
def _inner_adder(val=None):
"""
if val is None we return _inner_adder.v
else we increment and return ourselves
"""
if val is None:
return _inner_adder.v
_inner_adder.v += val
return _inner_adder
_inner_adder.v = v # save value
return _inner_adder
This continuously returns itself (_inner_adder) which, if a val is supplied, increments it (_inner_adder += val) and if not, returns the value as it is. Like I mentioned, it requires an extra () call in order to return the incremented value:
>>> add(1)(2)()
3
>>> add(1)(2)(3)() # and so on..
6
You can hate me, but here is a one-liner :)
add = lambda v: type("", (int,), {"__call__": lambda self, v: self.__class__(self + v)})(v)
Edit: Ok, how this works? The code is identical to answer of #Jim, but everything happens on a single line.
type can be used to construct new types: type(name, bases, dict) -> a new type. For name we provide empty string, as name is not really needed in this case. For bases (tuple) we provide an (int,), which is identical to inheriting int. dict are the class attributes, where we attach the __call__ lambda.
self.__class__(self + v) is identical to return CustomInt(self + v)
The new type is constructed and returned within the outer lambda.
If you want to define a function to be called multiple times, first you need to return a callable object each time (for example a function) otherwise you have to create your own object by defining a __call__ attribute, in order for it to be callable.
The next point is that you need to preserve all the arguments, which in this case means you might want to use Coroutines or a recursive function. But note that Coroutines are much more optimized/flexible than recursive functions, specially for such tasks.
Here is a sample function using Coroutines, that preserves the latest state of itself. Note that it can't be called multiple times since the return value is an integer which is not callable, but you might think about turning this into your expected object ;-).
def add():
current = yield
while True:
value = yield current
current = value + current
it = add()
next(it)
print(it.send(10))
print(it.send(2))
print(it.send(4))
10
12
16
Simply:
class add(int):
def __call__(self, n):
return add(self + n)
If you are willing to accept an additional () in order to retrieve the result you can use functools.partial:
from functools import partial
def add(*args, result=0):
return partial(add, result=sum(args)+result) if args else result
For example:
>>> add(1)
functools.partial(<function add at 0x7ffbcf3ff430>, result=1)
>>> add(1)(2)
functools.partial(<function add at 0x7ffbcf3ff430>, result=3)
>>> add(1)(2)()
3
This also allows specifying multiple numbers at once:
>>> add(1, 2, 3)(4, 5)(6)()
21
If you want to restrict it to a single number you can do the following:
def add(x=None, *, result=0):
return partial(add, result=x+result) if x is not None else result
If you want add(x)(y)(z) to readily return the result and be further callable then sub-classing int is the way to go.
The pythonic way to do this would be to use dynamic arguments:
def add(*args):
return sum(args)
This is not the answer you're looking for, and you may know this, but I thought I would give it anyway because if someone was wondering about doing this not out of curiosity but for work. They should probably have the "right thing to do" answer.
In Python, is it possible to redefine the default parameters of a function at runtime?
I defined a function with 3 parameters here:
def multiplyNumbers(x,y,z):
return x*y*z
print(multiplyNumbers(x=2,y=3,z=3))
Next, I tried (unsuccessfully) to set the default parameter value for y, and then I tried calling the function without the parameter y:
multiplyNumbers.y = 2;
print(multiplyNumbers(x=3, z=3))
But the following error was produced, since the default value of y was not set correctly:
TypeError: multiplyNumbers() missing 1 required positional argument: 'y'
Is it possible to redefine the default parameters of a function at runtime, as I'm attempting to do here?
Just use functools.partial
multiplyNumbers = functools.partial(multiplyNumbers, y = 42)
One problem here: you will not be able to call it as multiplyNumbers(5, 7, 9); you should manually say y=7
If you need to remove default arguments I see two ways:
Store original function somewhere
oldF = f
f = functools.partial(f, y = 42)
//work with changed f
f = oldF //restore
use partial.func
f = f.func //go to previous version.
