How to get the index of a maximum element in a NumPy array along one axis

Question:

I have a 2 dimensional NumPy array. I know how to get the maximum values over axes:

>>> a = array([[1,2,3],[4,3,1]])
>>> amax(a,axis=0)
array([4, 3, 3])

How can I get the indices of the maximum elements? I would like as output array([1,1,0]) instead.

Asked By: Peter Smit

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Answers:

>>> a.argmax(axis=0)

array([1, 1, 0])
Answered By: eumiro
v = alli.max()
index = alli.argmax()
x, y = index/8, index%8
Answered By: ahmed
>>> import numpy as np
>>> a = np.array([[1,2,3],[4,3,1]])
>>> i,j = np.unravel_index(a.argmax(), a.shape)
>>> a[i,j]
4
Answered By: blaz

argmax() will only return the first occurrence for each row.
http://docs.scipy.org/doc/numpy/reference/generated/numpy.argmax.html

If you ever need to do this for a shaped array, this works better than unravel:

import numpy as np
a = np.array([[1,2,3], [4,3,1]])  # Can be of any shape
indices = np.where(a == a.max())

You can also change your conditions:

indices = np.where(a >= 1.5)

The above gives you results in the form that you asked for. Alternatively, you can convert to a list of x,y coordinates by:

x_y_coords =  zip(indices[0], indices[1])
Answered By: SevakPrime

There is argmin() and argmax() provided by numpy that returns the index of the min and max of a numpy array respectively.

Say e.g for 1-D array you’ll do something like this

import numpy as np

a = np.array([50,1,0,2])

print(a.argmax()) # returns 0
print(a.argmin()) # returns 2

And similarly for multi-dimensional array

import numpy as np

a = np.array([[0,2,3],[4,30,1]])

print(a.argmax()) # returns 4
print(a.argmin()) # returns 0

Note that these will only return the index of the first occurrence.

Answered By: Hadi Mir
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