["python","list","indexing"]
Return the index of the first element of a list which makes a passed function true

The list.index(x) function returns the index in the list of the first item whose value is x. Is there a function, list_func_index(), similar to the index() function that has a function, f(), as a parameter. The function, f() is run on every element, e, of the list until f(e) returns True. Then list_func_index() returns the index of e. Codewise: >>> def list_func_index(lst, func): for i in range(len(lst)): if func(lst[i]): return i raise ValueError('no element making func True') >>> l = [8,10,4,5,7] >>> def is_odd(x): return x % 2 != 0 >>> list_func_index(l,is_odd) 3 Is there a more elegant solution? (and a better name for the function)
You could do that in a one-liner using generators: next(i for i,v in enumerate(l) if is_odd(v)) The nice thing about generators is that they only compute up to the requested amount. So requesting the first two indices is (almost) just as easy: y = (i for i,v in enumerate(l) if is_odd(v)) x1 = next(y) x2 = next(y) Though, expect a StopIteration exception after the last index (that is how generators work). This is also convenient in your "take-first" approach, to know that no such value was found --- the list.index() function would raise ValueError here.