javascript - 使用 jq 将元素数组分解为多个不同长度的数组

Question

我有一个 JSON 对象，我想使用 jq 将其从一种形式转换为另一种形式(当然，我可以使用 javascript 或 python 进行迭代，但 jq 会更好)。问题在于，输入包含长数组，每当数据在第一个数组中停止重复时，就需要将其分解为多个较小的数组。我不太确定如何描述这个问题，所以我只是在这里举一个例子，希望能更具解释性。一个安全的假设(如果有任何帮助的话)是输入数据始终在前两个元素上进行预排序(例如“row_x”和“col_y”):

输入:

{
  "headers": [ "col1", "col2", "col3" ],
  "data": [
    [ "row1","col1","b","src2" ],
    [ "row1","col1","b","src1" ],
    [ "row1","col1","b","src3" ],
    [ "row1","col2","d","src4" ],
    [ "row1","col2","e","src5" ],
    [ "row1","col2","f","src6" ],
    [ "row1","col3","j","src7" ],
    [ "row1","col3","g","src8" ],
    [ "row1","col3","h","src9" ],
    [ "row1","col3","i","src10" ],
    [ "row2","col1","l","src13" ],
    [ "row2","col1","j","src11" ],
    [ "row2","col1","k","src12" ],
    [ "row2","col3","o","src15" ]
  ]
}

期望的输出:

{
  "headers": [ "col1", "col2", "col3" ],
  "values": [
    [["b","b","b"],["d","e","f"],["g","h","i","j"]],
    [["j","k","l"],null,["o"]]
  ],
  "sources": [
    [["src1","src2","src3"],["src4","src5","src6"],["src7","src8","src9","src10"]],
    [["src11","src12","src13"],null,["src15"]]
  ]
}

这在 jq 中可行吗？

更新:它的一个变体是保留原始数据顺序，因此输出如下:

{
  "headers": [ "col1", "col2", "col3" ],
  "values": [
    [["b","b","b"],["d","e","f"],["j","g","h","i"]],
    [["l","j","k"],null,["o"]]
  ],
  "sources": [
    [["src2","src1","src3"],["src4","src5","src6"],["src7","src8","src9","src10"]],
    [["src13","src11","src12"],null,["src15"]]
  ]
}

Answer 1

可行吗？当然!

首先，您需要按行然后按列对数据进行分组。然后通过组构建您的值/源数组。

.headers as $headers | .data
    # make the data easier to access
    | map({ row: .[0], col: .[1], val: .[2], src: .[3] })
    # keep it sorted so they are in expected order in the end
    | sort_by([.row,.col,.src])
    # group by rows
    | group_by(.row)
    # create a map to each of the cols for easier access
    | map(group_by(.col)
        | reduce .[] as $col ({};
            .[$col[0].col] = [$col[] | {val,src}]
        )
    )
    # build the result
    | {
        headers: $headers,
        values: map([.[$headers[]] | [.[]?.val]]),
        sources: map([.[$headers[]] | [.[]?.src]])
    }

这将产生以下结果:

{
  "headers": [ "col1", "col2", "col3" ],
  "values": [
    [
      [ "b", "b", "b" ],
      [ "d", "e", "f" ],
      [ "i", "j", "g", "h" ]
    ],
    [
      [ "j", "k", "l" ],
      [],
      [ "o" ]
    ]
  ],
  "sources": [
    [
      [ "src1", "src2", "src3" ],
      [ "src4", "src5", "src6" ],
      [ "src10", "src7", "src8", "src9" ]
    ],
    [
      [ "src11", "src12", "src13" ],
      [],
      [ "src15" ]
    ]
  ]
}