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3th solution in Clear category for Min and Max by arek.kowalski107
import types
#PO
def parse_args(args, key):
ret = [key(i) for i in args]
return ret
def nomalise_args(*args):
new_args = [None]
while len(new_args) == 1:
new_args = list(args[0])
if isinstance(new_args, types.GeneratorType):
new_args = list(new_args)
if new_args == args[0] or type(new_args[0]) in [int, float]:
break
args = new_args
return new_args
def find_value(args, biggest=True):
index = 0
value = args[0]
for i, v in enumerate(args):
if (biggest and v > value) or (not biggest and v < value):
index = i
value = v
return index
def min(*args, **kwargs):
key = kwargs.get("key", None)
nomalised_args = nomalise_args(args)
if key:
parsed_args = parse_args(nomalised_args, key)
else:
parsed_args = nomalised_args
return nomalised_args[find_value(parsed_args, False)]
def max(*args, **kwargs):
key = kwargs.get("key", None)
nomalised_args = nomalise_args(args)
if key:
parsed_args = parse_args(nomalised_args, key)
else:
parsed_args = nomalised_args
return nomalised_args[find_value(parsed_args)]
if __name__ == '__main__': # pragma: no cover
# 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"
Dec. 10, 2016
