Python实现比较两个列表(list)范围

有一道题： 比较两个列表范围，如果包含的话，返回TRUE，否则FALSE。 详细题目如下：

Create a function, this function receives two lists as parameters, each list indicates a scope of numbers, the function judges whether list2 is included in list1.

Function signature:
differ_scope(list1, list2)

Parameters:
list1, list2 - list1 and list2 are constructed with strings,
each string indicates a number or a scope of
numbers. The number or scope are randomly, can
be overlapped. All numbers are positive.

E.g.
['23', '44-67', '12', '3', '20-90']
Return Values:
True - if all scopes and numbers indicated by list2 are included in list1.
False - if any scope or number in list2 is out of the range in list1.
Examples:
case1 - list1 = ['23', '44-67', '12', '3', '20-90']
list2 = ['22-34', '33', 45', '60-61']
differ_scope(list1, list2) == True
case2 - list1 = ['23', '44-67', '12', '3', '20-90']
list2 = ['22-34', '33', 45', '60-61', '100']
differ_scope(list1, list2) == False

贴上自己写的代码如下:(备注： python/' target='_blank'>python 2.7.6)

def differ_scope(list1, list2):   print "list1:" + str(list1)   print "list2:" + str(list2)   #设置临时存放列表   list1_not_ = [] #用于存放列表1正常的数字值，当然要用int()来转换   list1_yes_ = [] #用于存放列表1中范围值如 44-67   list1_final = [] #用于存放列表1中最终范围值 如:[1,2,3,4,5,6,7,8,9,10]   temp1    = []      list2_not_ = []  #用于存放列表2正常的数字值，当然要用int()来转换   list2_yes_ = []  #用于存放列表2中范围值如 44-67   list2_final= []  #用于存放列表2中最终范围值 如:[1,2,3,4,5,6,7,8,9,10]   temp2   = []    temp    = []  #用于存放列表1,与列表2比较后的列表,从而判断结果为True还是False.      #对列表1进行处理   for i in range(len(list1)): #用FOR循环对列表1进行遍历     tag = 0     if list1[i].find('-')>0:#对含范围的数字进行处理，放到list_yes_列表中        strlist = list1[i].split('-')     list1_yes_ = range(int(strlist[0]),int(strlist[1])+1)#让其生成一个范围列表     for each in list1_yes_:     #FOR循环遍历所有符合条件的.         [temp1.append(each)]     else:           #对列表1中正常的数字进行处理，放到list_not_列表中       list1_not_.append(int(list1[i]))#对列表1中进行处理，放到list_yes_      [temp1.append(i) for i in list1_not_ if not i in temp1]#去除重复项   list1_final = sorted(temp1) #比较后,排序,并放到list1_final列表中   print "list1_final value is:" + str(list1_final)#打印排序后最终list1_final列表       #对列表2进行处理   for i in range(len(list2)):     if list2[i].find('-')>0:       strlist = list2[i].split('-')     list2_yes_ = range(int(strlist[0]),int(strlist[1])+1)     for each in list2_yes_:         [temp2.append(each)]       print "Temp2:" + str(temp2)     else:       list2_not_.append(int(list2[i]))   [temp2.append(i) for i in list2_not_ if not i in temp2]   list2_final = sorted(temp2)   print "list2_final value is:" + str(list2_final)    #对两个列表进行比较,得出最终比较结果.   [temp.append(i) for i in list2_final if not i in list1_final]#比较两个列表差值.   print "In list2 but not in list1:%s" % (temp)#打印出列表1与列表2的差值   if len(temp)>=1 :     print "The result is: False"   else:     print "The result is: True"  if __name__ == '__main__':   list1 = ['23', '44-67', '12', '3','90-100']   list2 = ['22-34', '33', '45']   differ_scope(list1,list2) 

总结：
1. 这道题关键是想法，如果整成坐标的方式来比较，会很麻烦。
2. 列表转成范围后，如果消除重复项，同样是里面的关键所在。
3. 其次是对列表遍历的操作，同样挺重要。