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 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。