How does reduce
function work in python3 with three parameters instead of two.
So, for two,
tup = (1,2,3)
reduce(lambda x, y: x+y, tup)
<
If you omit the third parameter, then the first value from tup
is used as the initializer.
Or, to put it a different way, reduce()
places the optional 3rd parameter before the values of the second argument, if present.
Moreover, that means that if the second argument is an empty sequence, that third argument serves as the default, just as a second argument with only one element (and no explicit initializer argument), would be the default return value.
The functools.reduce() documentation includes a Python version of the function:
def reduce(function, iterable, initializer=None):
it = iter(iterable)
if initializer is None:
value = next(it)
else:
value = initializer
for element in it:
value = function(value, element)
return value
Note how the initializer
, when not None
, is used as the first value instead of a first value from iterable
.