初学python,以此题练习,不求财富。
# -*- coding: utf-8 -*-
dic = {'a':[1,2,3,4], 'b':[11,22,33]}
def gcd(m, n):
''' 求两个数的最大公约数 (greatest common divisor) '''
return m if n == 0 else gcd(n, m % n)
#三元表达式格式:true_part if condition else false_part
def lcm(m, n):
''' 求两个数的最小公倍数 (least common multiple) '''
return (m * n) / gcd(m, n)
lsLen = [len(dic['a']), len(dic['b'])]
lcmRet = lcm(lsLen[0], lsLen[1])
dicMultiple = {'a': lcmRet / lsLen[0], 'b': lcmRet / lsLen[1]}
dicRet = {'a':[], 'b':[]}
for key in dic:
for elem in dic[key]:
for n in range(dicMultiple[key]):
dicRet[key].append(elem)
print dicRet
#先求最小公倍数,再整形
dic = {'sports' :{'s.basketball':['a', 'b', 'c', 'd'], 's.football':['aa', 'bb', 'cc', 'dd'] },
'music':{'m.juzz':['e', 'f', 'g'], 'm.pop':['ee', 'ff', 'gg']}}
def gcd(a, b):
if b == 0:
return a
else:
return gcd(b, a % b)
def lcm(a, b):
return a * b / gcd(a, b)
myLcm = 1#最小公倍数
for k in dic:
for k2 in dic[k]:
myLcm = lcm(len(dic[k][k2]), myLcm)
newDiv = {}
for k in dic:
newDiv[k] = {}
for k2 in dic[k]:
newDiv[k][k2] = []
l = len(dic[k][k2])
for v in dic[k][k2]:
for i in range(myLcm / l):
newDiv[k][k2].append(v)
print newDiv