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I'm trying to get the indices of the maximum element in a Numpy array. This can be done using numpy.argmax. My problem is, that I would like to find the biggest element in the whole array and get the indices of that.

numpy.argmax can be either applied along one axis, which is not what I want, or on the flattened array, which is kind of what I want.

My problem is that using numpy.argmax with axis=None returns the flat index when I want the multi-dimensional index.

I could use divmod to get a non-flat index but this feels ugly. Is there any better way of doing this?

Seanny123
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Andreas Mueller
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3 Answers3

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You could use numpy.unravel_index() on the result of numpy.argmax():

>>> a = numpy.random.random((10, 10))
>>> numpy.unravel_index(a.argmax(), a.shape)
(6, 7)
>>> a[6, 7] == a.max()
True
Sven Marnach
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np.where(a==a.max())

returns coordinates of the maximum element(s), but has to parse the array twice.

>>> a = np.array(((3,4,5),(0,1,2)))
>>> np.where(a==a.max())
(array([0]), array([2]))

This, comparing to argmax, returns coordinates of all elements equal to the maximum. argmax returns just one of them (np.ones(5).argmax() returns 0).

eumiro
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    This will iterate the array three times, not only twice. One time to find the maximum, a second time to build the result of `==`, and a third time to extract the `True` values from this result. Note that there might be more than one item equal to the maximum. – Sven Marnach Feb 28 '12 at 14:40
1

To get the non-flat index of all occurrences of the maximum value, you can modify eumiro's answer slightly by using argwhere instead of where:

np.argwhere(a==a.max())

>>> a = np.array([[1,2,4],[4,3,4]])
>>> np.argwhere(a==a.max())
array([[0, 2],
       [1, 0],
       [1, 2]])
Richard Ludlow
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  • It's not effective since you get three passes and a matrix creation. Imagine we've got 9000x7000 image (A3@600dpi) - would you still insist on your solution? – Maksym Ganenko Nov 20 '18 at 10:14