Select a random element in a numpy matrix returning its index

Asked May 12 '21 at 17:46

Active May 12 '21 at 17:46

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I have a matrix which is shape=(3,3) and the rows represent probablity distributions. This is an example:

distribution_probablities=np.array([[1/3,2/3,0], 
                               [0.5,0.2,0.3],
                               [1,0,0]])

I want to obtain a vector which has length 3 and each value represent the index of the selected probability. A possible resulting vector is this one: result=[1,0,0]
In the first row the second element has bigger chance to be selected, in the second one it is the same for the second element and so on.. Another possible result is: result=[0,2,0]. Is it possible to do it without a for loop over the rows?

asked May 12 '21 at 17:46

Francesco De Santis

I don't fully understand. Does 1,0,0 means probability (row 2) 1, then 0 (row 1), then 0 (row 1)? – user_na May 12 '21 at 17:51
each row is a rpobability distribution. If you consider a matrix which has aonly a single row [0.5,0.3,0.2] i can get x=0,x=1,x=2 but the probability to obtain an index is proportional to the value which it contains. – Francesco De Santis May 12 '21 at 17:52
@MustafaAydın good solution – user_na May 12 '21 at 17:58
yes is a good one, thank you. But i was looking for a solution whitout for loops. For example using argwhere or something like that. – Francesco De Santis May 12 '21 at 18:00

Select a random element in a numpy matrix returning its index

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