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使用特定属性标记和删除对象数组中的重复项以进行比较

unique ruby arrays

Luke · 技术社区 · 10 年前

我有一个如下所示的对象数组:

[
  {
    "field name" => "Account number",
    "data type" => "number",
    "mneumonic" => "ACTNUM",
    "field number" => "027"
  },
  {
    "field name" => "Warning",
    "data type" => "code",
    "mneumonic" => "WARN1",
    "field number" => "034:01"
  },
  {
    "field name" => "Warning",
    "data type" => "code",
    "mneumonic" => "WARN2",
    "field number" => "034:02"
  },
  .....
]

我需要搜索数组并根据 "field name" 所有物为此,我可以使用类似的方法 uniq { |i| i["field name"] }

但是,对于找到的任何重复项,最终未删除的项需要添加一个属性: multiple => true 。我不在乎哪一个对象最终是留在数组中的对象,只要它标记有此属性即可。因此,在上面的示例中运行函数可能会产生:

[
  {
    "field name" => "Account number",
    "data type" => "number",
    "mneumonic" => "ACTNUM",
    "field number" => "027",
  },
  {
    "field name" => "Warning",
    "data type" => "code",
    "mneumonic" => "WARN1",
    "field number" => "034:01",
    "multiple" => true
  },

  .....
]

除了删除重复项之外,我还需要确保数组的顺序不受函数的影响。

解决这个问题的最佳方法是什么?

5 回复 | 直到 10 年前

sawa 10 年前

使用此阵列:

a = [
  {
    "field name" => "Account number",
    "data type" => "number",
    "mneumonic" => "ACTNUM",
    "field number" => "027",
  },
  {
    "field name" => "Warning",
    "data type" => "code",
    "mneumonic" => "WARN1",
    "field number" => "034:01",
  },
  {
    "field name" => "Warning",
    "data type" => "code",
    "mneumonic" => "WARN2",
    "field number" => "034:02",
  },
]

此代码:

file_names = {}
a.select do
  |h| k = h["field name"]
  if file_names[k]
    file_names[k]["multiple"] = true
    false
  else
    file_names[k] = h
    true
  end
end

将给出:

[
  {
    "field name"   => "Account number",
    "data type"    => "number",
    "mneumonic"    => "ACTNUM",
    "field number" => "027"
  },
  {
    "field name"   => "Warning",
    "data type"    => "code",
    "mneumonic"    => "WARN1",
    "field number" => "034:01",
    "multiple"     => true
  }
]

Piotr Kruczek Andrey Ivanov 10 年前

这里有一个非常简单的解决方案:

array # => your array of objects
used_names = []
multiple_names = []
array.each do |hash|
  name = hash['field name']
  if used_names.include? name
    multiple_names << name
    array.delete hash
  else
    used_names << name
  end
end
array.each do |hash|
  if multiple_names.include? hash['field name']
    hash['multiple'] = true
  end
end

vikram7 10 年前

这个版本只计算“域名”出现的次数,如果它大于或不大于1,它会根据需要更新哈希。

field_name_counts = Hash.new 0

array.each do |hash|
  field_name = hash["field name"]
  field_name_counts[field_name] += 1
end

array.each do |hash|
  field_name = hash["field name"]
  if field_name_counts[field_name] > 1
    hash["multiple"] = true
  else
    hash["multiple"] = false
  end
end

Matt Brictson 10 年前

此解决方案构建排除重复项的新阵列。对于原始数组中的每个项,它检查是否存在 existing 已看到同名的项目。如果是,则将该现有项标记为 existing["multiple"] = true 并跳过该迭代。

这具有省略新阵列中的重复项并标记原始项的预期效果。

unique_data = data.each_with_object([]) do |item, result|
  if (existing = result.find { |i| i["field name"] == item["field name"] })
    existing["multiple"] = true
    next
  end
  result << item
end

Cary Swoveland 10 年前

如果您使用的是Ruby v1.9+(其中哈希保证保持密钥插入顺序),您可以使用以下形式 Hash#update (又名 merge! )它使用一个块来确定合并的两个哈希中存在的键的值。 a 是@sawa给出的数组。

a.each_with_object({}) do |f,g|
  g.update(f["field name"]=>f) { |_,h| h.merge("multiple"=>true) }
end.values
  #=> [{"field name"=>"Account number", "data type"=>"number",
  #     "mneumonic"=>"ACTNUM", "field number"=>"027"},
  #    {"field name"=>"Warning", "data type"=>"code", "mneumonic"=>"WARN1",
  #     "field number"=>"034:01", "multiple"=>true}]