b10hazard - 1 year ago 132

Python Question

I would like to get the index of a 2 dimensional Numpy array that matches a row. For example, my array is this:

`vals = np.array([[0, 0],`

[1, 0],

[2, 0],

[0, 1],

[1, 1],

[2, 1],

[0, 2],

[1, 2],

[2, 2],

[0, 3],

[1, 3],

[2, 3],

[0, 0],

[1, 0],

[2, 0],

[0, 1],

[1, 1],

[2, 1],

[0, 2],

[1, 2],

[2, 2],

[0, 3],

[1, 3],

[2, 3]])

I would like to get the index that matches the row [0, 1] which is index 3 and 15. When I do something like

`numpy.where(vals == [0 ,1])`

`(array([ 0, 3, 3, 4, 5, 6, 9, 12, 15, 15, 16, 17, 18, 21]), array([0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0]))`

I want index array([3, 15]).

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Answer Source

You need the `np.where`

function to get the indexes:

```
>>> np.where((vals == (0, 1)).all(axis=1))
(array([ 3, 15]),)
```

To dissassemble that:

```
>>> vals == (0, 1)
array([[ True, False],
[False, False],
...
[ True, False],
[False, False],
[False, False]], dtype=bool)
```

and calling the `.all`

method on that array (with `axis=1`

) gives you `True`

where both are True:

```
>>> (vals == (0, 1)).all(axis=1)
array([False, False, False, True, False, False, False, False, False,
False, False, False, False, False, False, True, False, False,
False, False, False, False, False, False], dtype=bool)
```

and to get which indexes are `True`

:

```
>>> np.where((vals == (0, 1)).all(axis=1))
(array([ 3, 15]),)
```

I find my solution a bit more readable, but as unutbu points out, the following may be faster, and returns the same value as `(vals == (0, 1)).all(axis=1)`

:

```
>>> (vals[:, 0] == 0) & (vals[:, 1] == 1)
```

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