使用多个复杂对象对数组项进行分组

Question

我正在尝试重新组合一个复杂的对象数组。

这是我的数组：

[ 
  { scenario: "Treasury", diagnostic: "good results", action: "Manage Financial Recovery"},
  { scenario: "Treasury", diagnostic: "good results", action: "Analyze the impact of your investments"},
  { scenario: "Treasury", diagnostic: "Significant decline", action: "Ensure an adequate"},
  { scenario: "Treasury", diagnostic: "Significant decline", action: "Pilot your cash"},
  { scenario: "Turnover", diagnostic: "Improve trade efficiency of your business", action: "Valorize your labels"},
  { scenario: "Turnover", diagnostic: "Improve trade efficiency of your business", action: "Analyze the opportunity"}
  { scenario: "Turnover", diagnostic: "Improve trade efficiency of your firm", action: "Contacter un prestataire"}
];

我想将上面的数组统一到这个：

[ 
  { 
    scenario: "Treasury",
    diagnostics: [
        {
            diagnostic : "good results",
            actions: [
                "Manage Financial Recovery",
                "Analyze the impact of your investments"
            ]
        }
        {
            diagnostic : "Significant decline",
            actions: [
                "Ensure an adequate",
                "Pilot your cash"
            ]
        }
    ]
  },
  { 
    scenario: "Turnover",
    diagnostics: [
        {
            diagnostic : "Improve trade efficiency of your business",
            actions: [
                "Valorize your labels",
                "Analyze the opportunity"
            ]
        }
        {
            diagnostic : "Improve trade efficiency of your firm",
            actions: [
                "Contacter un prestataire"
            ]
        }
    ]
  }
];

所以我尝试使用 JSBin 来统一我的数组，但我没有得到预期的结果，那么获得没有重复对象的数组的最有效方法是什么。

Answer 1

您可以使用迭代方法并为 key 的分组项目使用辅助对象。

function getGrouped(array, keys, groupCB, children) {
    var result = [],
        hash = { _: result };

    groupCB = groupCB || function (o) { return o; };
    children = children || [];
    array.forEach(function (a) {
        keys.reduce(function (r, k, i) {
            var o = {};
            if (!r[a[k]]) {
                r[a[k]] = { _: [] };
                o[k] = a[k];
                o[children[i] || 'children'] = r[a[k]]._;
                r._.push(o);
            }
            return r[a[k]];
        }, hash)._.push(groupCB(a));
    });
    return result;
}

var data = [{ scenario: "Treasury", diagnostic: "good results", action: "Manage Financial Recovery" }, { scenario: "Treasury", diagnostic: "good results", action: "Analyze the impact of your investments" }, { scenario: "Treasury", diagnostic: "Significant decline", action: "Ensure an adequate" }, { scenario: "Treasury", diagnostic: "Significant decline", action: "Pilot your cash" }, { scenario: "Turnover", diagnostic: "Improve trade efficiency of your business", action: "Valorize your labels" }, { scenario: "Turnover", diagnostic: "Improve trade efficiency of your business", action: "Analyze the opportunity" }, { scenario: "Turnover", diagnostic: "Improve trade efficiency of your firm", action: "Contacter un prestataire" }],
    groupCB = function (o) { return o.action },
    keys = ['scenario', 'diagnostic'],
    children = ['diagnostics', 'actions'],
    result = getGrouped(data, keys, groupCB, children);

console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 2

您可以进行如下操作：;

