我一个txt内容如下:
3 1.234.65.197
6 120.38.244.29
6 183.221.184.29
6 222.129.57.61
请教下如何转为一个字典
3 1.234.65.197
6 120.38.244.29
6 183.221.184.29
6 222.129.57.61
请教下如何转为一个字典
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This topic created in 4194 days ago, the information mentioned may be changed or developed.
 | 1 VeryCB Nov 17, 2014 你期望的字典格式是什么样的？ |
 | 2 dbas OP 3: 1.234.65.197,6:120.38.244.29 |
 | 3 ChanneW Nov 17, 2014 3 和 6 是什么意思 |
 | 4 est Nov 17, 2014 dict(o.split(' ') for o in """ 3 1.234.65.197 6 120.38.244.29 6 183.221.184.29 6 222.129.57.61 """.splitlines()) |
| 5 est Nov 17, 2014  1 >>> dict(o.split(' ') for o in """ ... 3 1.234.65.197 ... 6 120.38.244.29 ... 6 183.221.184.29 ... 6 222.129.57.61 ... """.splitlines() if o) {'3': '1.234.65.197', '6': '222.129.57.61'} |
 | 6 irosyking Nov 17, 2014 写的不够好，你看一下 import re s='''3 1.234.65.197 6 120.38.244.29 6 183.221.184.29 6 222.129.57.61 ''' d=dict() for g in re.finditer(r'(\d) ((?:(?:2(?:[0-4][0-9]|5[0-5])|[0-1]?[0-9]?[0-9])\.){3}(?:(?:2(?:[0-4][0-9]|5[0-5])|[0-1]?[0-9]?[0-9])))',s): d[g.group(1)]=g.group(2) print d |
 | 7 SakuraSa Nov 17, 2014 |
| 8 SakuraSa Nov 17, 2014 如何markdown呢？  ?1 |
 | 10 chengdujin Nov 17, 2014 with open("input", "r") as f: data = [line.strip().split() for line in f.readlines()] print {d[0]:d[1] for d in data} |
| 11 dbas OP 想要est的结果。呵呵 |
| 12 chengdujin Nov 17, 2014 |
 | 13 xiaowangge Dec 3, 2014 《如何用好Google搜索引擎？》 http://www.zhihu.com/question/20161362 《十大高明的Google搜索技巧》 http://www.williamlong.info/archives/728.html 《提问的智慧》 http://wiki.woodpecker.org.cn/moin/AskForHelp |
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