Jack Twain - 9 months ago 38

Python Question

I have a large

`numpy`

`M`

`numpy`

Answer

Here's one way. I assume numpy has been imported using `import numpy as np`

.

```
In [20]: a
Out[20]:
array([[0, 1, 0],
[1, 0, 1],
[0, 0, 0],
[1, 1, 0],
[0, 0, 0]])
In [21]: np.where(~a.any(axis=1))[0]
Out[21]: array([2, 4])
```

It's a slight variation of this answer: How to check that a matrix contains a zero column?

Here's what's going on:

The `any`

method returns True if any value in the array is "truthy". Nonzero numbers are considered True, and 0 is considered False. By using the argument `axis=1`

, the method is applied to each row. For the example `a`

, we have:

```
In [32]: a.any(axis=1)
Out[32]: array([ True, True, False, True, False], dtype=bool)
```

So each value indicates whether the corresponding row contains a nonzero value. The `~`

operator is the binary "not" or complement:

```
In [33]: ~a.any(axis=1)
Out[33]: array([False, False, True, False, True], dtype=bool)
```

(An alternative expression that gives the same result is `(a == 0).all(axis=1)`

.)

To get the row indices, we use the `where`

function. It returns the indices where its argument is True:

```
In [34]: np.where(~a.any(axis=1))
Out[34]: (array([2, 4]),)
```

Note that `where`

returned a tuple containing a single array. `where`

works for n-dimensional arrays, so it always returns a tuple. We want the single array in that tuple.

```
In [35]: np.where(~a.any(axis=1))[0]
Out[35]: array([2, 4])
```

Source (Stackoverflow)