如何在数组中迭代两次

Question

如何通过在数组中迭代两次来从数组中获取数据。例如我有一个数据集：

var data = [
    {"Fruits ":"Apples","Fresh":"12","Rotten":"5","Total":"17"},
    {"Fruits ":"Oranges","Fresh":"34","Rotten":"6","Total":"40"},
    {"Fruits ":"Strawberries","Fresh":"67","Rotten":"8","Total":"75"},
    {"Fruits ":"Bananas","Fresh":"23","Rotten":"5","Total":"28"}
]

首先，我想在右侧进行迭代并获取所有键，直到第一个对象的末尾，然后向下迭代，以便获得键的所有值，因此我得到

预期输出

 [{ 'name': 'Fresh',
    'data': [12, 34, 67, 23]
    }, 
    {
    'name': 'Rotten',
    'data': [5, 6, 8, 5]
    },
    {
    'name': 'total',
    'data': [17, 40, 75, 28]
    }]

到目前为止我试过这个：

    var categorie = []
    var seriesNames = []
    var series = []

    for(var i=0; i<data.length; i++){
      categorie.push(_.values(data[i])[0])
    }

    for(i=1; i<data.length; i++){
        seriesNames.push(_.keys(data[0])[i])
    }

但我不知道如何获取数据数组并将其与 seriesName 连接。实时复制：plunker

编辑键、值和数据长度是可变的，因为我正在处理动态数据。

Answer 1

这是一种方法：

var output = data.reduce(function(acc, cur) {
    for(var k in cur) {
        if (!cur.hasOwnProperty(k) || k === 'Fruits ') {
            continue;
        }
        var container = acc.filter(function(item) {
            return item.name === k;
        })[0];
        if (!container) {
            container = {
                name: k,
                data: []
            };
            acc.push(container);
        }
        container.data.push(cur[k])
    }
    return acc;
},[]);

Answer 2

您可以使用组数组并将其用于结果数组中的正确分组。

var data = [{ "Fruits ": "Apples", "Fresh": "12", "Rotten": "5", "Total": "17" }, { "Fruits ": "Oranges", "Fresh": "34", "Rotten": "6", "Total": "40" }, { "Fruits ": "Strawberries", "Fresh": "67", "Rotten": "8", "Total": "75" }, { "Fruits ": "Bananas", "Fresh": "23", "Rotten": "5", "Total": "28" }],
    groups = ["Fresh", "Rotten", "Total"],
    result = [];

data.forEach(function (a) {
    groups.forEach(function (b, i) {
        result[i] = result[i] || { name: b, data: [] };
        result[i].data.push(+a[b]);
    });
});

console.log(result);

只是对动态属性的提案略有改动。它只接受有限数量的属性。

var data = [{ "Fruits ": "Apples", "Fresh": "12", "Rotten": "5", "Total": "17" }, { "Fruits ": "Oranges", "Fresh": "34", "Rotten": "6", "Total": "40" }, { "Fruits ": "Strawberries", "Fresh": "67", "Rotten": "8", "Total": "75" }, { "Fruits ": "Bananas", "Fresh": "23", "Rotten": "5", "Total": "28" }],
    result = [];

data.forEach(function (a) {
    Object.keys(a).forEach(function (k) {
        if (isFinite(a[k])) {
            if (!this[k]) {
                this[k] = { name: k, data: [] };
                result.push(this[k]);
            }
            this[k].data.push(a[k]);
        }
    }, this);
}, Object.create(null));

console.log(result);

Answer 3

我建议您直接使用输出对象，稍后将它们添加到数组中：

var data = [{"Fruits ": "Apples","Fresh": "12","Rotten": "5","Total": "17"}, {"Fruits ": "Oranges","Fresh": "34","Rotten": "6","Total": "40"}, {"Fruits ": "Strawberries","Fresh": "67","Rotten": "8","Total": "75"}, {"Fruits ": "Bananas","Fresh": "23","Rotten": "5","Total": "28"}];

var fresh = {name: 'Fresh', data: []};
var rotten = {name: 'Rotten', data: []};
var total = {name: 'Total', data: []};

data.forEach(function(fruit) {
  fresh.data.push(+fruit.Fresh);
  rotten.data.push(+fruit.Rotten);
  total.data.push(+fruit.Total)
});

var output = [fresh, rotten, total];
console.log(output);

Answer 4

由于您使用的是 underscore 它实际上是对集合的一个非常简单的 map。

首先，找到对象的键。

以后用map就可以得到需要的对象属性

var keys = _.keys(data[0]).reverse()
keys = _.without(keys, keys[0]) 


var c = _.map(keys, function(k) {
  return {
    name: k,
    data: _.map(data, k)
  };
});

