Numpy: search for the first matching string

Question

If I have a numpy array like this:

import numpy as np
x = np.array([[0,1],[0,2],[1,1],[0,2]])

How to return the index of the first row that matches [0,2]?

For lists it’s easy to use index:

[[0,1],[0,2],[1,1],[0,2]]

l.index([0,2])
> 1

And I know that it numpyhas a function numpy.where, but I'm not sure what to do with the output numpy.where:

np.where(x==[0,2])
> (array([0, 1, 1, 3, 3]), array([0, 0, 1, 0, 1]))

There also numpy.argmax, but it also does not return what I am looking for, it is just an index1

np.argmax(x == [0,2], axis = 1)

Answer 1

[0,2], broadcasting x, , x. , TRUE .all(1). , , np.where np.nonzero . -

In [132]: x
Out[132]: 
array([[0, 1],
       [0, 2],
       [1, 1],
       [0, 2]])

In [133]: search_list = [0,2]

In [134]: np.where((x == search_list).all(1))[0][0]
Out[134]: 1

In [135]: np.nonzero((x == search_list).all(1))[0][0]
Out[135]: 1