Question

我需要将git ls-remote的结果下拉到一个数组中，然后将该数组转换为这样的哈希：{commit_hash =＆gt;参考}。有时，两个提交哈希值是相同的（但可能有不同的引用）。所以我得到了这样的东西：

["19d97e408ee3f993745b053e281ac9dc69519e06","refs/heads/auto",
 "8f6f47c6e8023540b022586e368c68e1e814ce6d","refs/heads/callout_hooks",  
 "3cbdb4b2fcb85bc7f0ed08b62e2bf2445a7659e8","refs/heads/elab",
 "d38a9a26ef887c08b306bdab210b39882f58e587","refs/heads/elab_6.1",
 "19d97e408ee3f993745b053e281ac9dc69519e06","refs/heads/master",
 "906dfe6eebff832baf0f92683d751432fcc98ab7","refs/heads/regression"]

我要转换为：

{"19d97e408ee3f993745b053e281ac9dc69519e06" => "refs/heads/auto"...}

但是master和auto具有相同的哈希值，因此其中一个哈希值会在转换中被删除。

我如何1.）获取转换中丢弃的值的列表，或2.）通过向键添加特殊字符来使键唯一，如*？

Answer 1

我希望你能这样：

ary = [
       "19d97e408ee3f993745b053e281ac9dc69519e06","refs/heads/auto",
       "8f6f47c6e8023540b022586e368c68e1e814ce6d","refs/heads/callout_hooks",  
       "3cbdb4b2fcb85bc7f0ed08b62e2bf2445a7659e8","refs/heads/elab",
       "d38a9a26ef887c08b306bdab210b39882f58e587","refs/heads/elab_6.1",
       "19d97e408ee3f993745b053e281ac9dc69519e06","refs/heads/master",
       "906dfe6eebff832baf0f92683d751432fcc98ab7","refs/heads/regression"
     ]

array_hash = ary.each_slice(2).with_object(Hash.new { |h,k| h[k] = []}) do |(k,v),hash|
  hash[k] << v 
end

# the main advantage is here you wouldn't loose any data, all are with you. You can
# use it as per your need. I think it is a better approach to deal with your situation.
array_hash
# => {"19d97e408ee3f993745b053e281ac9dc69519e06"=>
#      ["refs/heads/auto", "refs/heads/master"],
#     "8f6f47c6e8023540b022586e368c68e1e814ce6d"=>["refs/heads/callout_hooks"],
#     "3cbdb4b2fcb85bc7f0ed08b62e2bf2445a7659e8"=>["refs/heads/elab"],
#     "d38a9a26ef887c08b306bdab210b39882f58e587"=>["refs/heads/elab_6.1"],
#     "906dfe6eebff832baf0f92683d751432fcc98ab7"=>["refs/heads/regression"]}

Answer 2

如果你做hash_value =＆gt;的哈希值参考数组，你会保留所有内容：

array = ["19d97e408ee3f993745b053e281ac9dc69519e06","refs/heads/auto",
 "8f6f47c6e8023540b022586e368c68e1e814ce6d","refs/heads/callout_hooks",  
 "3cbdb4b2fcb85bc7f0ed08b62e2bf2445a7659e8","refs/heads/elab",
 "d38a9a26ef887c08b306bdab210b39882f58e587","refs/heads/elab_6.1",
 "19d97e408ee3f993745b053e281ac9dc69519e06","refs/heads/master",
 "906dfe6eebff832baf0f92683d751432fcc98ab7","refs/heads/regression"
]

array.each_slice(2).reduce({}) { |h, (k, v)| (h[k] ||= []) << v; h }

看起来Arup和我的想法是一样的......

Answer 3

您为自己想做的事情提供了两个选项：

获取转化中丢弃的值列表
通过向密钥添加特殊字符使密钥唯一

我认为第二种方法是一个坏主意，原因如下：a）你必须有一种修改密钥的方法，这种方法可以使它们成为多个副本;和b）在原件和副本之间建立连接会很尴尬。而且，这简直太丑了。

我看到其他人提出了第三种可能性：更改生成的哈希的形式，以便值字符串数组。这可能对你有用，但它不是你要求的，所以我选择建立一个被删除的值的列表;即除了第一个以外的所有。

<强>代码

def create_hash_and_save_extras(arr)
  arr.each_slice(2).with_object([{},[]]) { |(k,v),(h,ex)|
    h.update({k=>v}) { |k, ov, nv| ex << {k=>nv}; ov } }
end