Technically, it is possible to do what you ask… but it's not a good idea. RiaD's answer is the Pythonic way to do this.
In Python 3:
>>> def f(x=1, y=2, z=3):
... print(x, y, z)
>>> f()
1 2 3
>>> f.__defaults__ = (4, 5, 6)
4 5 6
As with everything else that's under the covers and hard to find in the docs, the inspect module chart is the best place to look for function attributes.
The details are slightly different in Python 2, but the idea is the same. (Just change the pulldown at the top left of the docs page from 3.3 to 2.7.)
If you're wondering how Python knows which defaults go with which arguments when it's just got a tuple… it just counts backward from the end (or the first of *, *args, **kwargs—anything after that goes into the __kwdefaults__ dict instead). f.__defaults = (4, 5) will set the defaults to y and z to 4 and 5, and with default for x. That works because you can't have non-defaulted parameters after defaulted parameters.
There are some cases where this won't work, but even then, you can immutably copy it to a new function with different defaults:
>>> f2 = types.FunctionType(f.__code__, f.__globals__, f.__name__,
... (4, 5, 6), f.__closure__)
Here, the types module documentation doesn't really explain anything, but help(types.FunctionType) in the interactive interpreter shows the params you need.
The only case you can't handle is a builtin function. But they generally don't have actual defaults anyway; instead, they fake something similar in the C API.
yes, you can accomplish this by modifying the function's func.__defaults__ tuple
that attribute is a tuple of the default values for each argument of the function.
for example, to make pandas.read_csv always use sep='\t', you could do:
import inspect
import pandas as pd
default_args = inspect.getfullargspec(pd.read_csv).args
default_arg_values = list(pd.read_csv.__defaults__)
default_arg_values[default_args.index("sep")] = '\t'
pd.read_csv.__defaults__ = tuple(default_arg_values)
use func_defaults as in
def myfun(a=3):
return a
myfun.func_defaults = (4,)
b = myfun()
assert b == 4
check the docs for func_defaults here
UPDATE: looking at RiaD's response I think I was too literal with mine. I don't know the context from where you're asking this question but in general (and following the Zen of Python) I believe working with partial applications is a better option than redefining a function's defaults arguments
I find myself writing stuff like this too often and it seems too wordy:
obj = my_dict.get('obj')
if obj:
var = obj
Is there a better way to do this? Maybe in one line?
The get function takes a second argument, a default:
get(key[, default])
Return the value for key if key is in the dictionary, else default. If default is not given, it defaults to None, so that this method never raises a KeyError.
http://docs.python.org/library/stdtypes.html
So, you could use the below to replicate your question's code:
var = my_dict.get('obj', var)
There's no clean way to replace this code in one line, because it conditionally binds the name var. In the case that var was already defined, this one-liner is possible:
var = my_dict['obj'] if 'obj' in my_dict and my_dict['obj'] else var
However, this is still slightly different than the original code in case var was not already defined: the one-liner is raising a NameError and the original code just continues with var unbound.
Note that other answers behave differently when the value exists but is falsey.
obj = my_dict.get('obj')
if obj: # <-- test on truthiness of obj
# if we are here, it means:
# 1. my_dict has key 'obj', AND...
# 2. at least one of the following,
# my_dict['obj'].__bool__() returned True (__nonzero__ for Python 2)
# OR
# my_dict['obj'].__len__() returned result > 0
# OR
# my_dict['obj'] is not None, and has neither __bool__ nor __len__ defined
var = obj
I find this form to be the most explicitly readable and I use it all the time:
moo = animals["cow"] if "cow" in animals else None
Assuming that my_dict is what it says it is (a dictionary):
var = my_dict.get('obj') if 'obj' in my_dict else var
Edited to preserve intent with original question.
var = thedict.get('thekey', var)
I would like to convert a python variable name into the string equivalent as shown. Any ideas how?
var = {}
print ??? # Would like to see 'var'
something_else = 3
print ??? # Would print 'something_else'
TL;DR: Not possible. See 'conclusion' at the end.
