或

Question

我有一个数组：

[[1,"Location1"],[1,"Location2"],[1,"Location3"],[2,"Location4"],[2,"Location5"],[2,"Location6"]]

如何映射此数组以便我得到：

[[1,["Location1", "Location2", "Location3"]],[2,["Location4", "Location5", "Location6"]]]

Answer 1

执行以下操作： -

array = [[1,"Location1"],[1,"Location2"],[1,"Location3"],[2,"Location4"],[2,"Location5"],[2,"Location6"]]
array.group_by(&:first).map { |k, v_ary| [k, v_ary.map(&:last)] }
# => [[1, ["Location1", "Location2", "Location3"]], [2, ["Location4", "Location5", "Location6"]]]

或

array.each_with_object(Hash.new { |hsh, key| hsh[key] = [] }) { |(f,l), h| h[f] << l }.to_a
# => [[1, ["Location1", "Location2", "Location3"]], [2, ["Location4", "Location5", "Location6"]]]

Answer 2

[[1,"Location1"],[1,"Location2"],[1,"Location3"],[2,"Location4"],[2,"Location5"],[2,"Location6"]].each_with_object({}) do |element, array|
  array[element[0]] ||= []
  array[element[0]].push(array[element[1]])
end.to_a

Answer 3

代码或多或少会成为一个创建新数组的循环步行器，所以让它成为一个函数

def reOrgArr(arr)
  i = 0
  j = [[]]
  loop do
    group = arr[i]
    name = group[1]
    newgroup = group[0]
    j[newgroup].push(name)
    i += 1
    if i > arr.len break  # this will cause execution to exit the loop
  end
  return j
end

arr = [[1,"Location1"],[1,"Location2"],[1,"Location3"],[2,"Location4"],[2,"Location5"],[2,"Location6"]]
 newArr = reOrgArr(arr)

Answer 4

您可以执行以下操作（请注意，它会更改集合中的原始项目）：

arr.group_by(&:shift).transform_values(&:flatten).to_a

分步说明：

给定arr

[[1,"Location1"],[1,"Location2"],[1,"Location3"],[2,"Location4"],[2,"Location5"],[2,"Location6"]]

.group_by(&:shift)按第一个元素分组，将其从每个项目中删除（对集合中的项目进行突变）

{1=>[["Location1"], ["Location2"], ["Location3"]], 2=>[["Location4"], ["Location5"], ["Location6"]]}

.transform_values(&:flatten)嵌套数组

{1=>["Location1", "Location2", "Location3"], 2=>["Location4", "Location5", "Location6"]}

.to_a将哈希转换为数组

[[1, ["Location1", "Location2", "Location3"]], [2, ["Location4", "Location5", "Location6"]]]

将Ruby中的嵌套数组转换为具有相同id的项目

4 个答案:

或