判断一个列表是否属于另外一个列表

bxfqing

开始的想法时用issubset函数，但是这个不符合我的需求
我的需求是这样，有一组数字，比如 1,3,5,7,8,9,10,30,70,71,100
从小到大排列的，然后我要挑选出来连续的数字，
2个连续：7,8
三个连续：7,8,9

issubset会一口气都匹配成功 :wink:
所以我就用了下面这个
if ''.join([str(x) for x in checklist]) in ''.join([str(x) for x in continuelist]):
但是写法也有问题，居然会认为8158 8168是连续的数字。。。。
ERROR: >>> count: 5 floor ['8158', '8168']

有什么更好的方法完成这个功能吗？




参考代码如下

1. import urllib, urllib2, cookielib,re
   
2. import sys,os,random
   
3. import mimetypes
   
4. import mimetools
   
5. import codecs
   
6. import platform
   
7. import time
   
8. import datetime
   
9. import pdb
   
10. import getopt
    
11. from urllib2 import HTTPError, URLError
    
12. reload(sys)
    
13. sys.setdefaultencoding('utf8')
    
14. 
    
15. continuelist =[]
    
16. target_list = ['8086', '8143', '8145', '8149', '8158', '8168', '8174', '8177', '8181', '8188', '8190', '8195', '8203', '8207', '8209', '8211', '8213', '8217', '8224', '8229', '8444', '8453', '8456', '8458', '8462', '8471', '8475', '8481', '8490', '8496', '8507', '8518', '8529', '8538', '8559', '8574', '8591', '8606', '8620', '8950', '8959', '8963', '8970', '8975', '8980', '8984', '8988', '8991', '8996', '9001', '9003', '9011', '9024', '9032', '9048', '9059', '9072', '9084', '9092', '9415', '9419', '9424', '9426', '9432', '9435', '9439', '9445', '9447', '9451', '9454', '9456', '9461', '9463', '9468', '9471', '9477', '9481', '9487', '9490', '9605', '9608', '9677', '9680', '9682', '9689', '9693', '9697', '9699', '9701', '9705', '9709', '9713', '9717', '9720', '9723', '9730', '9742', '9759']
    
17. 
    
18. #print target_list
    
19. con_num = 2
    
20. #for i in range(int(target_list[0]),int(target_list[-1])+1):
    
21. for i in range(int(target_list[0]),int(target_list[-1])+1):
    
22.     continuelist.append(str(i))
    
23. print continuelist
    
24. for items in range(len(target_list)):
    
25.     checklist = target_list[items:items+int(con_num)]
    
26.     if len(checklist) < int(con_num):
    
27.         #print 'Skip',checklist
    
28.         continue
    
29.     if set(checklist).issubset(set(continuelist)):
    
30.     #if ''.join([str(x) for x in checklist]) in ''.join([str(x) for x in continuelist]):
    
31.         print u'ERROR: >>> count:',items+1,u'floor',checklist

复制代码

q1208c

回复 1# bxfqing

只针对你的标题, 我觉得用 set 可能会很容易吧? 只人判断 子集 就可以了.

但你问题中的连续数字, 我没搞明白的时, 你为啥不把它们变成数字, 而是弄些字符串在操作呢?
   

bxfqing

回复 2# q1208c


    谢谢你的提议，我使用你的方法试试

huangxiaohen

1. s= [1,3,4,5,66,88]
   
2. childList = [s[i:i+n] for i in range(len(s)) for n in range(1, len(s)-i+1)]
   
3. for child in childList:
   
4.         if len(child) == 2 and child[0]+1 == child[1]:
   
5.                 print "两个连续的:",child
   
6. 
   
7.         if len(child) == 3 and child[0]+1 == child[1] and child[1]+1 == child[2]:
   
8.                 print '三个连续的:',child

复制代码

bskay

傻了....

if ','.join([str(x) for x in checklist]) in ','.join([str(x) for x in continuelist]):


ls = map(int, target_list)
ret = []
part = []
for d in ls:
        if len(part) == 0:
                part.append(d)
                continue
        if part[-1]+1 == d:
                part.append(d)
                continue
        if len(part) < 2:
                part = []
                continue
        ret.append(part)
        part = []
       
#2个的
[a for a in ret if len(a) == 2]
#3个的
[a for a in ret if len(a) == 3]

lizhihui_kevin

比较笨的方法，欢迎指点：

1. target_list = ['8140', '8143', '8144', '8149', '8158', '8159', '8160', '8160', '8181', '8188', '8190', '8195', '8203', \
   
2.                '8207', '8209', '8211', '8213', '8217', '8224', '8229', '8444', '8453', '8456', '8458', '8462', '8471', \
   
3.                '8475', '8481', '8490', '8496', '8507', '8518', '8529', '8538', '8559', '8574', '8591', '8606', '8620', \
   
4.                '8950', '8959', '8963', '8970', '8975', '8980', '8984', '8988', '8991', '8996', '9001', '9002', '9003', \
   
5.                '9024', '9032', '9048', '9059', '9072', '9084', '9092', '9415', '9419', '9424', '9426', '9432', '9435', \
   
6.                '9439', '9445', '9447', '9451', '9454', '9456', '9461', '9463', '9468', '9471', '9477', '9481', '9487', \
   
7.                '9490', '9605', '9608', '9677', '9680', '9682', '9689', '9693', '9697', '9699', '9701', '9705', '9709', \
   
8.                '9713', '9717', '9720', '9723', '9730', '9742', '9759']
   
9. 
   
10. def findAllContinuousSubList(originalList = None):
    
11.     if originalList is None:
    
12.         originalList = []
    
13. 
    
14.     target_list = [int(i) for i in originalList]
    
15.     new_list=[]
    
16.     new_sub_list = []
    
17.     for i in range(0,len(target_list)-1):
    
18.         if target_list[i]-target_list[i+1]==-1:
    
19.             new_sub_list.append(target_list[i])
    
20.         elif target_list[i]-target_list[i+1]!=-1 and target_list[i-1] in new_sub_list:
    
21.             new_sub_list.append(target_list[i])
    
22.             new_list.append(new_sub_list)
    
23.             new_sub_list = []
    
24.     return new_list

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结果： [[8143, 8144], [8158, 8159, 8160], [9001, 9002, 9003]]

lizhihui_kevin

回复 4# huangxiaohen
强大， 但是如果list很大的话这个可能对内存消耗优点大，不妨试下iterator

1. def findAllContinuousSubListThroughIter(s):
   
2.     s = [int(i) for i in s]
   
3.     import itertools
   
4. 
   
5.     # iterator for two continuous sublists
   
6.     sub_list_com_2 = itertools.combinations(s, 2)
   
7.     for sub_list in sub_list_com_2:
   
8.         if sub_list[0]-sub_list[1]==-1:
   
9.             print ">>> 两个连续: %s" %list(sub_list)
   
10. 
    
11.     # iterator for three continuous sublists
    
12.     sub_list_com_3 = itertools.combinations(s, 3)
    
13.     for sub_list in sub_list_com_3:
    
14.         if sub_list[0]-sub_list[1]==-1 and sub_list[1]-sub_list[2]==-1:
    
15.             print ">>> 三个连续: %s" %list(sub_list)
    

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bxfqing

谢谢大家的帮助，我慢慢试试每种写法