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具有不可变js的递归分组

immutable.js grouping recursion javascript

rkralston · 技术社区 · 6 年前

var list = new List([
  { "col1": "1", "col2": "A", "col3": "cat", "col4": "dog", "col5": "blue"},
  { "col1": "1", "col2": "A", "col3": "cat", "col4": "dog", "col5": "blue"},
  { "col1": "1", "col2": "A", "col3": "bird", "col4": "dog", "col5": "blue"},
  { "col1": "1", "col2": "B", "col3": "cat", "col4": "dog", "col5": "blue"},
  { "col1": "1", "col2": "B", "col3": "cat", "col4": "dog", "col5": "blue"},
  { "col1": "1", "col2": "B", "col3": "bird", "col4": "dog", "col5": "blue"},
  { "col1": "1", "col2": "C", "col3": "cat", "col4": "dog", "col5": "blue"},
  { "col1": "1", "col2": "C", "col3": "bird", "col4": "dog", "col5": "blue"},
  { "col1": "1", "col2": "C", "col3": "cat", "col4": "dog", "col5": "blue"},
  { "col1": "2", "col2": "A", "col3": "cat", "col4": "dog", "col5": "blue"},
  { "col1": "2", "col2": "A", "col3": "bird", "col4": "dog", "col5": "blue"},
  { "col1": "2", "col2": "A", "col3": "cat", "col4": "dog", "col5": "blue"},
  { "col1": "2", "col2": "B", "col3": "cat", "col4": "dog", "col5": "blue"},
  { "col1": "2", "col2": "B", "col3": "bird", "col4": "dog", "col5": "blue"},
  { "col1": "2", "col2": "B", "col3": "cat", "col4": "dog", "col5": "blue"},
  { "col1": "2", "col2": "C", "col3": "cat", "col4": "dog", "col5": "blue"},
  { "col1": "2", "col2": "C", "col3": "bird", "col4": "dog", "col5": "blue"},
  { "col1": "2", "col2": "C", "col3": "cat", "col4": "dog", "col5": "blue"},
  { "col1": "3", "col2": "A", "col3": "cat", "col4": "dog", "col5": "blue"},
  { "col1": "3", "col2": "A", "col3": "cat", "col4": "dog", "col5": "blue"},
  { "col1": "3", "col2": "A", "col3": "bird", "col4": "dog", "col5": "blue"},
  { "col1": "3", "col2": "B", "col3": "cat", "col4": "dog", "col5": "blue"},
  { "col1": "3", "col2": "B", "col3": "cat", "col4": "dog", "col5": "blue"},
  { "col1": "3", "col2": "B", "col3": "bird", "col4": "dog", "col5": "blue"},
  { "col1": "3", "col2": "C", "col3": "cat", "col4": "dog", "col5": "blue"},
  { "col1": "3", "col2": "C", "col3": "cat", "col4": "dog", "col5": "blue"},
  { "col1": "3", "col2": "C", "col3": "bird", "col4": "dog", "col5": "blue"},
  { "col1": "4", "col2": "A", "col3": "cat", "col4": "dog", "col5": "blue"}
  ]);

let groupedData = list.groupBy(x => x['col1'])
  .map(row => row.groupBy(x => x['col2'])
       .map(row => row.groupBy(x => x['col3']))
   );

我该如何递归地对任何不

2 回复 | 直到 6 年前

rkralston 6 年前

一个递归分组的测试答案不可变.js表示数据表的列表对象:

recursiveGrouping = (columns, list) => {
    columns.forEach(column => {
        list = this.processIterable(column, list);
    })
    return list;
}

processIterable = (column, iterable) => {
    if(List.isList(iterable)) return iterable.groupBy(row => row[column]);
    else if(Map.isMap(iterable)) return iterable.map(value => this.processIterable(column, value));
    else return iterable;
}

基于以上示例,我将使用以下方法进行此调用:

let grouped data = this.recursiveGrouping(['col1', 'col2', 'col3'], list);

这将返回与手动调用子分组相同的分组。

×××¢× ××¨×§× 6 年前

一种方法是只分配第一个 groupBy 呼叫 groupedData ,然后每次分配 groupBy公司 呼叫你所在级别的每个单元格。这将改变初始结构的内部结构 分组数据

类似(未经测试):

function dfs(row, i, n){
  if (i > n)
    return;

  for (let j=0; j<row.length; j++){
    row[j] = row.groupBy(x => x['col' + i]);
    dfs(row[j], i + 1, n);
  }
}

let n = ???

let groupedData = list.groupBy(x => x['col1']);

for (row of groupedData)
  dfs(row, 2, n);