How to join the indexing in np.where?

Question

import numpy as np

a = np.array([1,2,3,4,5,6,7,8,9,10])
ind1 = np.where(a>8)
ind2 = np.where(a<3)

What I want is [1,2,9,10].

At this time, How to join the two index, 'ind1' and 'ind2'?

When I face the situation like this, I just wrote the code like below,

ind3 = np.where( (a>8) & (a<3) )

But if I face the more complex situation, I can not use the above code.

So I want to know the method which can find the index joining 'ind1' and 'ind2' directly, not fixing inside of 'np.where()'.

=================================

Sorry, I mistook but already there is a good answer, so I will not erase my original question.

What I mean is below,

import numpy as np

a = np.array([1,2,3,4,5,6,7,8,9,10])
ind1 = np.where(a>8)
ind2 = np.where(a>3)

What I want to expect is [9,10]. i.e. I want to intersection.

Answer 1

You can do it by using Boolean mask arrays:

ind1 = a > 8
ind2 = a < 3
ind3 = np.logical_or(ind1, ind2)
print(a[ind3])   # --> [ 1  2  9 10]

If you have more than two condition:

ind_n = np.logical_or.reduce((ind1, ind2, ind3, ...))

For using np.where, you must change your proposed code to:

ind3 = np.where((a > 8) | (a < 3))

Answer 2

Why not do a for loop?

idx = []
for i in range X:
  idx.append(np.where(some condition))

results = np.concat(idx)

Answer 3

In [41]: a = np.array([1, 2, 3, 4, 5, 6, 7, 8, 9, 10])
    ...: ind1 = np.where(a > 8)
    ...: ind2 = np.where(a > 3)
    ...: 
   

First, make sure you understand what each where/nonzero produces:

In [42]: ind1, ind2
Out[42]: ((array([8, 9]),), (array([3, 4, 5, 6, 7, 8, 9]),))

That's a tuple with one array (since a is 1d).

We can extract that array:

In [43]: ind1[0], ind2[0]
Out[43]: (array([8, 9]), array([3, 4, 5, 6, 7, 8, 9]))

numpy set operations aren't that great. But we can use Python sets:

In [45]: set(ind1[0]), set(ind2[0])
Out[45]: ({8, 9}, {3, 4, 5, 6, 7, 8, 9})
In [46]: set(ind1[0]).intersection(set(ind2[0]))
Out[46]: {8, 9}

Doing logical operations on the boolean inputs to where is usually a better choice.

We could test the 2 arrays for equality:

In [50]: ind1[0][:, None] == ind2[0]
Out[50]: 
array([[False, False, False, False, False,  True, False],
       [False, False, False, False, False, False,  True]])
In [52]: (ind1[0][:, None] == ind2[0]).any(axis=0)
Out[52]: array([False, False, False, False, False,  True,  True])
In [53]: ind2[0][(ind1[0][:, None] == ind2[0]).any(axis=0)]
Out[53]: array([8, 9])