How to check that variable is a lambda function

Question

I'm working on a project, which contains several modules. Simplifying the problem, there is some variable x. Sometimes it may be int or float or list. But it may be a lambda function, and should be treated in different way. How to check that variable x is a lambda?

For example

>>> x = 3
>>> type(x)
<type 'int'>
>>> type(x) is int
True
>>> x = 3.4
>>> type(x)
<type 'float'>
>>> type(x) is float
True
>>> x = lambda d:d*d
>>> type(x)
<type 'function'>
>>> type(x) is lambda
  File "<stdin>", line 1
    type(x) is lambda
                    ^
SyntaxError: invalid syntax
>>> type(x) is function
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
NameError: name 'function' is not defined
>>>

Closely related: How do I detect whether a Python variable is a function? — jpp
– jpp, Commented Jan 8, 2019 at 14:56

thefourtheye · Accepted Answer · 2014-05-25 06:01:38Z

31

You need to use types.LambdaType or types.FunctionType to make sure that the object is a function object like this

x = lambda d:d*d
import types
print type(x) is types.LambdaType
# True
print isinstance(x, types.LambdaType)
# True

and then you need to check the name as well to make sure that we are dealing with a lambda function, like this

x = lambda x: None
def y(): pass
print y.__name__
# y
print x.__name__
# <lambda>

So, we put together both these checks like this

def is_lambda_function(obj):
    return isinstance(obj, types.LambdaType) and obj.__name__ == "<lambda>"

As @Blckknght suggests, if you want to check if the object is just a callable object, then you can use the builtin callable function.

edited May 25, 2014 at 6:01

answered May 25, 2014 at 5:46

thefourtheye

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8 Comments

bereal Over a year ago

isinstance(x, types.LambdaType) looks more idiomatic.

Blckknght Over a year ago

It's worth noting that types.LambdaType is an alias for types.FunctionType. There's no (easy or reliable) way to tell the difference between a lambda function and one created with def.

thefourtheye Over a year ago

@Blckknght That is why we need to check the __name__ as well :-)

Blckknght Over a year ago

Ah, I see I hadn't payed enough attention to the end of the answer, and you've already addressed the issue of lambdas versus regular functions. I'd suggest though that there probably isn't much need to tell the difference. If your code really cares how a given function was defined, you're probably doing it wrong. Most of the time you should use "duck typing" rather than strict type checks. If you expect a function, accept anything that is callable (such as a class, or an instance of a class with a __call__ method). If you do need to check up front, use the builtin callable function.

Peter Nowee Over a year ago

__name__ can be changed. Alternatively, use __code__.co_name, which contains the name of the function/lambda at the time it was compiled and is read-only. See stackoverflow.com/questions/54094263/…

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Ricardo Decal · Accepted Answer · 2021-01-16 16:46:56Z

2

If you prefer the typing module, use Callable:

In [1]: from typing import Callable
In [2]: isinstance(lambda: None, Callable)
Out[2]: True

answered Jan 16, 2021 at 16:46

Ricardo Decal

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8 Comments

rth Over a year ago

what is the difference between your solution and @Blckknght suggestion to use callable built-in function?

Ricardo Decal Over a year ago

@rth you can't use callable() for typing hinting your signatures

rth Over a year ago

A regular function (not lambda) also returns True: def x(): pass, isinstance(x, Callable) ->True. So I don't see any advantage of your answer vs @thefourtheye one.

Ricardo Decal Over a year ago

1) your question should specify that you want to discriminate between lambda functions and ordinary functions if that is a requirement. 2) callable(lambda: None) == True

rth Over a year ago

I don't know how to make more specific, it is in the title How to check that variable is a lambda function :) Thank you for additional answer!

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Adriaan · Accepted Answer · 2023-03-21 07:50:31Z

0

Can consider the following:

type(x) == type(lambda x:x)

edited Mar 21, 2023 at 7:50

Adriaan

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answered Mar 21, 2023 at 7:49

user21445566

1

1 Comment

Adriaan Over a year ago

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