Python: Removing elements from list by index based on elements in another list

Question

So I have a list

a_list = [0,3,4,7,9,11,15]

and a list of outliers

outliers = [0,3,4]

The outliers is the list of indexes that need to be removed from a_list

So in this case, remove the elements in index 0, 3, 4 from a_list. Result should be:

remove these = [0,7,9]
a_list = [0,3,4,11,15]

If I use the del method, the index will change so once I remove 0th index, I will still remove the 3rd index but instead remove 9 and 4th index will remove 15.

How can I remove items from a_list with the indexes specified in a separate list?

Start from the largest one, i.e. remove 4th element first, then 3rd, then 0th. — Selcuk
– Selcuk, Commented Mar 24, 2021 at 4:18
first, get all the values in a different list and then remove it — nasc
– nasc, Commented Mar 24, 2021 at 4:18
Do something like new list = [j for i, j in enumerate(a_list) if i not in [0, 7, 9]] .... another solution is instead of removing items, replace them with a temporary number or character eg 0... and then add an extra line... list(filter((0).__ne__, a_list)) or list(filter(lambda a: a != 2, a_list)) — user6304394
– user6304394, Commented Mar 24, 2021 at 4:20

mhhabib · Accepted Answer · 2021-03-24 04:25:33Z

2

Compare index to outliers list. If the index is matching to outliers then skip it.

print([x for i, x in enumerate(a_list) if i not in outliers])

Better to use set(outliers) if outliers contain the duplicate elements.

edited Mar 24, 2021 at 4:25

answered Mar 24, 2021 at 4:21

mhhabib

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2 Comments

Francisco Over a year ago

A possible optimization is to convert outliers to a set

mhhabib Over a year ago

Could be better to use a set but I assumed outliers contains distinct element

JT Raber · Accepted Answer · 2021-03-24 04:21:34Z

-1

Here is a real quick implementation based on what the others were commenting about.

a_list = [0,3,4,7,9,11,15]
outliers = [0,3,4]
outliers.reverse()
for i in outliers:
    del a_list[i]
print(a_list)

Simply call outliers.reverse() before iterating and delete as normal.

answered Mar 24, 2021 at 4:21

JT Raber

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