How do I convert complex function to lambda in python?

Question

I'm thinking of ways to convert a more complex function to lambda and put it inside a map instead of f. The function is this:

#this function should be in lambda
def f(s):
    if (type(s) == int):
        return s*2
    else:
        return s.replace('r', '*')

lst = [4,5,6,'rer', 'tr',50,60]
lst = list(map( f, lst))
#result of lst is [8, 10, 12, '*e*', 't*', 100, 120]
#lst = list(map ( lambda x.... , lst))

Or lambda is only supposed to deal with short functions? Big ones have to be 'decomposed' out into separate functions?

U13-Forward · Accepted Answer · 2018-12-03 09:59:15Z

2

Use if else statement in a lambda:

print(list(map(lambda x: x*2 if type(x)==int else x.replace('r','*'),lst)))

Output:

[8, 10, 12, '*e*', 't*', 100, 120]

Even better, use isinstance:

print(list(map(lambda x: x*2 if isinstance(x,int) else x.replace('r','*'),lst)))

edited Dec 3, 2018 at 9:59

answered Dec 3, 2018 at 9:58

U13-Forward

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4 Comments

ERJAN Over a year ago

perfect... is it possible to create arbitrarily long lambda functions?

U13-Forward Over a year ago

@ERJAN What you mean?

ERJAN Over a year ago

how big can a lambda function be? it can be as complex and long as a separate function?

Stop harming Monica Over a year ago

@ERJAN What makes you think there are limits to the "size" of a lambda?

meW · Accepted Answer · 2018-12-03 09:59:57Z

1

Prefer isinstance in place of type

lst = [4,5,6,'rer', 'tr',50,60]
lst = list(map(lambda x: x*2 if isinstance(x, int) else x.replace('r', '*'), lst))

lst
[8, 10, 12, '*e*', 't*', 100, 120]

answered Dec 3, 2018 at 9:59

meW

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