Python: return the index of the first element of a list which makes a passed function true

Question

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)

Answer 1

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 throw ValueError here.