There is an usage scenario where you might need this. I'm not implying there are not better ways or achieving the same functionality.
This would be useful in order to 'dump' an arbitrary list of dictionaries in case of error, in debug modes and other similar situations.
What would be needed, is the reverse of the eval() function:
get_indentifier_name_missing_function()
which would take an identifier name ('variable','dictionary',etc) as an argument, and return a
string containing the identifier’s name.
Consider the following current state of affairs:
random_function(argument_data)
If one is passing an identifier name ('function','variable','dictionary',etc) argument_data to a random_function() (another identifier name), one actually passes an identifier (e.g.: <argument_data object at 0xb1ce10>) to another identifier (e.g.: <function random_function at 0xafff78>):
<function random_function at 0xafff78>(<argument_data object at 0xb1ce10>)
From my understanding, only the memory address is passed to the function:
<function at 0xafff78>(<object at 0xb1ce10>)
Therefore, one would need to pass a string as an argument to random_function() in order for that function to have the argument's identifier name:
random_function('argument_data')
Inside the random_function()
def random_function(first_argument):
, one would use the already supplied string 'argument_data' to:
serve as an 'identifier name' (to display, log, string split/concat, whatever)
feed the eval() function in order to get a reference to the actual identifier, and therefore, a reference to the real data:
print("Currently working on", first_argument)
some_internal_var = eval(first_argument)
print("here comes the data: " + str(some_internal_var))
Unfortunately, this doesn't work in all cases. It only works if the random_function() can resolve the 'argument_data' string to an actual identifier. I.e. If argument_data identifier name is available in the random_function()'s namespace.
This isn't always the case:
# main1.py
import some_module1
argument_data = 'my data'
some_module1.random_function('argument_data')
# some_module1.py
def random_function(first_argument):
print("Currently working on", first_argument)
some_internal_var = eval(first_argument)
print("here comes the data: " + str(some_internal_var))
######
Expected results would be:
Currently working on: argument_data
here comes the data: my data
Because argument_data identifier name is not available in the random_function()'s namespace, this would yield instead:
Currently working on argument_data
Traceback (most recent call last):
File "~/main1.py", line 6, in <module>
some_module1.random_function('argument_data')
File "~/some_module1.py", line 4, in random_function
some_internal_var = eval(first_argument)
File "<string>", line 1, in <module>
NameError: name 'argument_data' is not defined
Now, consider the hypotetical usage of a get_indentifier_name_missing_function() which would behave as described above.
Here's a dummy Python 3.0 code: .
# main2.py
import some_module2
some_dictionary_1 = { 'definition_1':'text_1',
'definition_2':'text_2',
'etc':'etc.' }
some_other_dictionary_2 = { 'key_3':'value_3',
'key_4':'value_4',
'etc':'etc.' }
#
# more such stuff
#
some_other_dictionary_n = { 'random_n':'random_n',
'etc':'etc.' }
for each_one_of_my_dictionaries in ( some_dictionary_1,
some_other_dictionary_2,
...,
some_other_dictionary_n ):
some_module2.some_function(each_one_of_my_dictionaries)
# some_module2.py
def some_function(a_dictionary_object):
for _key, _value in a_dictionary_object.items():
print( get_indentifier_name_missing_function(a_dictionary_object) +
" " +
str(_key) +
" = " +
str(_value) )
######
Expected results would be:
some_dictionary_1 definition_1 = text_1
some_dictionary_1 definition_2 = text_2
some_dictionary_1 etc = etc.
some_other_dictionary_2 key_3 = value_3
some_other_dictionary_2 key_4 = value_4
some_other_dictionary_2 etc = etc.
......
......
......
some_other_dictionary_n random_n = random_n
some_other_dictionary_n etc = etc.
