I am using scipy and want to create an an array of legnth n with a particular average.
Suppose I want an random arrays of length 3 with an average of 2.5 hence the possible options could be: [1.5, 3.5, 2.5] [.25, 7.2, .05] and so on and so forth...
I need to create many such arrays with varying lengths and different averages(specified) for each, so a generalized solution will be welcome.
Just generate numbers over the range you want (0...10 in this case)
>>> import random
>>> nums = [10*random.random() for x in range(5)]
Work out the average
>>> sum(nums)/len(nums)
4.2315222659844824
Shift the average to where you want it
>>> nums = [x - 4.2315222659844824 + 2.5 for x in nums]
>>> nums
[-0.628013346633133, 4.628537956666447, -1.7219257458163257, 7.617565127420011, 2.6038360083629986]
>>> sum(nums)/len(nums)
2.4999999999999996
You can use whichever distribution/range you like. By shifting the average this way it will always get you an average of 2.5 (or very close to it)
You haven't specified what distribution you want.
It's also not clear whether you want the average of the actual array to be 2.5, or the amortized average over all arrays to be 2.5.
The simplest solution—three random numbers in an even distribution from 0 to 2*avg—is this:
return 2*avg * np.random.rand(3)
If you want to guarantee that the average of the array is 2.5, that's a pretty simple constraint, but there are many different ways you could satisfy it, and you need to describe which way you want. For example:
n0 = random.random() * 2*avg
n1 = random.random() * (2*avg - n0)
n2 = random.random() * (2*avg - n0 - n1)
return np.array((n0, n1, n2))
I found a solution to the problem.
numpy.random.triangular(left, mode, right, size=None)
However, the minor problem is that it forces a triangular distribution on the samples.
[0,1)is 0.5, then you can shift the centre to 2.5 by adding 2...