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First solution in Clear category for Min and Max by WpoZoo
# migrated from python 2.7
def min(*args, **kwargs):
keyfun = None
if kwargs:
keyfun= kwargs['key']
items_data = []
mapped_data = []
argv_len = len(args)
if argv_len > 1:
items_data = [i for i in args]
mapped_data = list(map(keyfun, items_data))
else:
items_data = [i for i in args[0]]
mapped_data = list(map(keyfun, items_data))
deal_data = list(zip(items_data, mapped_data))
index = deal_data[0]
for i in deal_data:
if i[1] < index[1]:
index = i
return index[0]
def max(*args, **kwargs):
keyfun = None
if kwargs:
keyfun= kwargs['key']
items_data = []
mapped_data = []
argv_len = len(args)
if argv_len > 1:
items_data = [i for i in args]
mapped_data = list(map(keyfun, items_data))
else:
items_data = [i for i in args[0]]
mapped_data = list(map(keyfun, items_data))
deal_data = list(zip(items_data, mapped_data))
index = deal_data[0]
for i in deal_data:
if i[1] > index[1]:
index = i
return index[0]
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"
Oct. 9, 2014
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