红宝石比较2哈希

如果密钥的顺序不一致，有没有办法比较2哈希？例如红宝石比较2哈希

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}

每个散列具有相同的键。我将如何去比较它们并确保每个关键值是正确的。我所见过的所有问题和答案都以相同的顺序显示密钥。有关系吗？

我试着使用：

hash1_1 = hash1.select{|k,_| hash2.has_key? k}

它吐了出来：

{ "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}

请帮我解释什么，我需要做的。谢谢。

具有相同的键/值对

来源

2014-10-09 T0ny lombardi

两个哈希将等于不管键顺序：

a = {:x => 1, :y => 2} 
b = {:y => 2, :x => 1} 

a == b 
# => true

来源

2014-10-09 18:35:50 infused

感谢帮助。 – 2014-10-09 19:03:15

即使认为这种行为是有记录的，不同的关键顺序的两个哈希应该被认为是平等的吗？它可能是一个兼容性“功能”，但概念上看起来很可疑。 – tokland 2014-10-09 19:16:48

直到最近，哈希都没有保留插入顺序，所以为了兼容性的原因，比较成立。 – tadman 2014-10-09 19:23:05

你误解了你的结果。你的代码工作正常。你的算法虽然错了。

我相信你HASH1和HASH2并使用拒绝摆脱它们之间相匹配的一切创造差异：

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"}

你现在跟他们做什么，是你的。如果你因此得到空的散列，那么一切都是一样的。

还有一个gem called 'hashdiff'可以使用。

来源

2014-10-09 18:54:40

谢谢你解释说对我来说。这将有助于我的代码很多。谢谢。 – 2014-10-09 19:02:31

发现此解决方案存在问题。如果我从hash2中取出一个键值对，那么结果仍为空。 – 2014-10-09 19:14:55

Tony，你必须同时使用diff_in_hash1和diff_in_hash2。如果您从hash1中删除某些内容，它将显示在diff_in_hash2中，并反之亦然。 – 2014-10-09 19:37:06

红宝石比较2哈希

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