Replace elements of array with elements from another array based on a third array

[ibreeden]

It seems you are presenting a part of a mysterious algorithm.
You can do it the way you showed, but of course you must extend the "if" with a number of "elifs".

If I understand it correctly, the first array (input) and the third (control) must have the same number of items. Am I right?

The following construction is not easy to digest for a computer:

array(['No change', 'Change second element to 1',
       'Change first element to 1', 'Change fourth element to 1'],
      dtype='<U27')

I hope you can change this to something more suitable like I did in the following example.

input_list = [[0.05, 0.07 , 0.04, 0.03, 0.05, 0.72],
       [0.08, 0.083, 0.40 , 0.038, 0.05, 0.70],
       [0.40, 0.17, 0.07, 0.04, 0.03, 0.31],
       [0.16, 0.14, 0.08, 0.40, 0.08, 0.40]]

replacement = [[1, 0, 0, 0],
       [0, 1, 0, 0],
       [0, 0, 1, 0],
       [0, 0, 0, 1]]

# array(['No change', 'Change second element to 1',
#       'Change first element to 1', 'Change fourth element to 1'],
#       dtype='<U27')
# This array has to be coded in a useful manner. (What is dtype='<U27'? I ignore it.)
# I call it "control" because it controls the way "input_list" is
#     converted to "target".
# A value of "-1" means: 'No change'.
# Elements count from 0. So first element is 0, second element is 1 etc.
# Then the previous array looks like this:
control = [-1, 1, 0, 3]

target = []

if len(input_list) != len(control):
    raise Exception("ERROR: input_list does not have the same number of elements as control list.")

for i in range(len(input_list)):
    if control[i] == -1 :
        target.append(input_list[i])
    else:
        target.append(replacement[control[i]])

# Show result.
for item in target:
    print(item)

Output:
[0.05, 0.07, 0.04, 0.03, 0.05, 0.72]
[0, 1, 0, 0]
[1, 0, 0, 0]
[0, 0, 0, 1]