CODEWITHSUNDEEP
pythonnumpy
user308827
user308827

Reputation: 22031

Difference in shapes of numpy array

For the array:

import numpy as np
arr2d = np.array([[1,2,3],[4,5,6],[7,8,9]])
>>> arr2d
array([[1, 2, 3],
       [4, 5, 6],
       [7, 8, 9]])
>>> arr2d[2].shape
(3,)
>>> arr2d[2:,:].shape
(1, 3)

Why do I get different shapes when both statements return the 3rd row? and shouldn't the result be (1,3) in both cases since we are returning a single row with 3 columns?

Upvotes: 1

Views: 230

Answers (2)

Oliver W.
Oliver W.

Reputation: 13459

Why do I get different shapes when both statements return the 3rd row?

Because with the first operation you are indexing the rows, and selecting just ONE element, which -as mentioned in the single-element indexing paragraph of a multidimensional array- returns an array with a lower dimension (a 1D array).

In the 2nd example, you are using a slice as evident by the colon. Slicing operations do not reduce the dimensions of an array. This is also logical, because imagine the array would not have 3 but 4 rows. Then arr2d[2:,:].shape would be (2,3). The developers of numpy made slicing operations consistent and therefor they (slices) never reduce the number of dimensions of the array.

and shouldn't the result be (1,3) in both cases since we are returning a single row with 3 columns?

No, just because of the previous reasons.

Upvotes: 5

zw324
zw324

Reputation: 27210

When doing arr2d[2], you are taking a row out of the array;

While when doing arr2d[2:, :], you are taking a subset of rows out of the array ('slicing'), in this case being the rows starting from the 3rd to the end, which is only the 3rd, but it didn't change that you are taking a subset, not an element.

Upvotes: 1

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