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First solution in Clear category for Min and Max by apotokin
def min(*iter, **kwargs):
key = kwargs.get("key", lambda x: x)
if (len(iter) == 1): iter=iter[0] # in case there itere is 'iterator' in arguments
for index, item in enumerate(iter):
if not index:
m = item
continue
if key(m) > key(item): m = item
return m
def max(*iter, **kwargs):
key = kwargs.get("key", lambda x: x)
if (len(iter) == 1): iter=iter[0] # in case there itere is 'iterator' in arguments
for index, item in enumerate(iter):
if not index:
m = item
continue
if key(m) < key(item): m = item
return m
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"
# max(iterable, *[, key]) or min(iterable, *[, key])
#max(arg1, arg2, *args[, key]) or min(arg1, arg2, *args[, key])
July 21, 2015
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