如果最小值满足条件，则获取二维np.array中每一行的最小值索引。

Question

I've a 2D np.array with dimension 1000 (rows) x 12 (columns) .

I need to get the indices of those values that are below 1.5 .
If a row contains more than one value that satisfies this condition, then I need to keep only the indices of the lowest.

I'd be quite happy with using idx1,idx2=np.where(x < 1.5) , but this sometimes returns several indices that are in the same rows. I could of course loop over all repeated rows in idx1 and keep only the indices whose values in x where are the lowest, but I was wondering if there's a more pythonic way.

Thanks

Answer 1

One way would be to use a numpy masked array . Lets define the following random ndarray :

a = np.random.normal(1,2,(4,2))

print(a.round(2))
array([[ 1.41, -0.68],
       [-1.53,  2.74],
       [ 1.19,  2.66],
       [ 2.  ,  1.26]])

We can define a masked array with:

ma = np.ma.array(a, mask = a >= 1.5)

print(ma.round(2))
[[1.41 -0.68]
 [-1.53 --]
 [1.19 --]
 [-- 1.26]]

In order to deal with rows with no values bellow the threshold, you could do:

m = ma.mask.any(axis=1)
# array([ True,  True,  True,  True])

Which will contain a False if there are no valid values along a given row. And then take the np.argmin over the masked array to get the columns with the minimum values bellow 1.5:

np.argmin(ma, axis=1)[m]
# array([1, 0, 0, 1])

And for the rows you could do:

np.flatnonzero(m)
# array([0, 1, 2, 3])

Answer 2

You can just do this:

# First index is all rows
idx1 = np.arange(len(x))
# Second index is minimum values
idx2 = np.argmin(m, axis=1)
# Filter rows where minimum is not below threshold
valid = x[idx1, idx2] < 1.5
idx1 = idx1[valid]
idx2 = idx2[valid]