I know there is a method for a Python list to return the first index of something:
>>> l = [1, 2, 3]
>>> l.index(2)
1
Is
If you need the index of the first occurrence of only one value, you can use nonzero
(or where
, which amounts to the same thing in this case):
>>> t = array([1, 1, 1, 2, 2, 3, 8, 3, 8, 8])
>>> nonzero(t == 8)
(array([6, 8, 9]),)
>>> nonzero(t == 8)[0][0]
6
If you need the first index of each of many values, you could obviously do the same as above repeatedly, but there is a trick that may be faster. The following finds the indices of the first element of each subsequence:
>>> nonzero(r_[1, diff(t)[:-1]])
(array([0, 3, 5, 6, 7, 8]),)
Notice that it finds the beginning of both subsequence of 3s and both subsequences of 8s:
[1, 1, 1, 2, 2, 3, 8, 3, 8, 8]
So it's slightly different than finding the first occurrence of each value. In your program, you may be able to work with a sorted version of t
to get what you want:
>>> st = sorted(t)
>>> nonzero(r_[1, diff(st)[:-1]])
(array([0, 3, 5, 7]),)