判斷一個(gè)列表是否屬于另外一個(gè)列表

bxfqing

開始的想法時(shí)用issubset函數(shù)，但是這個(gè)不符合我的需求
我的需求是這樣，有一組數(shù)字，比如 1,3,5,7,8,9,10,30,70,71,100
從小到大排列的，然后我要挑選出來連續(xù)的數(shù)字，
2個(gè)連續(xù)：7,8
三個(gè)連續(xù)：7,8,9

issubset會(huì)一口氣都匹配成功 :wink:
所以我就用了下面這個(gè)
if ''.join([str(x) for x in checklist]) in ''.join([str(x) for x in continuelist]):
但是寫法也有問題，居然會(huì)認(rèn)為8158 8168是連續(xù)的數(shù)字。。。。
ERROR: >>> count: 5 floor ['8158', '8168']

有什么更好的方法完成這個(gè)功能嗎？




參考代碼如下

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

復(fù)制代碼

q1208c

回復(fù) 1# bxfqing

只針對(duì)你的標(biāo)題, 我覺得用 set 可能會(huì)很容易吧? 只人判斷 子集 就可以了.

但你問題中的連續(xù)數(shù)字, 我沒搞明白的時(shí), 你為啥不把它們變成數(shù)字, 而是弄些字符串在操作呢?
   

bxfqing

回復(fù) 2# q1208c


    謝謝你的提議，我使用你的方法試試

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 "兩個(gè)連續(xù)的:",child
   
6. 
   
7.         if len(child) == 3 and child[0]+1 == child[1] and child[1]+1 == child[2]:
   
8.                 print '三個(gè)連續(xù)的:',child

復(fù)制代碼

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個(gè)的
[a for a in ret if len(a) == 2]
#3個(gè)的
[a for a in ret if len(a) == 3]

lizhihui_kevin

比較笨的方法，歡迎指點(diǎn)：

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

復(fù)制代碼

結(jié)果： [[8143, 8144], [8158, 8159, 8160], [9001, 9002, 9003]]

lizhihui_kevin

回復(fù) 4# huangxiaohen
強(qiáng)大， 但是如果list很大的話這個(gè)可能對(duì)內(nèi)存消耗優(yōu)點(diǎn)大，不妨試下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 ">>> 兩個(gè)連續(xù): %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 ">>> 三個(gè)連續(xù): %s" %list(sub_list)
    

復(fù)制代碼

bxfqing

謝謝大家的幫助，我慢慢試試每種寫法