func = Lambda x = something: x

Question

I was doing some exercises on codewars and found this question:

Complete the function scramble(str1, str2) that returns true if a portion of str1 characters can be rearranged to match str2, otherwise returns false.

after completing it and looking at others answers I found:

scramble=lambda a,b,c=__import__('collections').Counter:not c(b)-c(a)

Can someone explain what is happening here? Why is there a = after the declaration of lambda and how are the arguments being assigned???

I did not found much explanation on this.

Beside the point, but just in case you weren't aware, [avoid named lambdas](/q/38381556/4518341), use a `def` instead. I suppose this is one reason why :) — wjandrea, May 17 '23 at 03:33
BTW, welcome to Stack Overflow! Check out the [tour], and [ask] if you want tips. — wjandrea, May 17 '23 at 03:34
Just to note, It's against PEP8 and not readable at all but a cool syntax useful for mini scripts in those websites. — S.B, May 17 '23 at 04:06

score 6 · Accepted Answer · edited May 17 '23 at 03:38

The equal sign is setting a default value for the c argument. The default value is __import__('collections').Counter.

The lambda is equivalent to

def scramble(a, b, c=__import__('collections').Counter):
    return not c(b)-c(a)

or, extracting the import,

import collections
def scramble(a, b, c=collections.Counter):
    return not c(b)-c(a)

except that __import__ doesn't assign the imported module to the collections variable.

func = Lambda x = something: x

1 Answers1