Rails：循环比较两个数组并将匹配推送到新数组

Question

我有两个数组需要相互检查。找到匹配项后，我需要取消第一个数组中的值并将该值推送到新数组。

这是一个例子：

第一个数组import_data

["mario", "blue", "SGC", "CJ672PA0"], 
["solid snake", "green", "NID", "VI6965KD"], 
["samus", "maroon", "TRUST", "DNYU6539"], 
["deckard cain", "purple", "JAFA", "SJW252MZ"],
["wedge", "yellow", "WALTER", "UJ28NVM1"]

第二个数组current_data

["gordon", "orange", "DRONE", "OGJR8C0D"], 
["samus", "black", "TRUST", "DNYU6539"], 
["commander shepard", "red", "WHX", "TH985OI3"]

请注意，"samus" 出现在两个数组中，数据完全相同，但颜色有所变化。

我的意图 是遍历 import_data 并检查数组中每个项目的最后一个索引（8 位数字 "id"）并进行比较反对 current_data。每次找到 "match"（基于 8 位 "id"）我想将该项目推入一个新数组 update_data 然后我想 unset 匹配在 import_data 数组中。

问题是两个数组的长度不同。因此，一旦我的循环超出一项的长度，它就无法再检查导致错误的索引。

我的代码基本上是这样的：

import_data.each_with_index do |data, index|
  if data[3] == current_data[index].id
    # @update_data << somehow push all of the data
  end
end

这不起作用的原因是 import_data 的长度比 current_data 的长度长。这导致错误：

"undefined method `id' for nil:NilClass"

理想情况下最终结果如下：

"samus" 已从 import_data 数组中删除并推入 update_data 数组

import_data 循环后的数组

["mario", "blue", "SGC", "CJ672PA0"], 
["solid snake", "green", "NID", "VI6965KD"], 
--- 
["deckard cain", "purple", "JAFA", "SJW252MZ"],
["wedge", "yellow", "WALTER", "UJ28NVM1"]

新填充的数组update_data

 ["samus", "maroon", "TRUST", "DNYU6539"]

Answer 1

完成您正在做的事情的一种方法：

import_data = [ 
                ["mario", "blue", "SGC", "CJ672PA0"], 
                ["solid snake", "green", "NID", "VI6965KD"], 
                ["samus", "maroon", "TRUST", "DNYU6539"], 
                ["deckard cain", "purple", "JAFA", "SJW252MZ"],
                ["wedge", "yellow", "WALTER", "UJ28NVM1"]
              ]

current_data = [
                 ["gordon", "orange", "DRONE", "OGJR8C0D"], 
                 ["samus", "black", "TRUST", "DNYU6539"], 
                 ["commander shepard", "red", "WHX", "TH985OI3"]
                ]

找到两个数组的键：

import_data_keys = import_data.map { |x| x[3] }
=> ["CJ672PA0", "VI6965KD", "DNYU6539", "SJW252MZ", "UJ28NVM1"]
current_data_keys = current_data.map { |x| x[3] }
=> ["OGJR8C0D", "DNYU6539", "TH985OI3"]

通过将这两个数组相交来找到公共键：

common_keys = import_data_keys & current_data_keys
=> ["DNYU6539"]

使用 common_keys 从 import_data 删除并推送到 update_data。

common_keys.each do |key|
  import_data.each_with_index do |data, index|
    if data[3] == key
      update_data << import_data[index]
      import_data.delete_at(index)
      break
    end
  end
end

注意：break 语句用于在找到并处理元素后停止迭代。

Answer 2

由于 Prakash 的答案有效，我认为这更直接：

import_data = [
  ["mario", "blue", "SGC", "CJ672PA0"], 
  ["solid snake", "green", "NID", "VI6965KD"], 
  ["samus", "maroon", "TRUST", "DNYU6539"], 
  ["deckard cain", "purple", "JAFA", "SJW252MZ"],
  ["wedge", "yellow", "WALTER", "UJ28NVM1"]
]
current_data = [
  ["gordon", "orange", "DRONE", "OGJR8C0D"], 
  ["samus", "black", "TRUST", "DNYU6539"], 
  ["commander shepard", "red", "WHX", "TH985OI3"]
]

current_ids = current_data.map {|c| c[3]}

update_data = import_data.inject([]) do |result, data|
  result << import_data.delete(data) if current_ids.include?(data[3])
  result
end

在运行之后你得到：

update_data = [
  ["samus", "maroon", "TRUST", "DNYU6539"]
]
import_data = [
  ["mario", "blue", "SGC", "CJ672PA0"], 
  ["solid snake", "green", "NID", "VI6965KD"], 
  ["deckard cain", "purple", "JAFA", "SJW252MZ"],
  ["wedge", "yellow", "WALTER", "UJ28NVM1"]
]

希望对您有所帮助。

Rails：循环比较两个数组并将匹配推送到新数组

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