How to avoid overwriting a value through for loop in python?

Question

Below is the one I coded. The problem here is I have values 1, 2, and 3 in the A matrix and hence at the output A has all the values 1.
The result I expect is:

A = np.matrix([[1, 2, 2, 1], 
               [1, 1, 3, 1],
               [1, 1, 1, 3]]).

Any help is appreciated. Sorry for my poor writing. Thank you!

A = np.matrix([[1, 15, 23, 2], [3, 2, 56, 7], [2, 6, 8, 25]])
bound = np.array([1, 15, 25, 56])
for i in range(3, 0, -1):
    A[np.logical_and(bound[i - 1] <= A, A <= bound[i])] = i

numpy.searchsorted ? np.searchsorted(bound, A, side='left') — akuiper
– akuiper, Commented Aug 27, 2020 at 3:09
Thank you for your prompt response. The first element of the output matrix is 0. I expect there 1 — Joyce
– Joyce, Commented Aug 27, 2020 at 3:12
if 1 maps to 1, why 15 also maps to 1 ? shouldn't it be 2 ? Since there's a 15 in bound ? — akuiper
– akuiper, Commented Aug 27, 2020 at 3:13
But anyways, you can try np.searchsorted(bound, A, side='right') to see if it gives what you need. — akuiper
– akuiper, Commented Aug 27, 2020 at 3:15
Oh sorry for that!. Yes 15 must be 2. I tried with np.searchsorted(bound, A, side='right'). It doesn't work — Joyce
– Joyce, Commented Aug 27, 2020 at 3:15

Ehsan · Accepted Answer · 2020-08-27 06:42:33Z

1

One way of doing it is saving the changed elements in a separate mask_:

mask_ = np.ones_like(A, dtype=bool)
for i in range(3,0,-1):
    mask = np.logical_and(bound[i - 1] <= A, A <= bound[i])
    A[np.multiply(mask_,mask)] = i
    mask_ = np.multiply(mask_,~mask)

output:

[[1 2 2 1]
 [1 1 3 1]
 [1 1 1 3]]

answered Aug 27, 2020 at 6:42

Ehsan

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