I am wondering how to duplicate each element in a list arbitrary of times, e.g.
l = ['a', 'b', 'c']
the duplicate elements in l result in a new list,
n = ['a', 'a', 'a', 'a', 'b', 'b', 'c', 'c', 'c']
so 'a' has been duplicated 3 times, 'b' once, 'c' twice. The number of duplicates for each element are decided by numpy.random.poisson e.g. numpy.random.poisson(2).
Here's a NumPy based vectorized approach using np.repeat to create an array -
np.repeat(l, np.random.poisson([2]*len(l)))
If you need a list as output, append .tolist() there -
np.repeat(l, np.random.poisson([2]*len(l))).tolist()
If you would like to keep at least one entry for each element, add a clipping there with np.random.poisson([2]*len(arr)).clip(min=1).
Multiply each element in the list with the value returned from numpy.random.poisson(2), join it and then feed it to list:
r = list(''.join(i*random.poisson(2) for i in l))
For one run, this randomly results in:
['a', 'b', 'b', 'b', 'b', 'b', 'b', 'c', 'c', 'c']
Since you use np either way, I'd go with Divakar's solution (which, for lists larger than your example, executes faster).
>>> l = ['a', 'b', 'c']
>>> n = []
>>> for e in l:
... n.extend([e] * numpy.random.poisson(2))
...
>>> n
['a', 'a', 'b', 'c']
