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.

• jpp: Your logic does work, just print(output) at the end.
• Jonas Wolff: 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
• idnavid: Your original code works fine.

Answer 1

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)] 

Answer 2

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.

Answer 3

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()

Answer 4

You can use the following solution to solve your problem:

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

Answer 5

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.