Python array/matrix dimension

Question

I create two matrices

import numpy as np
arrA = np.zeros((9000,3))
arrB = np.zerros((9000,6))

I want to concatenate pieces of those matrices. But when I try to do:

arrC = np.hstack((arrA, arrB[:,1]))

I get an error:

ValueError: all the input arrays must have same number of dimensions

I guess it's because np.shape(arrB[:,1]) is equal (9000,) instead of (9000,1), but I cannot figure out how to resolve it.

Could you please comment on this issue?

• Mo H.: I don't really see your issue here. You have a 9000x3 and a 9000x6 array, they are different sizes, so you cant concat them together... am I missing something?
• Leopoldo: I want to take 9000x3 array and concatenate it with 9000x1 array getting 9000x4 array. I hope this comment describe my issue completely.
• user3679917: Yes the dimensions/size format of the matrices is wrong they need to be equal. Try np.hstack((arrA, np.zeros((arrayB[1], 3))))
• Leopoldo: @Emyen it does not work.

Answer 1

You could preserve dimensions by passing a list of indices, not an index:

>>> arrB[:,1].shape
(9000,)
>>> arrB[:,[1]].shape
(9000, 1)
>>> out = np.hstack([arrA, arrB[:,[1]]])
>>> out.shape
(9000, 4)

Answer 2

This is easier to see visually.

Assume:

>>> arrA=np.arange(9000*3).reshape(9000,3)
>>> arrA
array([[ 0, 1, 2],
 [ 3, 4, 5],
 [ 6, 7, 8],
 ..., 
 [26991, 26992, 26993],
 [26994, 26995, 26996],
 [26997, 26998, 26999]])
>>> arrB=np.arange(9000*6).reshape(9000,6)
>>> arrB
array([[ 0, 1, 2, 3, 4, 5],
 [ 6, 7, 8, 9, 10, 11],
 [ 12, 13, 14, 15, 16, 17],
 ..., 
 [53982, 53983, 53984, 53985, 53986, 53987],
 [53988, 53989, 53990, 53991, 53992, 53993],
 [53994, 53995, 53996, 53997, 53998, 53999]])

If you take a slice of arrB, you are producing a series that looks more like a row:

>>> arrB[:,1]
array([ 1, 7, 13, ..., 53983, 53989, 53995])

What you need is a column the same shape as a column to add to arrA:

>>> arrB[:,[1]]
array([[ 1],
 [ 7],
 [ 13],
 ..., 
 [53983],
 [53989],
 [53995]])

Then hstack works as expected:

>>> arrC=np.hstack((arrA, arrB[:,[1]]))
>>> arrC
array([[ 0, 1, 2, 1],
 [ 3, 4, 5, 7],
 [ 6, 7, 8, 13],
 ..., 
 [26991, 26992, 26993, 53983],
 [26994, 26995, 26996, 53989],
 [26997, 26998, 26999, 53995]])

An alternate form is to specify -1 in one dimension and the number of rows or cols desired as the other in .reshape():

>>> arrB[:,1].reshape(-1,1) # one col
array([[ 1],
 [ 7],
 [ 13],
 ..., 
 [53983],
 [53989],
 [53995]])
>>> arrB[:,1].reshape(-1,6) # 6 cols
array([[ 1, 7, 13, 19, 25, 31],
 [ 37, 43, 49, 55, 61, 67],
 [ 73, 79, 85, 91, 97, 103],
 ..., 
 [53893, 53899, 53905, 53911, 53917, 53923],
 [53929, 53935, 53941, 53947, 53953, 53959],
 [53965, 53971, 53977, 53983, 53989, 53995]])
>>> arrB[:,1].reshape(2,-1) # 2 rows
array([[ 1, 7, 13, ..., 26983, 26989, 26995],
 [27001, 27007, 27013, ..., 53983, 53989, 53995]])

There is more on array shaping and stacking here

Answer 3

I would try something like this:

np.vstack((arrA.transpose(), arrB[:,1])).transpose()

• Leopoldo: It works and it's a little bit complicated. Thank you for your comment.

Answer 4

There several ways of making your selection from arrB a (9000,1) array:

np.hstack((arrA,arrB[:,[1]]))
np.hstack((arrA,arrB[:,1][:,None]))
np.hstack((arrA,arrB[:,1].reshape(9000,1)))
np.hstack((arrA,arrB[:,1].reshape(-1,1)))

One uses the concept of indexing with an array or list, the next adds a new axis (e.g. np.newaxis), the third uses reshape. These are all basic numpy array manipulation tasks.