CODEWITHSUNDEEP
pythonlambda
Peaceful
Peaceful

Reputation: 5470

Lambda function returns function name instead of value in Python

I am having difficulty in understanding the lambda function syntax in python. In particular, I don't understand why the following code shouldn't work:

def f(x):
    return lambda x:x**2

f(2)

The output that I expect is 4 but the actual output looks like this:

<function __main__.<lambda>>

What is going on? Thanks in advance

Upvotes: 4

Views: 8011

Answers (5)

SAGAR LIMBU
SAGAR LIMBU

Reputation: 1

Using the parameter with '()' outside in the function should return the value.

def my_color(x): return (lambda j: 100 if j== 0 else 200)(x)

output: my_color(0) => 100

Upvotes: 0

James
James

Reputation: 1238

f(2) returns the lambda. That is what <function __main__.<lambda>> is. Note that the x inside the scope of the lambda is not the same x that is passed in as the argument to f. So you could have defined your function with no arguments and it would have the same result.

To call the lambda, you can do f()(2).

Upvotes: 2

Kamehameha
Kamehameha

Reputation: 5488

You are doing it wrong.
Your f function returns a lambda function, which needs to be called.
So to make it work - 

>>> f(0)(2)
      ^ This can be anything
4

Try something like this -

>>> f = lambda x:x**2
>>> f(2) 
4

Upvotes: 1

Anand S Kumar
Anand S Kumar

Reputation: 90979

You need to call the lambda function to get the result. Not sure what you are doing with that though.

In your case -

f(2)(2)
>>> 4

If you just want f to refer to the lambda function, then do -

f = lambda x:x**2
f(2)
>>>> 4

Do not return it from a function.

Upvotes: 1

zhangxaochen
zhangxaochen

Reputation: 34037

You need to call the lambda using ():

In [1]: def f(x):
   ...:     return (lambda n:n**2)(x)
   ...: 

In [2]: f(3)
Out[2]: 9

Or assign the lambda to a var:

In [3]: f=lambda x:x**2

In [4]: f(4)
Out[4]: 16

Upvotes: 12

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