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== 方案3:fromkeys() ==
{{{
don li <[email protected]>
reply-to [email protected]
to [email protected]
date Fri, Dec 12, 2008 at 11:53
}}}

{{{#!python
alist = ['aaa', 'ccc', 'bbb', 'aaa', 'aaa', 'ccc']
adict = dict().fromkeys(alist, 0)

for a in alist:
    adict[a] += 1
}}}

TableOfContents

Include(ZPyUGnav)

统计列表重复项

提问

2008/12/11 卢熙 <[email protected]>

  • 要到达以下的效果： 

    alist = ['aaa', 'ccc', 'bbb', 'aaa', 'aaa', 'ccc']
    adict = fn(alist)
    print {'aaa': 3, 'bbb': 1, 'ccc': 2}
  • 在实际应用中，len(alist)很有可能超过10万,请问这个fn函数该如何写才能非常高效的完成这个任务？

方案1:for

萧萧 <[email protected]>
reply-to        [email protected]
to      [email protected]
date    Thu, Dec 11, 2008 at 22:51
subject [CPyUG:73576] Re: 如何高效的统计列表里面的重复项

>>> alist = ['aaa', 'ccc', 'bbb', 'aaa', 'aaa', 'ccc']
>>> adict = {}
>>> for i in alist:
...     try:
...             adict[i] += 1
...     except:
...             adict.setdefault(i, 1)
>>> adict

{'aaa': 3, 'bbb': 1, 'ccc': 2} ##endInc

方案2:count()

萧萧 <[email protected]>
reply-to        [email protected]
to      [email protected]
date    Fri, Dec 12, 2008 at 11:18

 alist = ['aaa', 'ccc', 'bbb', 'aaa', 'aaa', 'ccc']
 adict = dict([(i, alist.count(i) for i in list(set(alist))])

方案3:fromkeys()

don li <[email protected]>
reply-to        [email protected]
to      [email protected]
date    Fri, Dec 12, 2008 at 11:53

   1 alist = ['aaa', 'ccc', 'bbb', 'aaa', 'aaa', 'ccc']
   2 adict = dict().fromkeys(alist, 0)
   3 
   4 for a in alist:
   5     adict[a] += 1

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创建 by -- ZoomQuiet [DateTime(2008-12-12T01:33:16Z)]

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