Question

我有两个数组

arr1 = [[["name", "Alex"],["age", "4"], ["width", "55"], ["weight", "30"], ["species", "Alex"], ["female", "yes"], ["group"]], [["name", "All"],["age", "7"], ["width", "26"], ["weight", "3"], ["species", "cat"], ["female", "no"], ["group"]]]

arr2 = [["A23", "All", "Katy", "Max"], ["B23", "Sisi", "Alex"]]

如果名称，例如。 arr2的值Alex与arr1中的值Alex类似，arr2中的值B23被推送到子阵列[“group”]到第一个数组。所以我想得到

arr1 = [[["name", "Alex"],["age", "4"], ["width", "55"], ["weight", "30"], ["species", "Alex"], ["female", "yes"], ["group", "B23"]], [["name", "All"],["age", "7"], ["width", "26"], ["weight", "3"], ["species", "cat"], ["female", "no"], ["group", "A23"]]]

如何比较和组合这些数组？

Answer 1

您没有提供任何代码，因此我不会写出完整的解决方案。正如@EddeAlmeida评论的那样，使用哈希数组会更容易。这是一个将数据快速转换的基本结构：

require 'pp'

arr1 = [[["name", "Alex"],["age", "4"], ["width", "55"], ["weight", "30"], ["species", "Alex"], ["female", "yes"], ["group"]], [["name", "All"],["age", "7"], ["width", "26"], ["weight", "3"], ["species", "cat"], ["female", "no"], ["group"]]]

arr2 = [["A23", "All", "Katy", "Max"], ["B23", "Sisi", "Alex"]]

data = arr1.map { |a| a.tap { |x| x.last[1] = '' }.to_h }

pp data
# [{"name"=>"Alex",
#   "age"=>"4",
#   "width"=>"55",
#   "weight"=>"30",
#   "species"=>"Alex",
#   "female"=>"yes",
#   "group"=>""},
#  {"name"=>"All",
#   "age"=>"7",
#   "width"=>"26",
#   "weight"=>"3",
#   "species"=>"cat",
#   "female"=>"no",
#   "group"=>""}]

arr2.each do |code, *names|
  # add some logic here
end

# Coming back to (weird) nested arrays :
p data.map{|h| h.to_a}

Answer 2

我这样做了：

arr1.each do |data|
    a = data[0][1]
    b = ""
    arr2.each do |x|
        if (x.include?(a)) then
            b << "#{x[0]} "
        end
    end
data[6].push(b)

端

Answer 3

arr1 = [
        [["name", "Alex"],["age", "4"], ["width", "55"], ["weight", "30"],
         ["species", "Alex"], ["female", "yes"], ["group"]],
        [["name", "All"],["age", "7"], ["width", "26"], ["weight", "3"],
         ["species", "cat"], ["female", "no"], ["group"]]
       ]

arr2 = [["A23", "All", "Katy", "Max"], ["B23", "Sisi", "Alex"]]

这是一种允许arr1的每个元素的元素以任何顺序排列的方式，并在包含"group"的元素被修改后保持该顺序。

h2 = arr2.each_with_object({}) { |(first,*rest),h| rest.each { |s| h[s] = first } }
  #=> {"All"=>"A23", "Katy"=>"A23", "Max"=>"A23", "Sisi"=>"B23", "Alex"=>"B23"}

arr1.map do |a|
  h = a.each_with_object({}) { |(k,v),h| h[k]=v }
  h["group"] = h2[h["name"]]
  h.to_a
end
  #=> [[["name", "Alex"], ["age", "4"], ["width", "55"], ["weight", "30"],
  #     ["species", "Alex"], ["female", "yes"], ["group", "B23"]],
  #    [["name", "All"], ["age", "7"], ["width", "26"], ["weight", "3"],
  #     ["species", "cat"], ["female", "no"], ["group", "A23"]]]

如果Array#map不发生变异，请使用arr1。

arr1的两个元素中的第一个元素的块计算如下：

a = arr1.first
  #=> [["name", "Alex"], ["age", "4"], ["width", "55"], ["weight", "30"],
  #    ["species", "Alex"], ["female", "yes"], ["group"]]
h = a.each_with_object({}) { |(k,v),h| h[k]=v }
  #=> {"name"=>"Alex", "age"=>"4", "width"=>"55", "weight"=>"30",
  #    "species"=>"Alex", "female"=>"yes", "group"=>nil}
v = h["name"]
  #=> "Alex"
h["group"] = h2[v]
  #=> h2["Alex"] => "B23" 
h #=> {"name"=>"Alex", "age"=>"4", "width"=>"55", "weight"=>"30",
  #    "species"=>"Alex", "female"=>"yes", "group"=>"B23"} 
h.to_a
  #=> [["name", "Alex"], ["age", "4"], ["width", "55"], ["weight", "30"],
  #    ["species", "Alex"], ["female", "yes"], ["group", "B23"]]

arr1的第二个元素的计算类似。

比较两个数组并将它们组合起来

3 个答案: