replacing element np array

Question

I have this nparray :

[[0. 0. 0. 0. 1.]
 [1. 0. 0. 0. 0.]
 [0. 0. 1. 0. 0.]
 ...
 [0. 0. 1. 0. 0.]
 [0. 0. 0. 0. 1.]
 [1. 0. 0. 0. 0.]]

and I want to do something like this :

for item in array :
 if item[0] == 1:
 item=[0.8,0.20,0,0,0]
 elif item[1] == 1:
 item=[0.20,0.80,0,0,0]
 elif item[3] == 1:
 item=[0,0,0,0.8,0.2]
 elif item[4] == 1:
 item=[0,0,0,0.2,0.8]
 else:
 [0,0,1,0,0]

I try this :

def conver_probs2(arg):
 test= arg
 test=np.where(test==[1.,0.,0.,0.,0.], [0.8,0.20,0.,0.,0.],test)
 return test

but the result is this :

[[0. 0.2 0. 0. 1. ]
 [0.8 0.2 0. 0. 0. ]
 [0. 0.2 1. 0. 0. ]
 ...
 [0. 0.2 1. 0. 0. ]
 [0. 0.2 0. 0. 1. ]
 [0.8 0.2 0. 0. 0. ]]

not what I want ... any ideas?

Thanks!

Answer 1

A simple approach would be to iterate over the indices.

Then it'd be possible for you to reuse the same for loop that you showed like this:

for i in range(len(array)):
 if array[i][0] == 1:
 array[i] = [0.8, 0.2, 0, 0, 0] 
 ...

• rjpg: yes sometimes is better to keep it VERY simple :-) 'code' for item in array 'code' was not working ...

Answer 2

If your replacement array has the same shape as your target you can do this:

mask = target[target[:, 0] == 1]
target[mask] = replacements[mask]

here is a simple test

test_target = np.eye(4)
test_target[2:, 0] = 1
replacements = np.ones((4, 4)) * 42

Before using np.where try boolean indexing first. Usually it is what you want.