rand across np array dimensions

Question

How to compare values in val against arr based on operators o?

import numpy as np
from operator import gt, lt

val = np.array([[3,7,1], [4,8,5], [5,10,3]])
arr = np.array([[1,2,3,4,5], [6,7,8,9,10], [9,7,5,3,1]])
o = (gt, gt, lt)

# [3,7,1]
# (3 gt [1,2,3,4,5]) & (7 gt [6,7,8,9,10] & (1 lt [9,7,5,3,1])
# [True, True, False, False, False] & [True, False, False, False, False] & [True, True, True, True, False]
# => [True, False, False, False, False]

# [4,8,5]
# (4 gt [1,2,3,4,5]) & (8 gt [6,7,8,9,10] & (5 lt [9,7,5,3,1])
# => [True, True, False, False, False]

# [5,10,3]
# (5 gt [1,2,3,4,5]) & (10 gt [6,7,8,9,10]) & (3 lt [9,7,5,3,1])
# => [True, True, True, False, False]

result = np.array([[True, False, False, False, False],
                   [True, True, False, False, False],
                   [True, True, True, False, False]])

For each val_i, operator o_j is applied between val_i_j and all elements in arr_j. The resulting vectors are compared with __rand__.

Current solution seems too difficult.

out = []
for i in range(len(val)):
    m = o[0](val[i][0], arr[0])
    print(val[i][0], m)
    for j in range(len(o)):
        if j > 0:
            m &= o[j](val[i][j], arr[j])
            print(val[i][j], m)
    out.append(list(mask))

Asked By: aeiou

||

Source

Answer 1

You can try this:

import numpy as np

val = np.array([[3,7,1], [4,8,5], [5,10,3]])
arr = np.array([[1,2,3,4,5], [6,7,8,9,10], [9,7,5,3,1]])
o = (np.greater, np.greater, np.less)

out = np.logical_and.reduce([o[i](val[:, i, None], arr[i]) for i in range(len(o))])
print(out)

It gives:

[[ True False False False False]
 [ True  True False False False]
 [ True  True  True False False]]

Answered By: bb1

rand across np array dimensions

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__rand__ across np array dimensions

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rand across np array dimensions