# numpy get index where value is true

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### Question :

numpy get index where value is true
``````>>> ex=np.arange(30)
>>> e=np.reshape(ex,[3,10])
>>> e
array([[ 0,  1,  2,  3,  4,  5,  6,  7,  8,  9],
[10, 11, 12, 13, 14, 15, 16, 17, 18, 19],
[20, 21, 22, 23, 24, 25, 26, 27, 28, 29]])
>>> e>15
array([[False, False, False, False, False, False, False, False, False,
False],
[False, False, False, False, False, False,  True,  True,  True,
True],
[ True,  True,  True,  True,  True,  True,  True,  True,  True,
True]], dtype=bool)
``````

I need to find the rows that have true or rows in `e` whose value are more than 15. I could iterate using a for loop, however, I would like to know if there is a way numpy could do this more efficiently?

To get the row numbers where at least one item is larger than 15:

``````>>> np.where(np.any(e>15, axis=1))
(array([1, 2], dtype=int64),)
``````

You can use nonzero function. it returns the nonzero indices of the given input.

Easy Way

``````>>> (e > 15).nonzero()

(array([1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2]), array([6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]))
``````

to see the indices more cleaner, use `transpose` method:

``````>>> numpy.transpose((e>15).nonzero())

[[1 6]
[1 7]
[1 8]
[1 9]
[2 0]
...
``````

``````>>> numpy.nonzero(e > 15)

(array([1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2]), array([6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]))
``````

or the clean way:

``````>>> numpy.transpose(numpy.nonzero(e > 15))

[[1 6]
[1 7]
[1 8]
[1 9]
[2 0]
...
``````

A simple and clean way: use `np.argwhere` to group the indices by element, rather than dimension as in `np.nonzero(a)` (i.e., `np.argwhere` returns a row for each non-zero element).

``````>>> a = np.arange(10)
>>> a
array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])
>>> np.argwhere(a>4)
array([[5],
[6],
[7],
[8],
[9]])
``````

`np.argwhere(a)` is the same as `np.transpose(np.nonzero(a))`.

Note: You cannot use `a(np.argwhere(a>4))` to get the corresponding values in `a`. The recommended way is to use `a[(a>4).astype(bool)]` or `a[(a>4) != 0]` rather than `a[np.nonzero(a>4)]` as they handle 0-d arrays correctly. See the documentation for more details. As can be seen in the following example, `a[(a>4).astype(bool)]` and `a[(a>4) != 0]` can be simplified to `a[a>4]`.

Another example:

``````>>> a = np.array([5,-15,-8,-5,10])
>>> a
array([  5, -15,  -8,  -5,  10])
>>> a > 4
array([ True, False, False, False,  True])
>>> a[a > 4]
array([ 5, 10])
>>> a = np.add.outer(a,a)
>>> a
array([[ 10, -10,  -3,   0,  15],
[-10, -30, -23, -20,  -5],
[ -3, -23, -16, -13,   2],
[  0, -20, -13, -10,   5],
[ 15,  -5,   2,   5,  20]])
>>> a = np.argwhere(a>4)
>>> a
array([[0, 0],
[0, 4],
[3, 4],
[4, 0],
[4, 3],
[4, 4]])
>>> [print(i,j) for i,j in a]
0 0
0 4
3 4
4 0
4 3
4 4
``````