Finding the index of minimum value in a row in numpy array

Question

I can't seem to get my head around how to do this - I am new to Python and this kind of work with arrays. I have a large array, say:

array([[119., 323.,  42., 277., 401.],
       [122., 326.,  39., 278., 10.],
       [125., 329.,  36., 12., 407.],
       ...,
       [308., 314., 469., 188., 266.],
       [308., 314., 469., 188., 266.],
       [308., 314., 469., 188., 266.]])

I would like to find the column index of the minimum value in each row. For instance for the first 3 rows would give [2, 4, 3....]. I have experimented with .min() and np.where(), for instance:

np.where(array == array.min())

But I just can't seem to get the answer I'm looking for. Any help would be much appreciated, thanks

indexes = [x.index(min(x)) for x in the_array]

001
– 001

2019-03-11 18:46:07 +00:00
Commented Mar 11, 2019 at 18:46 — 001
– 001, Commented Mar 11, 2019 at 18:46

pho · Accepted Answer · 2023-01-09 20:00:45Z

12

Use numpy argmin():

np.argmin(a, axis=1)

where a is your numpy array.

edited Jan 9, 2023 at 20:00

pho

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answered Mar 11, 2019 at 18:41

Austin

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