George [...] data = [" 1.1.1.1"," 1.2.2.2"," 1.2.2.3"," ; 1.3.1.2"," 1.3.4.5"] 来自itertools import groupby datadict = \ dict((k,len( list(g)))for k,g in groupby(data,lambda s:s [:3])) print datadict 注意这个只有当版本已经排序时才能正常工作 按主要版本。 是的,我应该提到它。以下是一个更全面的例子。可能有更好的方法来排序版本号,但这就是我所做的。 data = [" 1.2.2.2"," 1.2。 2.3,1.3.1.2,1.1.1.1,1.3.14.5, " 1.3.21.6" ] 来自itertools import groupby import re RXBUILDSORT = re.compile(r''\ d + | [a-zA-Z]'') def versionsort(s): key = [] 部分在RXBUILDSORT.findall(s.lower()): 尝试: key.append(int(part)) 除外ValueError: key.append(ord(part)) 返回元组(键) data.sort(key =版本) 打印数据 datadict = \ dict((k,len(list(g)))for k,g in groupby(data,lambda s:s [:3])) print datadict 该代码可能正常运行,但它是完全不可读的,而良好的 Python程序员重视高可读性。所以正确的做法是将b / b 分成几行,给出有意义的名字,甚至可以添加评论 。 len(list(g)))看起来对我的小leniter()函数来说是个好工作 (或者更好的只是对len的语义的扩展)前一段时间 这里的一些人判断为无用,而我经常使用它来 Python和D ;-) 再见, bearophile Simon Mullis wrote:[...]data = [ "1.1.1.1", "1.2.2.2", "1.2.2.3", "1.3.1.2", "1.3.4.5"]from itertools import groupbydatadict = \dict((k, len(list(g))) for k,g in groupby(data, lambda s: s[:3]))print datadict 解决方案[...]data = [ "1.1.1.1", "1.2.2.2", "1.2.2.3", "1.3.1.2", "1.3.4.5"]from itertools import groupbydatadict = \ dict((k, len(list(g))) for k,g in groupby(data, lambda s: s[:3]))print datadictNote that this works correctly only if the versions are already sortedby major version.George[...]data = [ "1.1.1.1", "1.2.2.2", "1.2.2.3", "1.3.1.2", "1.3.4.5"]from itertools import groupbydatadict = \ dict((k, len(list(g))) for k,g in groupby(data, lambda s: s[:3]))print datadictNote that this works correctly only if the versions are already sortedby major version.Yes, I should have mentioned it. Here''s a fuller example below. There''smaybe better ways of sorting version numbers, but this is what I do.data = [ "1.2.2.2", "1.2.2.3", "1.3.1.2", "1.1.1.1", "1.3.14.5","1.3.21.6" ]from itertools import groupbyimport reRXBUILDSORT = re.compile(r''\d+|[a-zA-Z]'')def versionsort(s):key = []for part in RXBUILDSORT.findall(s.lower()):try:key.append(int(part))except ValueError:key.append(ord(part))return tuple(key)data.sort(key=versionsort)print datadatadict = \dict((k, len(list(g))) for k,g in groupby(data, lambda s: s[:3]))print datadictThat code may run correctly, but it''s quite unreadable, while goodPython programmers value high readability. So the right thing to do isto split that line into parts, giving meaningful names, and maybe evenadd comments.len(list(g))) looks like a good job for my little leniter() function(or better just an extension to the semantics of len) that time agosome people here have judged as useless, while I use it often in bothPython and D ;-)Bye,bearophile 这篇关于Re:dict生成器问题的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持! 上岸,阿里云!
05-27 05:31
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