Ruby 循环遍历 2 个哈希数组并比较键值

Question

我在尝试遍历两个哈希数组时遇到问题。 我的哈希看起来像：

h1= [
  {"name"=>"postgresql", "version"=>29},
  {"name"=>"java", "version"=>8},
  {"name"=>"python", "version"=>3}
]

h2= [
  {"name"=>"postgresql", "release"=>"postgresql-release", "url"=>"https://github.com/repo/postgresql-release"},
  {"name"=>"java", "release"=>"java-release", "url"=>"https://github.com/repo/java-release"}
]

combined = []

在检查 h1 和 h2 中“name”字段的哈希值后，我需要将这两个哈希组合到一个名为 combined 的数组中，这样：

• if h1["name"] == h2["name"] ---> 保存名称、版本、版本和网址。
• 如果h1["name"] != h2["name"]（例如：在h1中找到python但在h2中找不到）--->保存名称，版本，发布= nil，url= nil

combined 数组应如下所示：

combined = [
  {"name"=>"postgresql", "version"=>29, "release"=>"postgresql-release", "url"=>"https://github.com/repo/postgresql-release"},
  {"name"=>"java", "version"=>8, "release"=>"java-release", "url"=>"https://github.com/repo/java-release"},
  {"name"=> "python", "version"=>3, "release"=> nil, "url"=> nil}
]

我已经尝试了很多，并且总是有更多的迭代和/或在名称不同的情况下无法打印任何东西。

我的代码如下所示：

h1= [{"name"=>"postgresql", "version"=>29}, {"name"=>"java", "version"=>8}, {"name"=>"python", "version"=>3}]

h2= [{"name"=>"postgresql", "release"=>"postgresql-release", "url"=>"https://github.com/repo/postgresql-release"}, {"name"=>"java", "release"=>"java-release", "url"=>"https://github.com/repo/java-release"}]

combined = []

h1.each do |h1Key, h1Value|
  h2.each do |h2Key, h2Value|
    if h1["name"] == h2["name"]
      combined << {"name"=> h1["name"], "version"=> h1["version"], "release"=> h2["release"], "url"=> h2["url"]} 
    else
      combined << {"name"=> h1["name"], "version"=> h1["version"], "release"=> nil, "url"=> nil}
    end
  end
end
puts combined.to_s

我也无法正确比较 h1 和 h2 的值。

非常感谢您的帮助，因为我对 Ruby 还是很陌生，不幸的是我的逻辑并没有帮助我。

提前致谢！ ^^

Answer 1

首先，h1 和 h2 是哈希数组，而不是哈希本身。如果你遍历它们，你会得到数组中的每个散列，而不是某个散列的键和值。

为清楚起见，我们将它们重命名为 a1 和 a2。

a1= [{"name"=>"postgresql", "version"=>29}, {"name"=>"java", "version"=>8}, {"name"=>"python", "version"=>3}]
a2= [{"name"=>"postgresql", "release"=>"postgresql-release", "url"=>"https://github.com/repo/postgresql-release"}, {"name"=>"java", "release"=>"java-release", "url"=>"https://github.com/repo/java-release"}]

combined = []

现在，当我们遍历它们时，我们会依次得到每个哈希。

a1.each do |h1|

对于第一个数组中的每个散列，我们想在第二个数组中找到对应的散列。如果我们在第一个数组中查看 java 并在第二个数组中找到 postgresql，这并不意味着没有 java。我们现在只需要忽略它。如果我们没有找到匹配项，那么我们将分配 nil-s

h2 = a2.find { |h| h1["name"] == h["name"] } || {}

现在我们要么找到匹配项，要么没有找到匹配项，我们可以像你一样将结果添加到组合数组中。

combined << {"name"=> h1["name"], "version"=> h1["version"], "release"=> h2["release"], "url"=> h2["url"]}

大家一起：

a1= [{"name"=>"postgresql", "version"=>29}, {"name"=>"java", "version"=>8}, {"name"=>"python", "version"=>3}]
a2= [{"name"=>"postgresql", "release"=>"postgresql-release", "url"=>"https://github.com/repo/postgresql-release"}, {"name"=>"java", "release"=>"java-release", "url"=>"https://github.com/repo/java-release"}]

combined = []

a1.each do |h1|
  h2 = a2.find { |h| h1["name"] == h["name"] } || {}
  combined << {"name"=> h1["name"], "version"=> h1["version"], "release"=> h2["release"], "url"=> h2["url"]} 
end
puts combined.to_s #=> [{"name"=>"postgresql", "version"=>29, "release"=>"postgresql-release", "url"=>"https://github.com/repo/postgresql-release"}, {"name"=>"java", "version"=>8, "release"=>"java-release", "url"=>"https://github.com/repo/java-release"}, {"name"=>"python", "version"=>3, "release"=>nil, "url"=>nil}]

更简单的方法是使用map。然后你可以避免额外的数组和手动附加

a1= [{"name"=>"postgresql", "version"=>29}, {"name"=>"java", "version"=>8}, {"name"=>"python", "version"=>3}]
a2= [{"name"=>"postgresql", "release"=>"postgresql-release", "url"=>"https://github.com/repo/postgresql-release"}, {"name"=>"java", "release"=>"java-release", "url"=>"https://github.com/repo/java-release"}]

combined = a1.map do |h1|
  h2 = a2.find { |h| h1["name"] == h["name"] } || {}
  {"name"=> h1["name"], "version"=> h1["version"], "release"=> h2["release"], "url"=> h2["url"]} 
end
puts combined.to_s #=> [{"name"=>"postgresql", "version"=>29, "release"=>"postgresql-release", "url"=>"https://github.com/repo/postgresql-release"}, {"name"=>"java", "version"=>8, "release"=>"java-release", "url"=>"https://github.com/repo/java-release"}, {"name"=>"python", "version"=>3, "release"=>nil, "url"=>nil}]

另一个可能的改进是使用合并。

combined = a1.map do |h1|
  h2 = a2.find { |h| h1["name"] == h["name"] } || { "release" => nil, "url" => nil}
  h1.merge(h2)
end

Answer 2

你可以试试这个代码：

combined = h1
  .concat(h2)
  .group_by { |a| a['name'] }
  .values
  .map do |a|
    res = a.first.merge(a.last)
    res['release'] ||= nil
    res['url'] ||= nil
    res
  end

如果您不确定代码在做什么，请尝试逐步执行。

Answer 3

可能有很多方法可以做到这一点，但是由于您的内部哈希本质上具有一个键，因此首先想到的是将您的哈希数组转换为哈希的哈希，然后只需使用 Hash#deep_merge 即可正是这种东西。

hash1 = h1.index_by { |inner_hash| inner_hash['name'] }
hash2 = h2.index_by { |inner_hash| inner_hash['name'] }
combined_hash = hash1.deep_merge(hash2)
combined = combined_hash.values

这给出了除了 python nil URL 之外的预期输出。如果您无法在数据中修复该问题，要复制该结果，请尝试以下操作：

# ... as before
defaults = {'url' => nil, 'version' => nil}
combined_hash = combined_hash.transform_values { |v| defaults.merge(v) }
combined = combined_hash.values

如果您需要将默认值作为可能更改的数据中的确切键：

keys = (hash1.keys + hash2.keys).uniq
defaults = keys.map { |k| [k, nil] }.to_h