var data = [ { scenario: "Treasury", diagnostic: "good results", action: "Manage Financial Recovery"},
             { scenario: "Treasury", diagnostic: "good results", action: "Analyze the impact of your investments"},
             { scenario: "Treasury", diagnostic: "Significant decline", action: "Ensure an adequate"},
             { scenario: "Treasury", diagnostic: "Significant decline", action: "Pilot your cash"},
             { scenario: "Turnover", diagnostic: "Improve trade efficiency of your business", action: "Valorize your labels"},
             { scenario: "Turnover", diagnostic: "Improve trade efficiency of your business", action: "Analyze the opportunity"},
             { scenario: "Turnover", diagnostic: "Improve trade efficiency of your firm", action: "Contacter un prestataire"}
           ],
    hash = data.reduce(function(p,c){
                         var fd = null;
                         p[c.scenario] = p[c.scenario] ? (fd = p[c.scenario].diagnostics.find(d => d.diagnostic === c.diagnostic),
                                                          fd ? (fd.action.push(c.action),p[c.scenario])
                                                             : (p[c.scenario].diagnostics.push({diagnostic: c.diagnostic, action: [c.action]}),p[c.scenario]))
                                                       : {scenario: c.scenario, diagnostics: [{diagnostic: c.diagnostic, action: [c.action]}]};
                         return p;
                       },{});
  result = Object.keys(hash).map(k => hash[k]);
console.log(JSON.stringify(result,null,4));

Answer 3

也可以考虑实现一个单独的收集器函数，该函数仅使用查找本地注入的引用，因此，在单个 reduce iteration/cycle 中应用，它已经 aggregates/creates 所需的数据结构。

确实没有必要将 reduce 循环嵌套到 forEach 迭代中，正如 Nina 的方法所证明的那样，它也大大改变了后者 thisArgs 的用法。（不仅因为它很难阅读）。

Redu 的方法使用 reduce，但是在每个迭代步骤中都会有一个 find 也是一个迭代器...最终结果取决于 jet 另外两次迭代，keys 和 map.

人类可读的示例代码...

var diagnosticData = [
  { scenario: "Treasury", diagnostic: "good results", action: "Manage Financial Recovery"},
  { scenario: "Treasury", diagnostic: "good results", action: "Analyze the impact of your investments"},
  { scenario: "Treasury", diagnostic: "Significant decline", action: "Ensure an adequate"},
  { scenario: "Treasury", diagnostic: "Significant decline", action: "Pilot your cash"},
  { scenario: "Turnover", diagnostic: "Improve trade efficiency of your business", action: "Valorize your labels"},
  { scenario: "Turnover", diagnostic: "Improve trade efficiency of your business", action: "Analyze the opportunity"},
  { scenario: "Turnover", diagnostic: "Improve trade efficiency of your firm", action: "Contacter un prestataire"}
];


function collectGroupedScenarioDiagnostics(collector, diagnosticItem/*, idx, list*/) {
  var
    scenarioKey           = diagnosticItem.scenario,
    diagnosticKey         = diagnosticItem.diagnostic,

    groupedDiagnosticKey  = [scenarioKey, diagnosticKey].join(' : '),

    scenarioItem          = collector.scenarioStore[scenarioKey],
    groupedDiagnosticItem = collector.diagnosticStore[groupedDiagnosticKey];

  if (!scenarioItem) {
    scenarioItem          = collector.scenarioStore[scenarioKey] = {

      scenario    : scenarioKey,
      diagnostics : []
    };
    collector.diagnosticsList.push(scenarioItem);
  }
  if (!groupedDiagnosticItem) {
    groupedDiagnosticItem = collector.diagnosticStore[groupedDiagnosticKey] = {

      diagnostic: diagnosticKey,
      actions   : []
    };
    scenarioItem.diagnostics.push(groupedDiagnosticItem);
  }
  groupedDiagnosticItem.actions.push(diagnosticItem.action);

  return collector;
}


var groupedScenarioDiagnostics = diagnosticData.reduce(collectGroupedScenarioDiagnostics, {

  scenarioStore   : {},
  diagnosticStore : {},
  diagnosticsList : []

}).diagnosticsList;


console.log('groupedScenarioDiagnostics : ', groupedScenarioDiagnostics);

使用多个复杂对象对数组项进行分组

Group array items using multiple and complexe object

javascript

arrays

reduce

group-by

data-structures