Python lambda function in a list gives unexpected result

Question

So I’m trying to make a list where a lambda functions are the elements of the list. The lambda function calls
another function which I pass an argument to. The problem is that lambda function only ‘saves’ the last value for all other items in the list. See below

The question is what should I do to get the desired result?

Edit: The stated problem is simplified. I have to use lambda for the solution

This is the code I’m trying to understand the problem:

def f(B):
    print(B)
A = [lambda: f(a) for a in ["foo", "faz", "baz"]]
for a in A:
    a()

Desired result:

foo
faz
baz

Real result:

baz
baz
baz

Asked By: eiskaffe

||

Source

Answer 1

For loop in lambda saves last item of list so its better to write it way out lambda
like this:

def b():
    A = ["foo","faz","baz"]
    For a in A: 
         Print(a)

and output is:

foo
faz
baz

Answered By: parsa ataei

Answer 2

While lambda accesses the context it’s defined in, a for loop doesn#t create a new context each time it runs. That would be too inefficient.

Thus, by the time your code actually calls the lambda functions that context has ended and a contains the last value the loop assigned to it.

Correct code:

def f(B):
    print(B)
A = ["foo", "faz", "baz"]
for a in A:
    f(a)

If this answer is not sufficient, please clarify why do you need that lambda in the first place.

Answered By: Matthias Urlichs

Answer 3

if you need to create an array of function calls you can achieve what you are trying to do with the following:

def f(x):
     def g():
         print(x)
     return g

A = [f(a) for a in ["foo", "faz", "baz"]]

for a in A:
    a()

output

foo
faz
baz

Answered By: SR3142

Python lambda function in a list gives unexpected result

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