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How to remove multiple indexes from a list at the same time? [duplicate]

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python

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How do I remove multiple indexes from a list?

Suppose we want to remove multiple elements from a list by index range, then we can use del keyword i.e. It will delete the elements in list from index1 to index2 – 1.

How do I remove a list of lists indices in Python?

In Python, use list methods clear() , pop() , and remove() to remove items (elements) from a list. It is also possible to delete items using del statement by specifying a position or range with an index or slice.

How do I remove indexes from a list?

There are two options to remove an element by its index in list. Using del statement, and using pop() method. The del statement needs index of the element to remove.


You need to do this in a loop, there is no built-in operation to remove a number of indexes at once.

Your example is actually a contiguous sequence of indexes, so you can do this:

del my_list[2:6]

which removes the slice starting at 2 and ending just before 6.

It isn't clear from your question whether in general you need to remove an arbitrary collection of indexes, or if it will always be a contiguous sequence.

If you have an arbitrary collection of indexes, then:

indexes = [2, 3, 5]
for index in sorted(indexes, reverse=True):
    del my_list[index]

Note that you need to delete them in reverse order so that you don't throw off the subsequent indexes.


remove_indices = [1,2,3]
somelist = [i for j, i in enumerate(somelist) if j not in remove_indices]

Example:

In [9]: remove_indices = [1,2,3]

In [10]: somelist = range(10)

In [11]: somelist = [i for j, i in enumerate(somelist) if j not in remove_indices]

In [12]: somelist
Out[12]: [0, 4, 5, 6, 7, 8, 9]

There wasn't much hint on performance for the different ways so I performed a test on removing 5000 items from 50000 in all 3 generally different approaches, and for me numpy was the winner (if you have elements that fit in numpy):

  • 7.5 sec for the enumerated list comprehension [4.5 sec on another PC]
  • 0.08 sec for deleting items in reverse order [0.017 (!) sec]
  • 0.009 sec for numpy.delete [0.006 sec]

Here's the code I timed (in the third function conversion from/to list may be removed if working directly on numpy arrays is ok):

import time
import numpy as np
import random

def del_list_indexes(l, id_to_del):
    somelist = [i for j, i in enumerate(l) if j not in id_to_del]
    return somelist

def del_list_inplace(l, id_to_del):
    for i in sorted(id_to_del, reverse=True):
        del(l[i])

def del_list_numpy(l, id_to_del):
    arr = np.array(l, dtype='int32')
    return list(np.delete(arr, id_to_del))

l = range(50000)
random.shuffle(l)
remove_id = random.sample(range(len(l)), 5000) # 10% ==> 5000

# ...

If you can use numpy, then you can delete multiple indices:

>>> import numpy as np
>>> a = np.arange(10)
>>> np.delete(a,(1,3,5))
array([0, 2, 4, 6, 7, 8, 9])

and if you use np.r_ you can combine slices with individual indices:

>>> np.delete(a,(np.r_[0:5,7,9]))
array([5, 6, 8])

However, the deletion is not in place, so you have to assign to it.


If they're contiguous, you can just do

x[2:6] = []

If you want to remove noncontiguous indexes, it's a little trickier.

x = [v for i,v in enumerate(x) if i not in frozenset((2,3,4,5))] 

lst = ['a', 'b', 'c', 'd', 'e', 'f', 'g'];
lst = lst[0:2] + lst[6:]

This is a single step operation. It does not use a loop and therefore executes fast. It uses list slicing.