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pythonnumpy
Shan
Shan

Reputation: 19293

How do I get all the values from a NumPy array excluding a certain index?

I have a NumPy array, and I want to retrieve all the elements except a certain index. For example, consider the following array

a = [0,1,2,3,4,5,5,6,7,8,9]

If I specify index 3, then the resultant should be

a = [0,1,2,4,5,5,6,7,8,9]

Upvotes: 61

Views: 90490

Answers (5)

AlexDC
AlexDC

Reputation: 21

Another solution is to use the concatenate function of NumPy:

>>> x = np.arange(0,10)
array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])
>>> i = 3
>>> np.concatenate((x[:i],x[(i+1):]))
array([0, 1, 2, 4, 5, 6, 7, 8, 9])

Upvotes: 2

Aku
Aku

Reputation: 621

Here's a one-liner if a is a NumPy array:

>>> a[np.arange(len(a))!=3]
array([0, 1, 2, 4, 5, 5, 6, 7, 8, 9])

Upvotes: 18

Jdog
Jdog

Reputation: 10781

a_new = np.delete(a, 3, 0)

3 here is the index you wish to remove, and 0 is the axis (zero in this case if using 1D array). See np.delete

Upvotes: 31

unutbu
unutbu

Reputation: 880917

Like resizing, removing elements from an NumPy array is a slow operation (especially for large arrays since it requires allocating space and copying all the data from the original array to the new array). It should be avoided if possible.

Often you can avoid it by working with a masked array instead. For example, consider the array a:

import numpy as np

a = np.array([0,1,2,3,4,5,5,6,7,8,9])
print(a)
print(a.sum())
# [0 1 2 3 4 5 5 6 7 8 9]
# 50

We can mask its value at index 3 and can perform a summation which ignores masked elements:

a = np.ma.array(a, mask=False)
a.mask[3] = True
print(a)
print(a.sum())
# [0 1 2 -- 4 5 5 6 7 8 9]
# 47

Masked arrays also support many operations besides sum.

If you really need to, it is also possible to remove masked elements using the compressed method:

print(a.compressed())
# [0 1 2 4 5 5 6 7 8 9]

But as mentioned above, avoid it if possible.

Upvotes: 76

NPE
NPE

Reputation: 500923

This should do it:

In [9]: np.hstack((a[:3], a[4:]))
Out[9]: array([0, 1, 2, 4, 5, 5, 6, 7, 8, 9])

If performance is an issue, the following will do it in place:

In [22]: a[3:-1] = a[4:]; a = a[:-1]

Upvotes: 6

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