Hello NG,
I was wondering if there is a faster/nicer method (than a for loop) that will allow me to find the elements (AND their indices) in a list that verify a certain condition. For example, assuming that I have a list like:
mylist = [0, 1, 1, 1, 1, 5, 6, 7, 8, 1, 10]
I would like to find the indices of the elements in the list that are equal to 1 (in this case, the 1,2,3,4,9 elements are equal to 1). I could easily use a for loop but I was wondering if there is a faster method...
Everyone has already given you the answer (enumerate in a LC or GE), I'd just comment that it's easy enough to extend their answers to any given condition:
>>> def getindices(sequence, predicate): ... return [i for i, v in enumerate(sequence) if predicate(v)] ... >>> getindices([0,1,1,1,1,5,6,7,8,1,10], bool) [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] >>> def equalsone(v): ... return v == 1 ... >>> getindices([0,1,1,1,1,5,6,7,8,1,10], equalsone) [1, 2, 3, 4, 9] >>> def f(v): ... return pow(v, 3, 4) == 3 ... >>> getindices([0,1,1,1,1,5,6,7,8,1,10], f) [7]
Steve -- http://mail.python.org/mailman/listinfo/python-list
