## Issue

```
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.

## Solution

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))
```

Answered By – Ali_Sh

Answer Checked By – Cary Denson (BugsFixing Admin)