Unfortunately, get_indentifier_name_missing_function() would not see the 'original' identifier names (some_dictionary_,some_other_dictionary_2,some_other_dictionary_n). It would only see the a_dictionary_object identifier name.
Therefore the real result would rather be:
a_dictionary_object definition_1 = text_1
a_dictionary_object definition_2 = text_2
a_dictionary_object etc = etc.
a_dictionary_object key_3 = value_3
a_dictionary_object key_4 = value_4
a_dictionary_object etc = etc.
......
......
......
a_dictionary_object random_n = random_n
a_dictionary_object etc = etc.
So, the reverse of the eval() function won't be that useful in this case.
Currently, one would need to do this:
# main2.py same as above, except:
for each_one_of_my_dictionaries_names in ( 'some_dictionary_1',
'some_other_dictionary_2',
'...',
'some_other_dictionary_n' ):
some_module2.some_function( { each_one_of_my_dictionaries_names :
eval(each_one_of_my_dictionaries_names) } )
# some_module2.py
def some_function(a_dictionary_name_object_container):
for _dictionary_name, _dictionary_object in a_dictionary_name_object_container.items():
for _key, _value in _dictionary_object.items():
print( str(_dictionary_name) +
" " +
str(_key) +
" = " +
str(_value) )
######
In conclusion:
Python passes only memory addresses as arguments to functions.
Strings representing the name of an identifier, can only be referenced back to the actual identifier by the eval() function if the name identifier is available in the current namespace.
A hypothetical reverse of the eval() function, would not be useful in cases where the identifier name is not 'seen' directly by the calling code. E.g. inside any called function.
Currently one needs to pass to a function:
the string representing the identifier name
the actual identifier (memory address)
This can be achieved by passing both the 'string' and eval('string') to the called function at the same time. I think this is the most 'general' way of solving this egg-chicken problem across arbitrary functions, modules, namespaces, without using corner-case solutions. The only downside is the use of the eval() function which may easily lead to unsecured code. Care must be taken to not feed the eval() function with just about anything, especially unfiltered external-input data.
Totally possible with the python-varname package (python3):
from varname import nameof
s = 'Hey!'
print (nameof(s))
Output:
s
Install:
pip3 install varname
Or get the package here:
https://github.com/pwwang/python-varname
I searched for this question because I wanted a Python program to print assignment statements for some of the variables in the program. For example, it might print "foo = 3, bar = 21, baz = 432". The print function would need the variable names in string form. I could have provided my code with the strings "foo","bar", and "baz", but that felt like repeating myself. After reading the previous answers, I developed the solution below.
The globals() function behaves like a dict with variable names (in the form of strings) as keys. I wanted to retrieve from globals() the key corresponding to the value of each variable. The method globals().items() returns a list of tuples; in each tuple the first item is the variable name (as a string) and the second is the variable value. My variablename() function searches through that list to find the variable name(s) that corresponds to the value of the variable whose name I need in string form.
The function itertools.ifilter() does the search by testing each tuple in the globals().items() list with the function lambda x: var is globals()[x[0]]. In that function x is the tuple being tested; x[0] is the variable name (as a string) and x[1] is the value. The lambda function tests whether the value of the tested variable is the same as the value of the variable passed to variablename(). In fact, by using the is operator, the lambda function tests whether the name of the tested variable is bound to the exact same object as the variable passed to variablename(). If so, the tuple passes the test and is returned by ifilter().
The itertools.ifilter() function actually returns an iterator which doesn't return any results until it is called properly. To get it called properly, I put it inside a list comprehension [tpl[0] for tpl ... globals().items())]. The list comprehension saves only the variable name tpl[0], ignoring the variable value. The list that is created contains one or more names (as strings) that are bound to the value of the variable passed to variablename().
In the uses of variablename() shown below, the desired string is returned as an element in a list. In many cases, it will be the only item in the list. If another variable name is assigned the same value, however, the list will be longer.
>>> def variablename(var):
... import itertools
... return [tpl[0] for tpl in
... itertools.ifilter(lambda x: var is x[1], globals().items())]
...
