How to add every four rows as column

Question

I have a very large numpy array like this. How can I convert this into

array([[1, 1, 0, 0, 1, 0, 0, 1],
       [0, 1, 0, 0, 0, 1, 0, 0],
       [0, 0, 1, 0, 1, 1, 0, 0],
       [0, 0, 1, 0, 1, 1, 0, 0],
       [0, 0, 1, 1, 0, 1, 1, 0],
       [0, 0, 1, 1, 0, 1, 1, 0]])

into this every two rows into one row? Although I need every four rows as one row? This example would be helpful! I searched many places couldn't find a proper solution!

array([[1, 1, 0, 0, 1, 0, 0, 1 , 0, 1, 0, 0, 0, 1, 0, 0],
       [0, 0, 1, 0, 1, 1, 0, 0 , 0, 0, 1, 0, 1, 1, 0, 0],
       [0, 0, 1, 1, 0, 1, 1, 0 , 0, 0, 1, 1, 0, 1, 1, 0]])

and then back again to

array([[1, 1, 0, 0, 1, 0, 0, 1],
       [0, 1, 0, 0, 0, 1, 0, 0],
       [0, 0, 1, 0, 1, 1, 0, 0],
       [0, 0, 1, 0, 1, 1, 0, 0],
       [0, 0, 1, 1, 0, 1, 1, 0],
       [0, 0, 1, 1, 0, 1, 1, 0]])

score 4 · Answer 1 · answered Feb 09 '18 at 17:27

Use .reshape():

>>> import numpy as np
>>> a = np.array([[1, 1, 0, 0, 1, 0, 0, 1],
...        [0, 1, 0, 0, 0, 1, 0, 0],
...        [0, 0, 1, 0, 1, 1, 0, 0],
...        [0, 0, 1, 0, 1, 1, 0, 0],
...        [0, 0, 1, 1, 0, 1, 1, 0],
...        [0, 0, 1, 1, 0, 1, 1, 0]])
>>> b = a.reshape(3,16)
>>> b
array([[1, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0],
       [0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0],
       [0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0]])
>>> b.reshape(6,8)
array([[1, 1, 0, 0, 1, 0, 0, 1],
       [0, 1, 0, 0, 0, 1, 0, 0],
       [0, 0, 1, 0, 1, 1, 0, 0],
       [0, 0, 1, 0, 1, 1, 0, 0],
       [0, 0, 1, 1, 0, 1, 1, 0],
       [0, 0, 1, 1, 0, 1, 1, 0]])

what if the first array `a` has only five arrays? what can i do then..? — poda_badu, Feb 09 '18 at 18:03
That would be an incomplete matrix. What do you expect the missing elements to be? — JohanL, Feb 09 '18 at 18:23

How to add every four rows as column

1 Answers1