在 Ruby 中识别和更新具有索引值的数组中的重复项

一种方法是使用 Enumerable#tally 方法按值存储总计数的哈希值。

array = ["Ballerina", "Lagoon", "Black", "Space", "Golden", "Lagoon"]

# Generates hash for faster lookup
# {"Ballerina"=>1, "Lagoon"=>2, "Black"=>1, "Space"=>1, "Golden"=>1}
array_tally = array.tally

new_array = array.map.with_index do |value, index| 
    value_duplicated = array_tally[value] > 1
    
    if value_duplicated 
        "#{value} #{index}"
    else
       value 
    end
end

puts new_array.inspect
# output: ["Ballerina", "Lagoon 1", "Black", "Space", "Golden", "Lagoon 5"]

输入

array = %w[Ballerina Lagoon Black Space Golden Lagoon]

法典

# create a hash of element counts

counts = array.each_with_object(Hash.new(0)) { |element, counts| counts[element] += 1 }
# iterate through array and append index to duplicates
array.each_with_index do |element, index|
  if counts[element] > 1
    array[index] = "#{element} #{index}"
  end
end

p array

输出

#=> ["Ballerina", "Lagoon 1", "Black", "Space", "Golden", "Lagoon 5"]

我会做：

arr = %w[Ballerina Lagoon Black Space Golden Lagoon]

cnt=arr.tally
arr2=arr.map.with_index{|e, i| cnt[e]>1 ? "#{e} #{i}" : e }

p arr2
# ["Ballerina", "Lagoon 1", "Black", "Space", "Golden", "Lagoon 5"]

此答案适用于数组较大且重复项数量相对较少的情况，因为建议的计算只需要通过数组进行一次传递。

请注意，我并不是说它一定比第一步使用 Enumerable#tally 方法的方法更快，只是它只需要通过数组进行一次传递。

---

假设数组如下所示。

arr = ["Ballerina", "Space", "Lagoon", "Black", "Space", "Golden",
       "Lagoon", "Space"]

---

我们首先构造一个由索引组成的数组，对于索引的每个元素，要以所需的方式修改索引。indicesarrarr[i]iindices

indices = arr.each_with_index.
              with_object(Hash.new { |h,k| h[k] = [] }) do |(s,i),h|
                h[s] << i
              end.
              values.
              select { |a| a.size > 1 }.
              flatten
  #=> [1, 4, 7, 2, 6]

然后，我们根据需要通过遍历来修改 的副本。arrindices

indices.each_with_object(arr.dup) { |i,a| a[i] = "%s_%i" % [a[i], i] }
  #=> ["Ballerina", "Space_1", "Lagoon_2", "Black", "Space_4", "Golden",
  #    "Lagoon_6", "Space_7"]

---

通过检查 Kernel#dup 的源代码，不出所料地发现，该方法（用 C 语言实现）复制了一个内存块，因此不构成 元素的第二个枚举。arr

---

计算步骤如下。indices

h = arr.each_with_index.
        with_object(Hash.new { |h,k| h[k] = [] }) do |(s,i),h|
          h[s] << i
        end
  #=> {"Ballerina"=>[0], "Space"=>[1, 4, 7], "Lagoon"=>[2, 6],
       "Black"=>[3], "Golden"=>[5]}

a = h.values
  #=> [[0], [1, 4, 7], [2, 6], [3], [5]]

b = a.select { |a| a.size > 1 }
  #=> [[1, 4, 7], [2, 6]]

b.flatten
  #=> [1, 4, 7, 2, 6]