示例

create_hash_and_save_extras(arr) #=> [{"19d97e408ee3f993745b053e281ac9dc69519e06"=>"refs/heads/auto", # "8f6f47c6e8023540b022586e368c68e1e814ce6d"=>"refs/heads/callout_hooks", # "3cbdb4b2fcb85bc7f0ed08b62e2bf2445a7659e8"=>"refs/heads/elab", # "d38a9a26ef887c08b306bdab210b39882f58e587"=>"refs/heads/elab_6.1", # "906dfe6eebff832baf0f92683d751432fcc98ab7"=>"refs/heads/regression"}, # [{"19d97e408ee3f993745b053e281ac9dc69519e06"=>"refs/heads/master"}]]

<强>解释

Enumerable#each_slice发送给arr会返回一个枚举器：

enum1 = arr.each_slice(2) #=> #<Enumerator: [ # "19d97e408ee3f993745b053e281ac9dc69519e06", "refs/heads/auto", # "8f6f47c6e8023540b022586e368c68e1e814ce6d", "refs/heads/callout_hooks", # ... # "906dfe6eebff832baf0f92683d751432fcc98ab7", "refs/heads/regression" # ]:each_slice(2)>

Enumerator#with_object创建一个数组，由最初为空的哈希（由块变量h表示）和最初为空的数组（对于＆＃34; extras＆＃34;），表示通过块变量ex，然后将其发送到enum1以创建另一个枚举器（您可以将其视为＆＃34;复合枚举器＆＃34; - 请注意对{{1}的引用下面）。

each_slice(2)>:with_object({})

我们可以将enum2 = enum1.with_object([{},[]]) #=> #<Enumerator: #<Enumerator: [ # "19d97e408ee3f993745b053e281ac9dc69519e06", "refs/heads/auto", # "8f6f47c6e8023540b022586e368c68e1e814ce6d", "refs/heads/callout_hooks", # ... # "906dfe6eebff832baf0f92683d751432fcc98ab7", "refs/heads/regression" # ]:each_slice(2)>:with_object([{},[])>转换为数组，以查看它将传递到其块中的内容：

enum2

enum2.to_a #=> [[["19d97e408ee3f993745b053e281ac9dc69519e06", "refs/heads/auto"], # [{}, []]], # [["8f6f47c6e8023540b022586e368c68e1e814ce6d", "refs/heads/callout_hooks"], # [{}, []]], # [["3cbdb4b2fcb85bc7f0ed08b62e2bf2445a7659e8", "refs/heads/elab"], # [{}, []]], # [["d38a9a26ef887c08b306bdab210b39882f58e587", "refs/heads/elab_6.1"], # [{}, []]], # [["19d97e408ee3f993745b053e281ac9dc69519e06", "refs/heads/master"], # [{}, []]], # [["906dfe6eebff832baf0f92683d751432fcc98ab7", "refs/heads/regression"], # [{}, []]],传入其块的第一个元素是

enum2

因此，块变量分配如下：

[["19d97e408ee3f993745b053e281ac9dc69519e06", "refs/heads/auto"], [{}, []]]]]

我们现在使用Hash#update（又名k => "19d97e408ee3f993745b053e281ac9dc69519e06" v => "refs/heads/auto" h => {} ex = []）将Hash#merge!合并到{k,v}（h最初为空。）因此

h

变为

h.update({k=>v}) { |k, ov, nv| extras << {k=>nv}; ov }

后面是块

h.update({"19d97e408ee3f993745b053e281ac9dc69519e06"=>"refs/heads/auto"})

但该块仅适用于散列合并散列（{ |k, ov, nv| ex << {k=>nv}; ov }）和合并散列（h＆＃39; s参数）共享相同的键update，其中case k和ov分别是与nv和正在合并的哈希的键相关联的值。键h的合并值将是块返回的值。是的，当我们遇到重复时，这将适用。

所以现在

k

我们以这种方式继续h #=> {"19d97e408ee3f993745b053e281ac9dc69519e06"=>"refs/heads/auto"}的每个其他元素。当我们遇到

enum2

我们发现k = "19d97e408ee3f993745b053e281ac9dc69519e06" v = "refs/heads/master" h = {"19d97e408ee3f993745b053e281ac9dc69519e06"=>"refs/heads/auto", "8f6f47c6e8023540b022586e368c68e1e814ce6d"=>"refs/heads/callout_hooks", "3cbdb4b2fcb85bc7f0ed08b62e2bf2445a7659e8"=>"refs/heads/elab", "d38a9a26ef887c08b306bdab210b39882f58e587"=>"refs/heads/elab_6.1"}已经在合并散列k中，因此会对该块进行评估，以确定合并散列h中k的值。我们希望保留当前值h，即h[k]，这就是块返回的内容。但是，首先，我们将（仍为空的）数组ov附加到重复值，表示为散列。

ex

ruby转换数组以散列保存重复键

3 个答案: