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Given an array of size n I want to generate random probabilities for each index such that Sigma(a[0]..a[n-1])=1
One possible result might be:
0 1 2 3 4
0.15 0.2 0.18 0.22 0.25
Another perfectly legal result can be:
0 1 2 3 4
0.01 0.01 0.96 0.01 0.01
How can I generate these easily and quickly? Answers in any language are fine, Java preferred.
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(1, 0, 0), an equally valid triple, must come from (1, 0, 0). One solution: The set of positive numbers that add to 1 form a n equilateral triangle in three-space, with coordinates (1,0,0), (0,1,0), (0,0,1).
You could use the RAND() function to generate N numbers (8 in your case) in column A. Then, in column B you could use the following formula B1=A1/SUM(A:A)*320 , B2=A2/SUM(A:A)*320 and so on (where 320 is the sum that you are interested into). So you can just enter =RAND() in A1, then drag it down to A8.
Method #2 : Using random.sample() + sum() This single utility function performs the exact required as asked by the problem statement, it generated N no. of random numbers in a list in the specified range and returns the required list. The task of performing sum is done using sum().
The randint() method to generates a whole number (integer). You can use randint(0,50) to generate a random number between 0 and 50. To generate random integers between 0 and 9, you can use the function randrange(min,max) .
Get n random numbers, calculate their sum and normalize the sum to 1 by dividing each number with the sum.
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