np.unique([1, 3, 0, 3, 1, 1], return_counts=True)
returns
(array([0, 1, 3]), array([1, 3, 2]))
which excludes missing elements, in this case, 2.
Is there an easy and efficient way to get all counts, for example:
count(ar=[1, 3, 0, 3, 1, 1], from=0, to=4) # returns [1, 3, 0, 2]
?
You could use np.bincount, passing it a minlength of the maximum value in the array plus 1:
ar = np.array([1, 3, 0, 3, 1, 1])
np.bincount(ar, minlength=np.amax(ar)+1)
# array([1, 3, 0, 2], dtype=int64)
You could use np.add.at:
ar = np.array([1, 3, 0, 3, 1, 1])
count = np.zeros(np.amax(ar)+1)
np.add.at(count, ar, 1)
Output in count:
array([1., 3., 0., 2.])
But using np.bincount will be much faster