Find all index position in list based on partial string inside item in list

Question

mylist = ["aa123", "bb2322", "aa354", "cc332", "ab334", "333aa"]

I need the index position of all items that contain 'aa'. I'm having trouble combining enumerate() with partial string matching. I'm not even sure if I should be using enumerate.

I just need to return the index positions: 0,2,5

score 164 · Accepted Answer · edited Feb 24 '22 at 16:48

164

You can use enumerate inside a list-comprehension:

indices = [i for i, s in enumerate(mylist) if 'aa' in s]

edited Feb 24 '22 at 16:48

Tomerikoo

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answered Feb 13 '13 at 08:29

StoryTeller - Unslander Monica

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how can you make this be case insensitive? – abbood Feb 14 '17 at 06:45
1

@abbood - By comparing to `s.lower()` – StoryTeller - Unslander Monica Feb 14 '17 at 11:22
what if I just want to get single index for the first match only? – shashwat Oct 06 '17 at 15:48
@shashwat - You don't have to use the generator expression in the brackets to created a list. You can use it to create a generator `gen = ()` which you then call `next` on. – StoryTeller - Unslander Monica Oct 06 '17 at 18:25
if you could please give an example for how I use this? I am completely new to Python. – shashwat Oct 07 '17 at 05:24
@StoryTeller-UnslanderMonica I am fairly new to python. Could you provide a simple explanation how read you code indices = [i for i, s in enumerate(mylist) if 'aa' in s] Or could you write the code in different way? – user716255 Oct 25 '20 at 03:19
It does not run if there are NA's. How to consider this case? – Julien Sep 21 '22 at 15:56

score 24 · Answer 2 · answered Feb 13 '13 at 08:28

24

Your idea to use enumerate() was correct.

indices = []
for i, elem in enumerate(mylist):
    if 'aa' in elem:
        indices.append(i)

Alternatively, as a list comprehension:

indices = [i for i, elem in enumerate(mylist) if 'aa' in elem]

answered Feb 13 '13 at 08:28

pemistahl

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How might the list comprehension get extended to instead of "if 'aa'" to be "if ["aa","bb"] in elem" ?? – GrantRWHumphries Apr 17 '18 at 12:46
@GrantRWHumphries You go one level lower with second for loop: `indices = [i for i, elem in enumerate(mylist) if any(a in elem for a in ["aa","bb"])]`. – Daniel Jan 29 '22 at 13:33

score 12 · Answer 3 · answered Feb 13 '13 at 08:30

12

Without enumerate():

>>> mylist = ["aa123", "bb2322", "aa354", "cc332", "ab334", "333aa"]
>>> l = [mylist.index(i) for i in mylist if 'aa' in i]
>>> l
[0, 2, 5]

answered Feb 13 '13 at 08:30

TerryA

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This is likely to run in O(n^2), whereas using enumerate will be O(n). – johnsyweb Feb 13 '13 at 08:33
6

This will return a wrong result if one of the strings containing 'aa' is duplicated. – John La Rooy Feb 13 '13 at 10:01

score 2 · Answer 4 · answered May 12 '22 at 10:57

Based on this answer, I'd like to show how to "early exit" the iteration once the first item containing the substring aa is encountered. This only returns the first position.

import itertools
first_idx = len(tuple(itertools.takewhile(lambda x: "aa" not in x, mylist)))

This should be much more performant than looping over the whole list when the list is large, since takewhile will stop once the condition is False for the first time.

I know that the question asked for all positions, but since there will be many users stumbling upon this question when searching for the first substring, I'll add this answer anyways.

score -4 · Answer 5 · edited Mar 07 '19 at 20:06

-4

spell_list = ["Tuesday", "Wednesday", "February", "November", "Annual", "Calendar", "Solstice"]

index=spell_list.index("Annual")
print(index)

edited Mar 07 '19 at 20:06

Andronicus

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answered Mar 07 '19 at 19:49

user11167492

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This does not answer the question. A value error will be thrown when you use this method but only specifies list.index('an'). The method will be unable to find annual. – mastersom Nov 26 '19 at 11:37
1

Right. This is an incorrect answer to the question that has been asked. – Ombrophile May 19 '20 at 13:39

Find all index position in list based on partial string inside item in list

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