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
– void, Commented Jun 13, 2017 at 6:17
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
– void, Commented Jun 13, 2017 at 6:18
Split your code into several lines. I don't understand what kind of output you are after in relation to the input — Sam
– Sam, Commented Jun 13, 2017 at 6:20
Just. Write. A. For. Loop. And use enumerate instead of .index which makes your algorithm inefficient. — juanpa.arrivillaga
– juanpa.arrivillaga, Commented Jun 13, 2017 at 6:20

juanpa.arrivillaga · Accepted Answer · 2023-11-24 20:16:30Z

13

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)
...
>>> 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.

edited Nov 24, 2023 at 20:16

answered Jun 13, 2017 at 6:29

juanpa.arrivillaga

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1 Comment

GeeTransit Over a year ago

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.

Community · Accepted Answer · 2020-06-20 09:12:55Z

1

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.

edited Jun 20, 2020 at 9:12

CommunityBot

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answered Jun 13, 2017 at 6:31

DexJ

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Comments

SiHa · Accepted Answer · 2017-06-13 08:37:32Z

1

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.

edited Jun 13, 2017 at 8:37

SiHa

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answered Jun 13, 2017 at 6:57

Edwin van Mierlo

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1 Comment

SiHa Over a year ago

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.

Peter Mortensen · Accepted Answer · 2019-07-23 00:12:25Z

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, 2019 at 0:12

Peter Mortensen

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answered Jun 13, 2017 at 6:22

codelessbugging

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