Query about picking values from numpy array

It's important to understand what each thing does. To do this break up the action left to right. Perhaps rewriting will make this more clear:

x[:][:][0] -> ( ( x[:] )[:] )[0]  # Both are valid and equivalent Python syntax

So basically, we apply [:] to x, then [:] to the result, then [0] to this result. What does x[:]? Just return a copy of x! Thus

( (x[:])[:] )[0] == ( (x)[:] )[0] == (x[:])[0] == x[0]

This is of course, not what you expected. On the other hand,

x[:, :, 0]

returns at once the 0 column of all rows of all frames (I'm treating the index as [frame, row, col]).

The short answer: because thats the syntax (see Numpy basics indexing).

arr[:]       == arr # full slice of all dimensions of the array 
arr[:][:]    == arr # full slice of a full slice of all dimensions 
arr[:][:][0] == arr # equal to arr[0] because the first 2 [:] slice all

vs

arr[:,:,0]  # slice all of 1st dim, slice all of 2nd dim, get 0th of 3rd arr

One way to figure things like this out yourself is to make a simplified example and experiment (heeding How to debug small programs):

import numpy as np

res = np.arange(4 * 3 * 2).reshape(4,3,2)


print(":,:,:")
print(res[:, :, :])  

print("\n1:2,1:2,:")
print(res[1:2, 1:2, :])

print("\n:,:,0")
print(res[:, :, 0])

print("\n:,:,1")
print(res[:,:,1])

Output:

# :,:,:   == all of it  
[[[ 0  1]
  [ 2  3]
  [ 4  5]]

[[ 6  7]
  [ 8  9]
  [10 11]]

[[12 13]
  [14 15]
  [16 17]]

[[18 19]
  [20 21]
  [22 23]]]

# 1:2,1:2,:
[[[8 9]]]

# :,:,0
[[ 0  2  4]
[ 6  8 10]
[12 14 16]
[18 20 22]]

# :,:,1
[[ 1  3  5]
[ 7  9 11]
[13 15 17]
[19 21 23]]

There are lots of questions about numpy-slicing on SO, some of which are worthwhile studying to advance your knowledge (suggested as probably dupes but they do not address the confusion correctly):

• Numpy extract submatrix
• Selecting specific rows and columns from NumPy array