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)
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Answer
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.