Ruby比较2个散列
Ruby比较2个散列
提问于 2014-10-10 02:32:19
如果键的顺序不同，有没有办法比较两个哈希值？例如
hash1 = { "it"=> 10,"was"=>11,"the"=>14,"best"=>1,"of"=>12,"times"=>2,
"worst"=>1,"age"=>2,"wisdom"=>1,"foolishness"=>1,"epoch"=>2,
"belief"=>1,"incredulity"=>1,"season"=>2,"light"=>1,"darkness"=>1,
"spring"=>1,"hope"=>1,"winter"=>1,"despair"=>1,"we"=>4,"had"=>2,
"everything"=>1,"before"=>2,"us"=>2,"nothing"=>1,"were"=>2,"all"=>2,
"going"=>2,"direct"=>2,"to"=>1,"heaven"=>1,"other"=>1,
"way"=>1,"in"=>2,"short"=>1,"period"=>2,"so"=>1,"far"=>1,"like"=>1,
"present"=>1,"that"=>1,"some"=>1,"its"=>2,"noisiest"=>1,
"authorities"=>1,"insisted"=>1,"on"=>1,"being"=>1,
"received"=>1,"for"=>2,"good"=>1,"or"=>1,"evil"=>1,"superlative"=>1,
"degree"=>1,"comparison"=>1,"only"=>1 }
hash2 = {"superlative"=>1, "it"=>10, "going"=>2, "spring"=>1, "age"=>2,
"despair"=>1, "received"=>1, "good"=>1, "some"=>1, "worst"=>1, "was"=>11,
"only"=>1,"us"=>2, "evil"=>1, "belief"=>1, "for"=>2, "darkness"=>1,
"comparison"=>1, "short"=>1, "in"=>2, "present"=>1, "direct"=>2, "were"=>2,
"way"=>1, "degree"=>1, "or"=>1, "of"=>12, "epoch"=>2, "incredulity"=>1,
"period"=>2, "heaven"=>1, "other"=>1, "being"=>1, "its"=>2, "so"=>1,
"authorities"=>1, "times"=>2, "we"=>4, "noisiest"=>1, "light"=>1, "hope"=>1,
"foolishness"=>1, "everything"=>1, "far"=>1, "wisdom"=>1, "season"=>2, "like"=>1,
"before"=>2, "had"=>2, "the"=>14, "nothing"=>1, "winter"=>1, "best"=>1,
"that"=>1, "all"=>2, "insisted"=>1, "to"=>1, "on"=>1}每个散列都有相同的键。我如何比较它们，并确保每个键值都是正确的。我看到的所有问题和答案都以相同的顺序显示了带有键的散列。有关系吗？
我试着使用：
hash1_1 = hash1.select{|k,_| hash2.has_key? k}
它会吐出来：
{ "it"=>10, "was"=>11, "the"=>14, "best"=>1, "of"=>12, "times"=>2,
"worst"=>1, "age"=>2, "wisdom"=>1, "foolishness"=>1, "epoch"=>2,
"belief"=>1, "incredulity"=>1, "season"=>2, "light"=>1,
"darkness"=>1, "spring"=>1, "hope"=>1, "winter"=>1, "despair"=>1,
"we"=>4, "had"=>2, "everything"=>1, "before"=>2, "us"=>2, "nothing"=>1,
"were"=>2, "all"=>2, "going"=>2, "direct"=>2, "to"=>1, "heaven"=>1,
"other"=>1, "way"=>1, "in"=>2, "short"=>1, "period"=>2, "so"=>1, "far"=>1,
"like"=>1, "present"=>1, "that"=>1, "some"=>1, "its"=>2, "noisiest"=>1,
"authorities"=>1, "insisted"=>1, "on"=>1, "being"=>1, "received"=>1,
"for"=>2, "good"=>1, "or"=>1, "evil"=>1, "superlative"=>1, "degree"=>1,
"comparison"=>1, "only"=>1}请帮我解释一下我需要做什么。谢谢。
回答 3
Stack Overflow用户
回答已采纳
发布于 2014-10-10 02:35:50
无论键的顺序如何，具有相同键/值对的两个散列将是相等的：
a = {:x => 1, :y => 2}
b = {:y => 2, :x => 1}
a == b
# => trueStack Overflow用户
发布于 2014-10-10 02:54:41
你曲解了你的结果。你的代码运行得很好。但是你的算法是错误的。
我将使用您的hash1和hash2，并使用reject创建一个差异，以删除它们之间匹配的所有内容：
hash1["it"] = 9
hash1["Tony"] = "great"
hash2["Jeff"] = "awesome"
hash1.delete "was"
diff_in_hash1 = hash1.reject{|k,v| hash2[k] == v}
# => {"it"=>9, "Tony"=>"great"}
diff_in_hash2 = hash2.reject{|k,v| hash1[k] == v}
# => {"it"=>10, "was"=>11, "Jeff"=>"awesome"}你现在怎么处理它们，由你决定。如果你得到的结果是空的哈希，那么一切都是一样的。
还有一个可能有用的gem called 'hashdiff'。
Stack Overflow用户
发布于 2018-05-31 19:00:05
据我正确理解，您的主要目标是回答两个散列之间是否有任何差异。因此，如果完全匹配(两个散列的键和相应值)，则比较结果应为true，否则为false。正如前面提到的here，从Ruby2.3开始。您可以使用：
{ a: 1 } <= { a: 1 } = true
{ a: 1 } <= { a: 2 } = false
{ a: 2 } <= { a: 1 } = false
{ a: 2 } <= { a: 2 } = true
{ a: 1 } >= { a: 1 } = true
{ a: 1 } >= { a: 2 } = false
{ a: 2 } >= { a: 1 } = false
{ a: 2 } >= { a: 2 } = true
{ a: 1 } < { a: 1 } = false
{ a: 1 } < { a: 2 } = false
{ a: 2 } < { a: 1 } = false
{ a: 2 } < { a: 2 } = false
{ a: 1 } > { a: 1 } = false
{ a: 1 } > { a: 2 } = false
{ a: 2 } > { a: 1 } = false
{ a: 2 } > { a: 2 } = false
{ a: 1 } == { a: 1 } = true
{ a: 1 } == { a: 2 } = false和
{ a: 1, b: 2 } > { a: 1 } = true
{ a: 1, b: 2 } == { a: 1 } = false
{ a: 1, b: 2 } >= { a: 1 } = true
{ a: 1, b: 2 } > { a: 1, b: 5 } = false页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/26285603
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