Accessing second index in 3-d numpy array

Question

I have a 2-d numpy array that I convert into a 3-d array by using:

trainSet = numpy.reshape(trainSet, (trainSet.shape[0], 1, trainSet.shape[1]))

My problem is I'm trying to adjust values in the 1st axis (the new one). However, no matter what I try I can not actually figure out how to access the 1's in the new axis. I've tried a multitude of print statements such as

print(trainSet[0][0][0])
print(trainSet[1][0][1])

but no matter what I try I can not print out any 1's, just the contents of the 2-d array. I know it's a 3d array because when I use

print(trainSet.shape)

I get

(12, 1, 793)

Any help would be immensely appreciated!

You index a 3d array with expressions like M[i, j, k], M[:, 0, :], etc — hpaulj
– hpaulj, Commented Mar 8, 2017 at 23:40

Paul Panzer · Accepted Answer · 2017-03-08 23:53:41Z

1

The one refers to the size of that particular dimension, not the content of the array. The array [[5 5]] has shape (1, 2) but, still, you don't really expect its values must include 1 for that reason, do you?

What inserting a dimension of length one does is increasing the nesting level by one, so a 2d array [[a b] [c d]] will become [[[a b] [c d]]] or [[[a b]] [[c d]]] or [[[a] [b]] [[c] [d]]] depending on where you insert the new axis.

In particular, there are no new elements.

edited Mar 8, 2017 at 23:53

answered Mar 8, 2017 at 23:46

Paul Panzer

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gzc · Accepted Answer · 2017-03-09 05:01:20Z

0

If I didn't mistake your question, what you want is

trainSet[:,0,:]

answered Mar 9, 2017 at 5:01

gzc

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Dennis Dam · Accepted Answer · 2017-03-09 07:18:44Z

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the amount of values remains the same, you're just creating a 3D structure instead of the 2D structure.

answered Mar 9, 2017 at 7:18

Dennis Dam

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