How to slice numpy array with rows of varying length

Question

I have a numpy array arr of numpy arrays each with varying length. I can get the shape of arr:

arr.shape

>>> (9,)

I can get the shape of one of the elements of arr:

arr[0].shape

>>> (6, 1, 2)

And I know that all such elements have shape (n, 1, 2).

I want to slice arr to get a 1 dimensional result as follows:

arr[:,:,:,0]

But I get the following error:

IndexError: too many indices for array

EDIT: My original question was misleading. I want to do this slice so that I can assign values to the slice. So getting the slice in a new variable is not useful for my case. Essentially I want to do something like this in a simple one liner:

arr[:,:,:,0] = arr[:,:,:,0] - np.min(arr[:,:,:,0])

The element arrays have to be operated on individually. It is basically a list of arrays. — hpaulj
– hpaulj, Commented Jan 26, 2020 at 14:37

Georgina Skibinski · Accepted Answer · 2020-01-26 13:47:52Z

1

You can fix your first (in fact all varying ones) dimension, and apply your transformation per static-shaped elements of arr

import numpy as np
from random import randint

arr=np.array([np.random.randint(3,15, size=(randint(3,9),randint(3,7),randint(6,19))) for el in range(9)])

print(arr.shape)
print(arr[0].shape)
for i in range(arr.shape[0]):
    arr[i][:,:,0]-=arr[i][:,:,0].min()
    print(arr[i][:,:,0])

edited Jan 26, 2020 at 13:47

answered Jan 26, 2020 at 13:31

Georgina Skibinski

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Comments

Mihai Alexandru-Ionut · Accepted Answer · 2020-01-26 13:07:22Z

1

You could use list comprehension version of your solution.

desired_result = np.array([el[:,:,0] for el in arr])

answered Jan 26, 2020 at 13:07

Mihai Alexandru-Ionut

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1 Comment

Alexander Soare Over a year ago

Thank you. Although I made a mistake with the way I asked the question and now I've reframed it.

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