how to select nth element of array in a tuple?

Question

I have a tuple like this:

(array([0, 0, 0]), array([0, 1, 2]))

I would like to select the first element of each array in this tuple and want to achieve this result: [0,0] or [0,1] or [0,2]

How can I do this?

Actually, this tuple is obtained from a function in numpy.where what I do is like this:

import numpy as np

x=np.array([[1,2,3],[4,5,6],[7,8,9]])

xi=np.where(x<4)

any suggestion?

uhhhh, I just get the answer after I asked it XD zip(*xi)[0] or zip(*xi)[1] or zip(*xi)[2] — Jen, Oct 26 '17 at 02:11
getXi = lambda X, i: (X[0][i], X[1][i]); print(getXi(xi, 1)) — Kinght 金, Oct 26 '17 at 02:21
You may want to look at [`np.argwhere`](https://docs.scipy.org/doc/numpy-1.13.0/reference/generated/numpy.argwhere.html). — user2357112, Oct 26 '17 at 02:27
@BradSolomon I am not sure if it is the same situation: I am considering array in tuple, but the issue you posted is tuple in list. but I found the answer is alike. thank you — Jen, Oct 26 '17 at 02:45

score 0 · Accepted Answer · answered Oct 26 '17 at 02:18

0

Is this what you want ?

list(zip(tp[0],tp[1]))
Out[689]: [(0, 0), (0, 1), (0, 2)]

Data input

tp=(np.array([0, 0, 0]), np.array([0, 1, 2]))

If you want to list of list

l=list(zip(tp[0],tp[1]))
list(map(list, l))
Out[691]: [[0, 0], [0, 1], [0, 2]]

answered Oct 26 '17 at 02:18

BENY

score 0 · Answer 2 · answered Oct 26 '17 at 02:18

0

You can iterate over the numpy array through nditer and pair them:

np.array([(x,y) for x,y in np.nditer(xi)]) 

array([[0, 0],
   [0, 1],
   [0, 2]], dtype=int32)

answered Oct 26 '17 at 02:18

skrubber

2 Answers2