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This topic created in 5056 days ago, the information mentioned may be changed or developed.
求问哪种方法快一点儿。。目前只能想到手工 for 。。
其实这个 list 或者 tuple 是从 xml 转换过来的。
直接处理字符串的话结果好像不对⋯⋯
>>> a = "<name>123</name><name>345</name><name>456</name>"
>>> b = "<name>123</name><name>333</name>"
>>> a.split(b)
['<name>123</name><name>345</name><name>456</name>']
>>> b.split(a)
['<name>123</name><name>333</name>']
求问这样的怎么处理比较好⋯⋯
其实这个 list 或者 tuple 是从 xml 转换过来的。
直接处理字符串的话结果好像不对⋯⋯
>>> a = "<name>123</name><name>345</name><name>456</name>"
>>> b = "<name>123</name><name>333</name>"
>>> a.split(b)
['<name>123</name><name>345</name><name>456</name>']
>>> b.split(a)
['<name>123</name><name>333</name>']
求问这样的怎么处理比较好⋯⋯
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1
lisztli Jul 6, 2012  2
sa, sb = set(a), set(b)
return list(sa - sb) |
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2
sethverlo OP @lisztli 不对啊⋯⋯咦你不是饭否 api 的那个维护者么。。。
>>> a = "<id>123</id><id>345</id><id>456</id>" >>> b = "<id>123</id><id>333</id>" >>> sa, sb = set(a), set(b) >>> list(sa - sb) ['5', '4', '6'] >>> a = ['123', '456', '567'] >>> b = ['123,', '555'] >>> sa, sb = set(a), set(b) >>> list(sa - sb) ['123', '456', '567'] |
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4
lisztli Jul 6, 2012 1
>>> a = ['123', '456', '567']
>>> b = ['123,', '555'] 再看一遍代码…… |
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5
lisztli Jul 6, 2012 1
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10
fanzeyi Jul 6, 2012
a = [1,2,3,4,5]
b = [5,6,7,8] c = [xx for xx in b if xx not in a] |
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17
ch_linghu Jul 12, 2012 1
try difflib
http://docs.python.org/library/difflib.html |
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