How to change content of numpy array when indexing with a list?

Question

could anyone explain me the reson why indexing the array using a list and using [x:x] lead to a very different result when manipulating numpy arrays?

Example:

a = np.array([[1,2,3,4],[3,4,5,5],[4,5,6,3], [1,2,5,5], [1, 2, 3, 4]]) 
print(a, '\n')

print(a[[3, 4]][:1][:, 1])
a[[3, 4]][:1][:, 1] = 99
print(a, '\n')


print(a[3:4][:1][:, 1])
a[3:4][:1][:, 1] = 99
print(a, '\n')

Output:

[[1 2 3 4]
 [3 4 5 5]
 [4 5 6 3]
 [1 2 5 5]
 [1 2 3 4]] 

[2]
[[1 2 3 4]
 [3 4 5 5]
 [4 5 6 3]
 [1 2 5 5]
 [1 2 3 4]] 

[2]
[[ 1  2  3  4]
 [ 3  4  5  5]
 [ 4  5  6  3]
 [ 1 99  5  5]
 [ 1  2  3  4]]

Is there a way to modify the array when indexing with a list?

Have you read about copy versus view in indexing? Advanced versus basic indexing? a[[3, 4]] creates a copy. a[3:4] a view. Try when indexing to use the form that indexes all dimensions at once, rather than chaining them. Especially when trying to set values. — hpaulj
– hpaulj, Commented Feb 18, 2021 at 18:29
@hpaulj thanks for the comment, ok, it makes much more sense now. However, I would still like to use advanced/fancy indexing, and modify the underlying array, like what I would get with a view. Is there a way to do that? — ClonedOne
– ClonedOne, Commented Feb 18, 2021 at 21:44

hpaulj · Accepted Answer · 2021-02-18 21:51:54Z

1

Create an index that selects the desired elements without chaining:

In [114]: a[[3,4],1]=90
In [115]: a
Out[115]: 
array([[ 1,  2,  3,  4],
       [ 3,  4,  5,  5],
       [ 4,  5,  6,  3],
       [ 1, 90,  5,  5],
       [ 1, 90,  3,  4]])

answered Feb 18, 2021 at 21:51

hpaulj

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