>>> var = {}
>>> variablename(var)
['var']
>>> something_else = 3
>>> variablename(something_else)
['something_else']
>>> yet_another = 3
>>> variablename(something_else)
['yet_another', 'something_else']
as long as it's a variable and not a second class, this here works for me:
def print_var_name(variable):
for name in globals():
if eval(name) == variable:
print name
foo = 123
print_var_name(foo)
>>>foo
this happens for class members:
class xyz:
def __init__(self):
pass
member = xyz()
print_var_name(member)
>>>member
ans this for classes (as example):
abc = xyz
print_var_name(abc)
>>>abc
>>>xyz
So for classes it gives you the name AND the properteries
This is not possible.
In Python, there really isn't any such thing as a "variable". What Python really has are "names" which can have objects bound to them. It makes no difference to the object what names, if any, it might be bound to. It might be bound to dozens of different names, or none.
Consider this example:
foo = 1
bar = 1
baz = 1
Now, suppose you have the integer object with value 1, and you want to work backwards and find its name. What would you print? Three different names have that object bound to them, and all are equally valid.
In Python, a name is a way to access an object, so there is no way to work with names directly. There might be some clever way to hack the Python bytecodes or something to get the value of the name, but that is at best a parlor trick.
If you know you want print foo to print "foo", you might as well just execute print "foo" in the first place.
EDIT: I have changed the wording slightly to make this more clear. Also, here is an even better example:
foo = 1
bar = foo
baz = foo
In practice, Python reuses the same object for integers with common values like 0 or 1, so the first example should bind the same object to all three names. But this example is crystal clear: the same object is bound to foo, bar, and baz.
Technically the information is available to you, but as others have asked, how would you make use of it in a sensible way?
>>> x = 52
>>> globals()
{'__builtins__': <module '__builtin__' (built-in)>, '__name__': '__main__',
'x': 52, '__doc__': None, '__package__': None}
This shows that the variable name is present as a string in the globals() dictionary.
>>> globals().keys()[2]
'x'
In this case it happens to be the third key, but there's no reliable way to know where a given variable name will end up
>>> for k in globals().keys():
... if not k.startswith("_"):
... print k
...
x
>>>
You could filter out system variables like this, but you're still going to get all of your own items. Just running that code above created another variable "k" that changed the position of "x" in the dict.
But maybe this is a useful start for you. If you tell us what you want this capability for, more helpful information could possibly be given.
By using the the unpacking operator:
>>> def tostr(**kwargs):
return kwargs
>>> var = {}
>>> something_else = 3
>>> tostr(var = var,something_else=something_else)
{'var' = {},'something_else'=3}
You somehow have to refer to the variable you want to print the name of. So it would look like:
print varname(something_else)
There is no such function, but if there were it would be kind of pointless. You have to type out something_else, so you can as well just type quotes to the left and right of it to print the name as a string:
print "something_else"
What are you trying to achieve? There is absolutely no reason to ever do what you describe, and there is likely a much better solution to the problem you're trying to solve..
The most obvious alternative to what you request is a dictionary. For example:
>>> my_data = {'var': 'something'}
>>> my_data['something_else'] = 'something'
>>> print my_data.keys()
['var', 'something_else']
>>> print my_data['var']
something
Mostly as a.. challenge, I implemented your desired output. Do not use this code, please!
#!/usr/bin/env python2.6
class NewLocals:
"""Please don't ever use this code.."""
def __init__(self, initial_locals):
self.prev_locals = list(initial_locals.keys())
def show_new(self, new_locals):
output = ", ".join(list(set(new_locals) - set(self.prev_locals)))
self.prev_locals = list(new_locals.keys())
return output
# Set up
eww = None
eww = NewLocals(locals())
# "Working" requested code
var = {}
print eww.show_new(locals()) # Outputs: var
something_else = 3
print eww.show_new(locals()) # Outputs: something_else
# Further testing
another_variable = 4
and_a_final_one = 5
print eww.show_new(locals()) # Outputs: another_variable, and_a_final_one
Does Django not do this when generating field names?
http://docs.djangoproject.com/en/dev//topics/db/models/#verbose-field-names
Seems reasonable to me.
