第一题:
def normalize(name):
name = name.lower()
return name[0].upper() + name[1:]
第二题:
def prod(L):
return reduce(lambda x,y : x*y , L)
第三题:
dict = {'0':0,'1':1,'2':2,'3':3,'4':4,'5':5,'6':6,'7':7,'8':8,'9':9}
num = len(s) - s.index('.') - 1
s1 = [x for x in s if x in dict.keys()]
return (reduce(lambda x,y : x*10 + y ,map(lambda x:dict[x],s1)))/(10**num)
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赏侬十巴掌
第一题:
def normalize(name):
name = name.lower()
return name[0].upper() + name[1:]
第二题:
def prod(L):
return reduce(lambda x,y : x*y , L)
第三题:
dict = {'0':0,'1':1,'2':2,'3':3,'4':4,'5':5,'6':6,'7':7,'8':8,'9':9}
num = len(s) - s.index('.') - 1
s1 = [x for x in s if x in dict.keys()]
return (reduce(lambda x,y : x*10 + y ,map(lambda x:dict[x],s1)))/(10**num)