List comprehension with if condition in python

Question

I have a list which has elements of type int and str. I need to copy the int elements to another list. I tried a list comprehension which is not working. My equivalent loop is working.

input = ['a',1,'b','c',2,3,'c','d']
output = []
[output.append(a) for a in input if type(a) == int]

[None, None, None]

same logic in loop works.

output = []
for a in input:
    if type(a) == int:
        output.append(a)
print(output)

[1,2,3,]

Can I know what made the difference.

well your list comprehension never got any values, it just kept giving them to output, just write a if instead of 'output.append(a)' if you want them to go the list comprehensoion — Jonas Wolff, Sep 12 '18 at 01:08

U13-Forward · Answer 1 · 2018-09-12T01:12:18.343

When you're doing:

input = ['a',1,'b','c',2,3,'c','d']
output = []
[output.append(a) for a in input if type(a) == int]

append returns None, so that's why lot of none's

Btw,:

print(output)

Will be desired result.

Your logic actually works!

But it's not good to use list comprehension as side-effects, so best way would be:

output=[a for a in input if type(a) == int]

Also final best would be isinstance:

output=[a for a in input if isinstance(a,int)]

John Hall · Answer 2 · 2018-09-12T01:36:23.357

An idiomatic solution would be:

    input = ['a', 1, 'b', 'c', 2, 3, 'c', 'd']
    output = [a for a in input if type(a) == int]

You do not want to use output.append. That is implied by the list comprehension. You can change the first a in the comprehension to an expression containing a. output.append(a) is a method call that returns NONE

Use list comprehensions when you want to collect values into a list you are assigning to a variable or as part of larger expression. Do not use them for side effects as a different format for a 'for loop'. Instead use the for loop.

Spaces after the commas are not required but are considered good style.

score 2 · Answer 3 · answered Sep 12 '18 at 01:14

2

input = ['a',1,'b','c',2,3,'c','d']
output = [a for a in input if type(a) == int]

The list comprehension automatically creates the list - no need to use append()

answered Sep 12 '18 at 01:14

Dan Swain

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score 0 · Answer 4 · edited Sep 12 '18 at 02:32

0

You can use the following solution to solve your problem:

output = [a for a in input if type(a) == int]

edited Sep 12 '18 at 02:32

Grant Miller

27,532
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answered Sep 12 '18 at 01:15

Michael Smith

1
1

score 0 · Answer 5 · answered Sep 12 '18 at 02:39

A solution would be:

input = ['a',1,'b','c',2,3,'c','d']
output=[item for item in input if type(item) is int]  or
output=[item for item in input if isinstance(item,int)]

and the last one is quicker than the first.

but when you use:

input = ['a',1,'b','c',2,3,'c','d']
output=[]
[output.append(a) for a in input if type(a) == int]

it means the code execute two results ,the one is output.append(a) append object in end and the list comprehension with if condition creates a new list object after executing ,as a.append()(Look_in_IDLE) return None , so there are three None.

List comprehension with if condition in python

5 Answers5

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