I think this is a cool solution and I suppose the best you can get. But do you see any way to handle the ambigious results, your function may return?
As "is" operator behaves unexpectedly with integers shows, low integers and strings of the same value get cached by python so that your variablename-function might priovide ambigous results with a high probability.
In my case, I would like to create a decorator, that adds a new variable to a class by the varialbename i pass it:
def inject(klass, dependency):
klass.__dict__["__"+variablename(dependency)]=dependency
But if your method returns ambigous results, how can I know the name of the variable I added?
var any_var="myvarcontent"
var myvar="myvarcontent"
#inject(myvar)
class myclasss():
def myclass_method(self):
print self.__myvar #I can not be sure, that this variable will be set...
Maybe if I will also check the local list I could at least remove the "dependency"-Variable from the list, but this will not be a reliable result.
Here is a succinct variation that lets you specify any directory.
The issue with using directories to find anything is that multiple variables can have the same value. So this code returns a list of possible variables.
def varname( var, dir=locals()):
return [ key for key, val in dir.items() if id( val) == id( var)]
I don't know it's right or not, but it worked for me
def varname(variable):
for name in list(globals().keys()):
expression = f'id({name})'
if id(variable) == eval(expression):
return name
it is possible to a limited extent. the answer is similar to the solution by #tamtam .
The given example assumes the following assumptions -
You are searching for a variable by its value
The variable has a distinct value
The value is in the global namespace
Example:
testVar = "unique value"
varNameAsString = [k for k,v in globals().items() if v == "unique value"]
#
# the variable "varNameAsString" will contain all the variable name that matches
# the value "unique value"
# for this example, it will be a list of a single entry "testVar"
#
print(varNameAsString)
Output : ['testVar']
You can extend this example for any other variable/data type
I'd like to point out a use case for this that is not an anti-pattern, and there is no better way to do it.
This seems to be a missing feature in python.
There are a number of functions, like patch.object, that take the name of a method or property to be patched or accessed.
Consider this:
patch.object(obj, "method_name", new_reg)
This can potentially start "false succeeding" when you change the name of a method. IE: you can ship a bug, you thought you were testing.... simply because of a bad method name refactor.
Now consider: varname. This could be an efficient, built-in function. But for now it can work by iterating an object or the caller's frame:
Now your call can be:
patch.member(obj, obj.method_name, new_reg)
And the patch function can call:
varname(var, obj=obj)
This would: assert that the var is bound to the obj and return the name of the member. Or if the obj is not specified, use the callers stack frame to derive it, etc.
Could be made an efficient built in at some point, but here's a definition that works. I deliberately didn't support builtins, easy to add tho:
Feel free to stick this in a package called varname.py, and use it in your patch.object calls:
patch.object(obj, varname(obj, obj.method_name), new_reg)
Note: this was written for python 3.
import inspect
def _varname_dict(var, dct):
key_name = None
for key, val in dct.items():
if val is var:
if key_name is not None:
raise NotImplementedError("Duplicate names not supported %s, %s" % (key_name, key))
key_name = key
return key_name
def _varname_obj(var, obj):
key_name = None
for key in dir(obj):
val = getattr(obj, key)
equal = val is var
if equal:
if key_name is not None:
raise NotImplementedError("Duplicate names not supported %s, %s" % (key_name, key))
key_name = key
return key_name
def varname(var, obj=None):
if obj is None:
if hasattr(var, "__self__"):
return var.__name__
caller_frame = inspect.currentframe().f_back
try:
ret = _varname_dict(var, caller_frame.f_locals)
except NameError:
ret = _varname_dict(var, caller_frame.f_globals)
else:
ret = _varname_obj(var, obj)
if ret is None:
raise NameError("Name not found. (Note: builtins not supported)")
return ret
This will work for simnple data types (str, int, float, list etc.)
