>>> 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?
4 Answers
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 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] ...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
>>>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.
