Send dictionary to function that does not accept **kwargs? - python

I just recently have started learning about the wonders of **kwargs but I've hit a stumbling block. Is there a way of sending keywords of a dictionary to a function that does not accept keyword arguments?
Consider the following simple setup:
def two(**kwargs):
return kwargs['second']
def three(**kwargs):
return kwargs['third']
parameterDict = {}
parameterDict['first'] = 1
parameterDict['second'] = 2
parameterDict['third'] = 3
I use some external code that interfaces in the following style:
fitObject = externalCode(two, first=1, second=2)
The problem is: "externalCode" does not accept **kwargs, so is there a smart way of getting the dictionary information into an acceptable form?
Also, the different functions take as parameters different subsets of the parameterDict. So, the function "two" might accept the "first" and "second" parameters, it rejects the "third". And "three" accepts all three.
------------------------------------- EDIT -------------------------------------
People have correctly commented that the above code won't fail -- so I've figured out my problem, and I'm not sure if it's worth a repost or not. I was doing something like this:
def printHair(**kwargs):
if hairColor == 'Black':
print 'Yep!'
pass
personA = {'hairColor':'blue'}
printHair(**personA)
NameError: global name 'hairColor' is not defined
And, apparently the fix is to explicitly include hairColor when defining: printHair(hairColor, **kwargs) in the first place.

>>> def externalCode(two, first=1, second=2):
... print two, first, second
...
>>> params = {'two': 9, 'first': 8, 'second': 7}
>>> externalCode(**params)
9 8 7

You can use the keyword expansion operator (**) to unpack a dictionary into function arguments.
fitObject = externalCode(two, **parameterDict)

How about using get
def printHair(**kwargs):
#check if hair colour is in kwargs
#if it is not you can set a default, otherwise default is None
personA['hairColor'] = kwargs.get('hairColor','Blue')
printHair(**personA)

Related

Use an object (e.g., an Enum) as a key in **kwargs?

Very simple question from a Python newbie:
My understanding is that the keys in a dict are able to be just about any immutable data type. Is it possible to pass an immutable object (e.g., a member of an enum class) as a key in the **kwargs dictionary for a function or a class? I have tried it and the answer seems to be "no":
from enum import Enum
class MyEnum(Enum):
X= 'X'
Y= 'Y'
def func(*args,**kwargs):
pass
func(MyEnum.X = 1)
Output:
"SyntaxError: keyword can't be an expression"
However, there may be something I am missing.
EDIT: Note that I am not trying to make the key equal to MyEnum.X.value (which is a string in this case); I want the key to be the actual Enum object, e.g. MyEnum.X.
You're doing:
func(MyEnum.X = 1)
Here, the problem is MyEnum.X = 1 -- Your keyword (MyEnum.X) is actually an expression (getattr(MyEnum, 'X')), and expressions can't be used as keywords in function calls. In fact, only identifiers can be used as keywords.
To get your call to work, you'll need to use dictionary unpacking like this:
func(**{MyEnum.X.name: 1})
Note, to get the name of the attribute, I needed to do MyEnum.X.name or MyEnum.X.value, depending on how you set up your enum -- In your case, I they are the same thing.
>>> from enum import Enum
>>> class Foo(Enum):
... X = 'X'
...
>>> Foo.X.value
'X'
>>> Foo.X.name
'X'
This won't work, because of the way keyword arguments are being processed. The documentation says:
[...] Next, for each keyword argument, the identifier is used to determine the corresponding slot (if the identifier is the same as the first formal parameter name, the first slot is used, and so on) [...]
So there must be a way to match the key from the dictionary to the formal parameter name. The exception:
keywords must be strings
when you try to pass something that's not a string:
func(**{MyEnum.X: 1})
suggest the simplest case is required: keys must be strings.
A possible workaround is to make implicit things explicit: just create a class that contains all the necessary information you want to pass in its attributes and pass it. The code will surely be more readable.
The answer to my original question is indeed "no". However, thanks to the input from mgilson and BartoszKP and others, the following work around I came up with is not a bad solution, and solves my current problem. I offer it for others to look at who are trying to do something similar:
from enum import Enum
class MyEnum(Enum):
X= 'X'
Y= 'Y'
def func(*args,**kwargs):
#replace kwargs with kwargsNew
kwargsNew = {}
for kwkey, kwvalue in kwargs.items():
try: kwargsNew[MyEnum(kwkey)] = kwvalue
except ValueError: kwargsNew[kwkey] = kwvalue
doStuffWithKwargs(kwargsNew)
def doStuffWithKwargs(k):
for K in k:
print(K)
#Pass the name X or Y as the key;
#all other keys not found in `MyEnum` are treated normally
func(X = 1, Y = 2, Z = 3)
Output:
Z
MyEnum.X
MyEnum.Y
(no errors)
Do you actually want to create an instnace of MyEnum?
myenum = MyEnum()
func(myenum.X = 1)
One alternative I have found is to pass a dict into *args instead of **kwargs, or to assign a dict to kwargs[0] directly:
func({MyEnum.X: 1})
func(kwargs = {MyEnum.X: 1})
(No errors produced)
However, I really don't like either of these methods.
EDIT: See my second answer for a much better solution.

Is it possible to change a function's default parameters in Python?

