Let’s suppose I have a `np.array`

like:

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

I would like to know if there is a pythonic way to find all the columns that contain at least one occurence of 0. In the example I would like to retrieve the indexes 2 and 4.

I need to remove those columns, but I also need to know how many columns I have removed (the indexes are not strictly necessary).

So in the end I simply need the result

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

### >Solution :

If you want to simply remove the columns, you can use `np.all`

(or its `ndarray`

) variant to find the columns you want to keep. Use the resulting boolean mask to index the 2nd axis:

```
>>> arr[:, arr.all(axis=0)]
array([[1., 1., 1.],
[1., 1., 1.],
[1., 1., 1.],
[1., 1., 1.]])
```

If you want to find the indices of those columns with at least one zero, you can use `np.any`

in conjunction with `np.nonzero`

(or `np.flatnonzero`

if you prefer):

```
>>> np.any(arr == 0, axis=0).nonzero()
(array([2, 4], dtype=int64),)
```

If you want to count them, you can sum the boolean mask directly:

```
>>> np.any(arr == 0, axis=0).sum()
2
```