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First solution in Clear category for Min and Max by jacekgrycza
def min(*args, **kwargs):
key = kwargs.get("key", None)
# print (len(args))
if(len(args)==1):
argsl=list(args[0])
x=argsl[0]
for i in argsl:
if(key!=None):
if key(i)key(x):
x=i
else:
if i>x:
x=i
else:
x=args[0]
for i in args:
if(key!=None):
if key(i)>key(x):
x=i
else:
if i>x:
x=i
print (x)
return (x)
min(2.2,3.5,4.4,1.2, key=int)
max(2.2,3.5,4.4,1.2, key=int)
max(3, 2)
min(abs(i) for i in range(-10, 10))
#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. 18, 2016
