numpy get index where value is true

Question

>>> 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?

Answer 1

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),)

Answer 2

You can use the 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]
 ...

Not Bad Way

>>> 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]
 ...

Answer 3

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 almost the same as np.transpose(np.nonzero(a)), but it produces a result of the correct shape for a 0-d array.

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]])
>>> for i,j in a: print(i,j)
... 
0 0
0 4
3 4
4 0
4 3
4 4

Answer 4

I prefer np.flatnonzero(arr) to the nonzero() option when you only need the row idx. arr.nonzero() works, but it returns a tuple instead of an array. flatnonzero() is equivalent to np.nonzero(np.ravel(arr))[0].

As mentioned in the comments, np.where() is discouraged by the NumPy docs.