Reshape np.array(n, m) to np.array(m, n)

Question

I have an array like this

ar = np.array([[0, 1, 0, 1, 0, 1, 0, 1],
       [0, 1, 0, 0, 0, 1, 1, 1]])

I want to get

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

If I do as below I get

ar.reshape(8,2)
array([[0, 1],
       [0, 1],
       [0, 1],
       [0, 1],
       [0, 1],
       [0, 0],
       [0, 1],
       [1, 1]])

My decision is

npar = np.zeros((np.asarray(ar).shape[1],np.asarray(ar).shape[0]))
npar[:, 0] = ar[0]
npar[:, 1] = ar[1]
npar
array([[0., 0.],
       [1., 1.],
       [0., 0.],
       [1., 0.],
       [0., 0.],
       [1., 1.],
       [0., 1.],
       [1., 1.]])

But is there more fast & convinient way?

how about simply taking the transpose? ```arr.T``` – grodzi Sep 23 '18 at 11:42 — grodzi, Sep 23 '18 at 11:42

score 2 · Accepted Answer · answered Sep 23 '18 at 11:42

2

You simply need to take the transpose I think

np.transpose(ar)

answered Sep 23 '18 at 11:42

Atul Shanbhag

636
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Reshape np.array(n, m) to np.array(m, n)

1 Answers1