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

how to make my own mapping type in python

I have created a class MyClassthat contains a lot of simulation data. The class groups simulation results for different simulations that have a similar structure. The results can be retreived with a MyClass.get(foo) method. It returns a dictionary with simulationID/array pairs, array being the value of foo for each simulation.
Now I want to implement a method in my class to apply any function to all the arrays for foo. It should return a dictionary with simulationID/function(foo) pairs.
For a function that does not need additional arguments, I found the following solution very satisfying (comments always welcome :-) ):
def apply(self, function, variable):
result={}
for k,v in self.get(variable).items():
result[k] = function(v)
return result
However, for a function requiring additional arguments I don't see how to do it in an elegant way. A typical operation would be the integration of foo with bar as x-values like np.trapz(foo, x=bar), where both foo and bar can be retreived with MyClass.get(...)
I was thinking in this direction:
def apply(self, function_call):
"""
function_call should be a string with the complete expression to evaluate
eg: MyClass.apply('np.trapz(QHeat, time)')
"""
result={}
for SID in self.simulations:
result[SID] = eval(function_call, locals=...)
return result
The problem is that I don't know how to pass the locals mapping object. Or maybe I'm looking in a wrong direction. Thanks on beforehand for your help.
Roel
You have two ways. The first is to use functools.partial:
foo = self.get('foo')
bar = self.get('bar')
callable = functools.partial(func, foo, x=bar)
self.apply(callable, variable)
while the second approach is to use the same technique used by partial, you can define a function that accept arbitrary argument list:
def apply(self, function, variable, *args, **kwds):
result={}
for k,v in self.get(variable).items():
result[k] = function(v, *args, **kwds)
return result
Note that in both case the function signature remains unchanged. I don't know which one I'll choose, maybe the first case but I don't know the context on you are working on.
I tried to recreate (the relevant part of) the class structure the way I am guessing it is set up on your side (it's always handy if you can provide a simplified code example for people to play/test).
What I think you are trying to do is translate variable names to variables that are obtained from within the class and then use those variables in a function that was passed in as well. In addition to that since each variable is actually a dictionary of values with a key (SID), you want the result to be a dictionary of results with the function applied to each of the arguments.
class test:
def get(self, name):
if name == "valA":
return {"1":"valA1", "2":"valA2", "3":"valA3"}
elif name == "valB":
return {"1":"valB1", "2":"valB2", "3":"valB3"}
def apply(self, function, **kwargs):
arg_dict = {fun_arg: self.get(sim_args) for fun_arg, sim_args in kwargs.items()}
result = {}
for SID in arg_dict[kwargs.keys()[0]]:
fun_kwargs = {fun_arg: sim_dict[SID] for fun_arg, sim_dict in arg_dict.items()}
result[SID] = function(**fun_kwargs)
return result
def joinstrings(string_a, string_b):
return string_a+string_b
my_test = test()
result = my_test.apply(joinstrings, string_a="valA", string_b="valB")
print result
So the apply method gets an argument dictionary, gets the class specific data for each of the arguments and creates a new argument dictionary with those (arg_dict).
The SID keys are obtained from this arg_dict and for each of those, a function result is calculated and added to the result dictionary.
The result is:
{'1': 'valA1valB1', '3': 'valA3valB3', '2': 'valA2valB2'}
The code can be altered in many ways, but I thought this would be the most readable. It is of course possible to join the dictionaries instead of using the SID's from the first element etc.

Deal with undefined arguments more elegantly

The accepted paradigm to deal with mutable default arguments is:
def func(self, a = None):
if a is None:
a = <some_initialisation>
self.a = a
As I might have to do this for several arguments, I would need to write very similar 3 lines over and over again. I find this un-pythonically a lot of text to read for a very very standard thing to do when initialising class objects or functions.
Isn't there an elegant one-liner to replace those 3 lines dealing with the potentially undefined argument and the standard required copying to the class instance variables?
If a "falsy" value (0, empty string, list, dict, etc.) is not a valid value for a, then you can cut down the initialization to one line:
a = a or <initialize_object>
Another way of doing the same thing is as follows:
def func(self,**kwargs):
self.a=kwargs.get('a',<a_initialization>)
...
This has the added bonus that the value of a passed to the function could be None and the initialization won't overwrite it. The disadvantage is that a user using the builtin help function won't be able to tell what keywords your function is looking for unless you spell it out explicitly in the docstring.
EDIT
One other comment. The user could call the above function with keywords which are not pulled out of the kwargs dictionary. In some cases, this is good (if you want to pass the keywords to another function for instance). In other cases, this is not what you want. If you want to raise an error if the user provides an unknown keyword, you can do the following:
def func(self,**kwargs):
self.a=kwargs.pop('a',"Default_a")
self.b=kwargs.pop('b',"Default_b")
if(kwargs):
raise ... #some appropriate exception...possibly using kwargs.keys() to say which keywords were not appropriate for this function.
You could do this
def func(self, a=None):
self.a = <some_initialisation> if a is None else a
But why the obsession with one liners? I would usually use the 3 line version even if it gets repeated all over the place because if makes your code very easy for experienced Python programmers to read
just a little solution I came up by using an extra function, can be improved of course:
defaultargs.py:
def doInit(var, default_value,condition):
if condition:
var = default_value
return var
def func(a=None, b=None, c=None):
a = doInit(a,5,(a is None or not isinstance(a,int)))
b = doInit(b,10.0,(a is None or not isinstance(a,float)))
c = doInit(c,"whatever",(a is None or not isinstance(c, str)))
print a
print b
print c
if __name__ == "__main__":
func(10)
func(None,12341.12)
func("foo",None,"whowho")
output:
10
10.0
whatever
5
10.0
whatever
5
10.0
whowho
I like your question. :)
Edit: If you dont care about the variables type, please dont use isinstance().

Convert Variable Name to String?

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?

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