python remove list index base on another lists

Question

I have a 2 lists as detailed below:

a = [[0, 1, 1, 2], [0, 5, 0, 0], [2, 0, 3, 3]]
remove_a_index = [[0], [0, 2, 3], [1]]

What is the best solution to remove list index of a base on the number from remove_a_index for e.g. for a[0] I need to remove number 0

Can you provide the desired output based on your example please? — Nuageux, May 20 '17 at 10:49

Moinuddin Quadri · Accepted Answer · 2017-05-20T11:01:10.457

You may use a nested list comprehension expression using zip() and enumerate() to filter the content as:

>>> a = [[0, 1, 1, 2], [0, 5, 0, 0], [2, 0, 3, 3]]
>>> remove_a_index = [[0], [0, 2, 3], [1]]

>>> a = [[j for i, j  in enumerate(x) if i not in y] for x, y in zip(a, remove_a_index)]
# where new value of `a` will be:
# [[1, 1, 2], [5], [2, 3, 3]]

Based on your desired result, in case if you just want to remove zeroes from the a list then you don't need the intermediate remove_a_index list. You may use a list comprehension expression to skip the zeroes from the new list as:

>>> a = [[0, 1, 1, 2], [0, 5, 0, 0], [2, 0, 3, 3]]

>>> [[j for j in i if j!=0] for i in a]
[[1, 1, 2], [5], [2, 3, 3]]

For longer lists, you could also use `zip(a, map(set, remove_a_index))` to speed up the lookup. — tobias_k, May 20 '17 at 11:08

score 1 · Answer 2 · answered May 20 '17 at 10:53

1

If I understood correctly the question, this should work:

for i, to_remove in enumerate(remove_a_index):
    for j in reversed(to_remove):
        del a[i][j]

answered May 20 '17 at 10:53

user1620443

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score 1 · Answer 3 · edited Mar 07 '21 at 11:03

1

Python has a language feature called List Comprehensions that is perfectly suited to making this sort of thing extremely easy. The following statement does exactly what you want and stores the result in l3:

As an example, if I have l1 = [1,2,6,8] and l2 = [2,3,5,8], l1 - l2 should return [1,6]:

l3 = [x for x in l1 if x not in l2]
l3 will contain [1, 6].

edited Mar 07 '21 at 11:03

halfer

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answered May 20 '17 at 10:55

Aditya

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Amit Wolfenfeld · Answer 4 · 2017-05-21T05:48:21.267

0

You can do the following:

Create a new list.
Modify the all the items that you wish to remove to 'remove' on the original list.
Populate the new list with all the items that are not 'remove'.

The code:

a_new=[]
for i, item in enumerate(a):
    for element_to_remove in remove_a_index[i]:
        item[element_to_remove]='remove'

    new_item = [element  for element in item if element!='remove']
    a_new.append(new_item)
a=a_new

edited May 21 '17 at 05:48

answered May 20 '17 at 11:07

Amit Wolfenfeld

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I think you should provide at least a brief explanation to your answer. – lmiguelvargasf May 21 '17 at 02:04

score 0 · Answer 5 · answered May 20 '17 at 11:14

The shortest one-liner is below:

a = [[0, 1, 1, 2], [0, 5, 0, 0], [2, 0, 3, 3]]
remove_a_index = [[0], [0, 2, 3], [1]]

b = [[y for y in original_tuple if y not in remove_a_index[index]] for index, original_tuple in enumerate(a)]
print b

To explain, it's using list comprehension to loop through a and use the index:

[? for index, original_tuple in enumerate(a)]

At this point the index is (0, 1, 2...) and original_tuple is each tuple.

Then for each tuple, you can access the subtracting tuple (remove_a_index[x]) by checking if it's in it or not.

[y for y in original_tuple if y not in remove_a_index[index]] ...

python remove list index base on another lists

5 Answers5