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Reduce solution in Clear category for Min and Max by hencoappel
def reduce(func, seq):
it = iter(seq)
val = next(it)
for x in it:
val = func(val, x)
return val
def min(*args, **kwargs):
f = kwargs.get("key", lambda x: x)
l = args if len(args) > 1 else args[0]
return reduce(lambda x, y: x if f(x) <= f(y) else y, l)
def max(*args, **kwargs):
f = kwargs.get("key", lambda x: x)
l = args if len(args) > 1 else args[0]
return reduce(lambda x, y: x if f(x) >= f(y) else y, l)
if __name__ == '__main__':
#These "asserts" using only for self-checking and not necessary for auto-testing
assert max(3, 2) == 3, "Simple case max"
assert min(3, 2) == 2, "Simple case min"
assert max([1, 2, 0, 3, 4]) == 4, "From a list"
assert min("hello") == "e", "From string"
assert max(2.2, 5.6, 5.9, key=int) == 5.6, "Two maximal items"
assert min([[1, 2], [3, 4], [9, 0]], key=lambda x: x[1]) == [9, 0], "lambda key"
Jan. 7, 2015
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