Multiple If/else in list comprehension in Python

Question

I have this

s = ['son','abc','pro','bro']
b = ['son','bro']
c = ['pro','quo']

The expected output is this. Where items in the output are index(item_in_s) if it is present in list b. Or index(item_in_s)+10 if an item is in c.

[0,12,3]

I tried this:

index_list = [s.index(item) if item in b else s.index(item)+10 if item in c for item in s]
print(index)

But apparently this is a syntax error. So I tried this:

index_list = [s.index(item) if item in b else s.index(item)+10 for item in s if item in c]
    print(index)

Output:

[12]

This just changes the whole logic. Although I could do this

fin = [s.index(item) if item in b else s.index(item)+10 if item in c  else '' for item in s]
fin = [item for item in fin if item!='']
print(fin)

Desired output obtained:

[0, 12, 3]

But how do I obtain what I want in list comprehension itself or is there something like else continue in list comprehensions?

simply put something like `[x if x in list1 else x+10 if x in list2 for x in my_list]` which gives a syntax error. — void, Jun 13 '17 at 06:17
I want something like `[x if x in list1 else x+10 if x in list2 else pass for x in my_list]` but i know pass gives syntax error too. Any alternatives? — void, Jun 13 '17 at 06:18
Split your code into several lines. I don't understand what kind of output you are after in relation to the input — Sam, Jun 13 '17 at 06:20
Just. Write. A. For. Loop. And use `enumerate` instead of `.index` which makes your algorithm inefficient. — juanpa.arrivillaga, Jun 13 '17 at 06:20
Oh thanks for the tip first i used enumerate i thought that would be inefficient. You are right it would search everytime yes. — void, Jun 13 '17 at 06:22

juanpa.arrivillaga · Answer 1 · 2022-10-10T21:40:32.353

Fundamentally, a list-comprehension forces you to be very inefficient:

>>> [i if item in b else i + 10 if item in c else None for i, item in enumerate(s) if item in b or item in c]
[0, 12, 3]

This has to check the membership item in b and c twice each in the worst-case if you want that output. Instead, just use a for-loop:

>>> index_list = []
>>> for i, item in enumerate(s):
...     if item in b:
...         index_list.append(i)
...     elif item in c:
...         index_list.append(i + 10)
...     else item in c:
...         index_list.append(None)
...
>>> index_list
[0, 12, 3]
>>>

Simple, readable, straight-forward and Pythonic.

Edit

You can do something close to this with assignment expressions:

[
    i if item_in_b else i + 10 if item_in_c else None 
    for i, item in enumerate(s) 
    if (item_in_b:=item in b) or (item_in_c:=item in c)
]

but I still prefer the loop.

The condition `item in b` could be removed and only checked once when Python 3.8 comes and we can use `(item_in_b := item in b)` and do `if item_in_b` later on. — GeeTransit, Jun 26 '19 at 19:38

score 1 · Answer 2 · edited Jul 23 '19 at 00:12

1

index_list = [s.index(item) if item in b else s.index(item) + 10 for item in s if item in b or item in c]

We make sure it's in either b or c, and then the index will be either of the cases.

edited Jul 23 '19 at 00:12

Peter Mortensen

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answered Jun 13 '17 at 06:22

codelessbugging

2,849
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score 1 · Answer 3 · edited Jun 20 '20 at 09:12

your solution glitch can be solved by avoiding the elements that doesn't exist in list b&c.

you can do this by creating new list and applying simple set operation

check this little change in your solution.

fin = [s.index(item) if item in b else s.index(item)+10 if item in c  else '' for item in list(set(b+c)&set(s))]

by doing this your conditinal statement else is never going to execute cause list on which you are iterating only has element that either list in b or c only.

score 1 · Answer 4 · edited Jun 13 '17 at 08:37

already good answers given, here is one which is not mentioned yet:

fin = [item for item in ([s.index(item) if item in b else s.index(item)+10 if item in c  else '' for item in s]) if item!='']
print(fin)

basically it is a combination of the original 2 lines of code:

fin = [s.index(item) if item in b else s.index(item)+10 if item in c  else '' for item in s]
fin = [item for item in fin if item!='']

Not necessarily "faster or better", just a combination which was not given before. When doing list comprehensions you always have the risk that you iterate more than you really need to.

A better solution would be a for loop, which is given in another answer.

It would be more useful if you could explain how your solution works, and how it is better than the other answers. Code-only answers are not terribly helpful. — SiHa, Jun 13 '17 at 07:01

Multiple If/else in list comprehension in Python

4 Answers4

Edit

check this little change in your solution.

Linked

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