Using the Numpy matrix.argmax ()
method, we can find the index value of the smallest element in a given matrix that has one or more dimensions.
Syntax:
matrix.argmin()
Return: Return index number of minimum elements in matrix
Example # 1:
In this example, we see that using the matrix.argmax ()
we can find the index of the minimum element in the given matrix.

Exit :
0
Example # 2:

Exit :
4
import numpy as np
def find_nearest(array, value):
array = np.asarray(array)
idx = (np.abs(array  value)).argmin()
return array[idx]
array = np.random.random(10)
print(array)
# [ 0.21069679 0.61290182 0.63425412 0.84635244 0.91599191 0.00213826
# 0.17104965 0.56874386 0.57319379 0.28719469]
value = 0.5
print(find_nearest(array, value))
# 0.568743859261
Say that you have a list values = [3,6,1,5]
, and need the index of the smallest element, i.e. index_min = 2
in this case.
Avoid the solution with itemgetter()
presented in the other answers, and use instead
index_min = min(range(len(values)), key=values.__getitem__)
because it doesn"t require to import operator
nor to use enumerate
, and it is always faster(benchmark below) than a solution using itemgetter()
.
If you are dealing with numpy arrays or can afford numpy
as a dependency, consider also using
import numpy as np
index_min = np.argmin(values)
This will be faster than the first solution even if you apply it to a pure Python list if:
numpy
arrayI have run the benchmark on my machine with python 2.7 for the two solutions above (blue: pure python, first solution) (red, numpy solution) and for the standard solution based on itemgetter()
(black, reference solution).
The same benchmark with python 3.5 showed that the methods compare exactly the same of the python 2.7 case presented above
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