I have a numpy array of shape (206, 482, 3). I wanted to pick the 1st channel so I used name_of_array[:][:][0] but apparently that doesn't select the 1st channel.
I think name_of_array[:,:,0] picks the 1st channel. I don't understand why. Why name_of_array[:][:][0] != name_of_array[:,:,0]?
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):