Removing items from a nested list Python

Question

I am trying to remove items from a nested list in Python. I have a nested list as follows:

families = [[0, 1, 2],[0, 1, 2, 3],[0, 1, 2, 3, 4],[1, 2, 3, 4, 5],[2, 3, 4, 5, 6]]

I want to remove the entries in each sublist that coorespond to the indexed position of the sublist in the master list. So, for example, I need to remove 0 from the first sublist, 1 from second sublist, etc. I am trying to use a list comrehension do do this. This is what I have tried:

familiesNew = [ [ families[i][j] for j in families[i] if i !=j ] for i in range(len(families)) ]

This works for range(len(families)) up to 3, however beyond that I get IndexError: list index out of range. I am not sure why. Can somebody give me an idea of how to do this. Preferably a one-liner (list comprehension).

Thanks.

Niklas B. · Accepted Answer · 2012-03-23 23:43:58Z

10

You almost got it right. Just replace families[i][j] with j and it works:

>>> [ [ j for j in families[i] if i !=j ] for i in range(len(families)) ]
[[1, 2], [0, 2, 3], [0, 1, 3, 4], [1, 2, 4, 5], [2, 3, 5, 6]]

It can be written a bit cleaner using the enumerate function:

>>> [[f for f in family if f != i] for i, family in enumerate(families)]
[[1, 2], [0, 2, 3], [0, 1, 3, 4], [1, 2, 4, 5], [2, 3, 5, 6]]

Or even using remove if you don't mind changing the original list:

>>> for i, family in enumerate(families): family.remove(i)

edited Mar 23, 2012 at 23:43

answered Mar 23, 2012 at 23:15

Niklas B.

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1 Comment

johntfoster Over a year ago

Thanks, I decided to go with the last option as it was in fact okay and desirable to change the list in place.

sberry · Accepted Answer · 2012-03-24 00:32:04Z

5

Edited question, removing my answer which was solving the wrong problem. Also, added additional answer by @Ashwini:

For comparison's sake:

root# python -m timeit 'families = [[0, 1, 2],[0, 1, 2, 3],[0, 1, 2, 3, 4],[1, 2, 3, 4, 5],[2, 3, 4, 5, 6]]' '[x.remove(ind) for ind,x in enumerate(families) ]'
100000 loops, best of 3: 3.42 usec per loop    

root# python -m timeit -s 'families = [[0, 1, 2],[0, 1, 2, 3],[0, 1, 2, 3, 4],[1, 2, 3, 4, 5],[2, 3, 4, 5, 6]]' '[[f for f in family if f != i] for i, family in enumerate(families)]'
100000 loops, best of 3: 4.87 usec per loop

root# python -m timeit -s 'families = [[0, 1, 2],[0, 1, 2, 3],[0, 1, 2, 3, 4],[1, 2, 3, 4, 5],[2, 3, 4, 5, 6]]' '[ filter(lambda x:x!=i,j) for i,j in enumerate(families) ]'
100000 loops, best of 3: 7.99 usec per loop

These are micro-second, so I think whatever you want to do is fine unless you are going to be doing this a lot of times.

edited Mar 24, 2012 at 0:32

answered Mar 23, 2012 at 23:19

sberry

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3 Comments

Niklas B. Over a year ago

I think the semantics of this is not quite right. It's not that the element at the index i should be removed from every sublist, but the item i itself.

mgilson Over a year ago

This is a clever trick, but I think it doesn't quite do what it is intended to do. (It would work if the latter elements of the outer list were sequences starting at 0). +1 for timeit though.

sberry Over a year ago

My mistake for misreading the intent of the question (and subsequently for not checking the output to match)

mgilson · Accepted Answer · 2012-03-23 23:42:29Z

3

Does this do what you want?

familiesNew=[ filter(lambda x:x!=i,j) for i,j in enumerate(families) ]

EDIT

Also note, the reason yours failed is because at the third element of the outer list ([1, 2, 3, 4, 5]) you're trying to get the fifth element in your for loop (for j in families[i] == for j in [1,2,3,4,5]), but families[i] has a length of 5, meaning the largest index is 4. Sorry if that explanation is a little unclear...perhaps the following will help clear it up a little:

families = [[0, 1, 2],[0, 1, 2, 3],[0, 1, 2, 3, 4],[1, 2, 3, 4, 5],[2, 3, 4, 5, 6]]

def f(i,j):
    print i,j,families[i]
    return families[i][j]
#THIS DOES NOT WORK -- but it will tell you where it failed.
familiesNew = [ [ f(i,j) for j in families[i] if i !=j ] for i in range(len(families)) ]

edited Mar 23, 2012 at 23:42

answered Mar 23, 2012 at 23:17

mgilson

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Comments

Ashwini Chaudhary · Accepted Answer · 2012-03-23 23:43:24Z

2

If you want to modify the original list then try this:

>>>[x.remove(ind) for ind,x in enumerate(families) ]
>>>families
[[1, 2], [0, 2, 3], [0, 1, 3, 4], [1, 2, 4, 5], [2, 3, 5, 6]]

edited Mar 23, 2012 at 23:43

answered Mar 23, 2012 at 23:33

Ashwini Chaudhary

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3 Comments

mgilson Over a year ago

This answer is interesting because it does the calculation in place. Presumably it would return a list of None ... but that might be OK for some applications.

Ashwini Chaudhary Over a year ago

yes! this list comprehension will return a list of None of length == len(families) and will modify the original list.

Li-aung Yip Over a year ago

good solution, though usually it's considered impolite to use a list comprehension for side effects.

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