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⇤ ← Revision 1 as of 2008-12-12 01:33:16
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| == 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))]) }}} |
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))])
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