>>> def my_print(var_str) :
print var_str+':', globals()[var_str]
>>> a = 5
>>> b = ['hello', ',world!']
>>> my_print('a')
a: 5
>>> my_print('b')
b: ['hello', ',world!']
It's not very Pythonesque but I was curious and found this solution. You need to duplicate the globals dictionary since its size will change as soon as you define a new variable.
def var_to_name(var):
# noinspection PyTypeChecker
dict_vars = dict(globals().items())
var_string = None
for name in dict_vars.keys():
if dict_vars[name] is var:
var_string = name
break
return var_string
if __name__ == "__main__":
test = 3
print(f"test = {test}")
print(f"variable name: {var_to_name(test)}")
which returns:
test = 3
variable name: test
To get the variable name of var as a string:
var = 1000
var_name = [k for k,v in locals().items() if v == var][0]
print(var_name) # ---> outputs 'var'
Thanks #restrepo, this was exactly what I needed to create a standard save_df_to_file() function. For this, I made some small changes to your tostr() function. Hope this will help someone else:
def variabletostr(**df):
variablename = list(df.keys())[0]
return variablename
variabletostr(df=0)
The original question is pretty old, but I found an almost solution with Python 3. (I say almost because I think you can get close to a solution but I do not believe there is a solution concrete enough to satisfy the exact request).
First, you might want to consider the following:
objects are a core concept in Python, and they may be assigned a variable, but the variable itself is a bound name (think pointer or reference) not the object itself
var is just a variable name bound to an object and that object could have more than one reference (in your example it does not seem to)
in this case, var appears to be in the global namespace so you can use the global builtin conveniently named global
different name references to the same object will all share the same id which can be checked by running the id builtin id like so: id(var)
This function grabs the global variables and filters out the ones matching the content of your variable.
def get_bound_names(target_variable):
'''Returns a list of bound object names.'''
return [k for k, v in globals().items() if v is target_variable]
The real challenge here is that you are not guaranteed to get back the variable name by itself. It will be a list, but that list will contain the variable name you are looking for. If your target variable (bound to an object) is really the only bound name, you could access it this way:
bound_names = get_variable_names(target_variable)
var_string = bound_names[0]
Possible for Python >= 3.8 (with f'{var=}' string )
Not sure if this could be used in production code, but in Python 3.8(and up) you can use f' string debugging specifier. Add = at the end of an expression, and it will print both the expression and its value:
my_salary_variable = 5000
print(f'{my_salary_variable = }')
Output:
my_salary_variable = 5000
To uncover this magic here is another example:
param_list = f'{my_salary_variable=}'.split('=')
print(param_list)
Output:
['my_salary_variable', '5000']
Explanation: when you put '=' after your var in f'string, it returns a string with variable name, '=' and its value. Split it with .split('=') and get a List of 2 strings, [0] - your_variable_name, and [1] - actual object of variable.
Pick up [0] element of the list if you need variable name only.
my_salary_variable = 5000
param_list = f'{my_salary_variable=}'.split('=')
print(param_list[0])
Output:
my_salary_variable
or, in one line
my_salary_variable = 5000
print(f'{my_salary_variable=}'.split('=')[0])
Output:
my_salary_variable
Works with functions too:
def my_super_calc_foo(number):
return number**3
print(f'{my_super_calc_foo(5) = }')
print(f'{my_super_calc_foo(5)=}'.split('='))
Output:
my_super_calc_foo(5) = 125
['my_super_calc_foo(5)', '125']
Process finished with exit code 0
This module works for converting variables names to a string:
https://pypi.org/project/varname/
Use it like this:
from varname import nameof
variable=0
name=nameof(variable)
print(name)
//output: variable
Install it by:
pip install varname
print "var"
print "something_else"
Or did you mean something_else?