找出symmetric_difference的结果来自哪个集合的最佳实践是什么?
intersect = s1.symmetric_difference(s2)结果应该是
{'34':'s1', '66':'s2'} 其中'34','66‘是唯一的项目。
发布于 2013-08-07 11:42:44
为了最干净地做到这一点,以下几点应该有效:
intersect = s1.symmetric_difference(s2)
result = dict([(i, ("s1" if i in s1 else "s2")) for i in intersect])发布于 2013-08-07 11:42:29
{x : 's1' for x in intersect if x in s1} + {x : 's2' for x in intersect if x in s2}或
{x : ('s1' if x in s1 else 's2') for x in intersect}https://stackoverflow.com/questions/18102319
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