还有一个笨蛋：https://plnkr.co/edit/KtJPMu?p=preview

Answer 5

使用array.prototype.map:

var arr = [
{"Fruits ":"Apples","Fresh":"12","Rotten":"5","Total":"17"},
{"Fruits ":"Oranges","Fresh":"34","Rotten":"6","Total":"40"},
{"Fruits ":"Strawberries","Fresh":"67","Rotten":"8","Total":"75"},
{"Fruits ":"Bananas","Fresh":"23","Rotten":"5","Total":"28"}
];

var newArr = ["Fresh", "Rotten", "Total"].map(function(item) {
  return {
    name: item,
    data: arr.map(function(innerItem) {
      return parseInt(innerItem[item],10);
    })
  }
});

console.log(newArr);

Answer 6

您可以尝试这样的操作：

您可以有一个需要排除的键列表，并根据这个列表处理键。

var data = [{"Fruits ":"Apples","Fresh":"12","Rotten":"5","Total":"17"},
{"Fruits ":"Oranges","Fresh":"34","Rotten":"6","Total":"40"},
{"Fruits ":"Strawberries","Fresh":"67","Rotten":"8","Total":"75"},
{"Fruits ":"Bananas","Fresh":"23","Rotten":"5","Total":"28"}
]
var keys_to_exclude = ["Fruits"];
var _tmp = {};

data.forEach(function(o) {
  for (var k in o) {
    if (keys_to_exclude.indexOf(k.trim()) < 0) {
      _tmp[k] = _tmp[k] || [];
      _tmp[k].push(o[k]);
    }
  }
});

var result = Object.keys(_tmp).map(function(k) {
  return {
    name: k,
    data: _tmp[k]
  }
});

document.getElementById("r").innerHTML = JSON.stringify(result, 0, 4);

<pre id="r"></pre>

Answer 7

试试这个：

var result = [
  {'name':'Fresh', 'data':[]},
  {'name':'Rotten', 'data':[]},
  {'name':'Total', 'data':[]}
];

for (d of data) {
  result[0].data.push(d['Fresh']);
  result[1].data.push(d['Rotten']);
  result[2].data.push(d['Total']);
}

输出如下所示：

[{"name":"Fresh","data":["12","34","67","23"]},
{"name":"Rotten","data":["5","6","8","5"]},
{"name":"Total","data":["17","40","75","28"]}]

Answer 8

我会在上面推荐@nina-scholz 的回答。然而，如果有人想看看它是如何用简单的 JS 完成的，请参考下面的代码：

var data = [{"Fruits ":"Apples","Fresh":"12","Rotten":"5","Total":"17"},
{"Fruits ":"Oranges","Fresh":"34","Rotten":"6","Total":"40"},
{"Fruits ":"Strawberries","Fresh":"67","Rotten":"8","Total":"75"},
{"Fruits ":"Bananas","Fresh":"23","Rotten":"5","Total":"28"}]

//list of items that needs to be in the final list
var categorie = ["Fresh", "Rotten", "Total"];
var seriesNames = {};
var series = [];

// iterate through the initial array
for(var i=0; i<data.length; i++){
  // iterate through the category
  for(var j=0; j<categorie.length; j++) {
    if(data[i].hasOwnProperty(categorie[j])) {
        // seriesNames will hold the category name and corresponding value will be an array of values(total, rotten, fresh) from all objects
        if(seriesNames[categorie[j]]) { // array already exists
            var arr = seriesNames[categorie[j]];
            arr.push(data[i][categorie[j]]);
        } else { // create a new array
            seriesNames[categorie[j]] = new Array();
        }
    }
  }
}

// create the required output object
for(var attr in seriesNames) {
    var obj = {};
    obj['name'] = attr;
    obj['data'] = seriesNames[attr];
    series.push(obj);
}

// expected output is in series.
console.debug(series);

Answer 9

var i, j, ref, output = { Fresh: [], Rotten: [], Total: [] }; //ignore anything that isn't here.
for (i of data) {
  for (j in i) {
    if (typeof (ref = output[j]) !== 'undefined')
      ref.push(i[j]);
  }
}

output 看起来像这样 (JSON)

{
  "Fresh":["12","34","67","23"],
  "Rotten":["5","6","8","5"],
  "Total":["17","40","75","28"]
}

这不是您想要的格式，所以我们可以运行再看一遍：

var output2 = [];
for (k in output) {
  output2.push({ name: k, data: output[k] });
}

output2 现在看起来应该像您想要的那样 (JSON)

[{
  "name":"Fresh",
  "data":["12","34","67","23"]
},{
  "name":"Rotten",
  "data":["5","6","8","5"]
},{
  "name":"Total",
  "data":["17","40","75","28"]
}]

为了完善它，我们可以将它们放在一个函数中

function nameMeSomething(data) {
  var i, j, k, l, m = [], n = { Fresh: [], Rotten: [], Total: [] }; //ignore anything that isn't here.
  for (i of data) for (j in i) if (typeof (l = n[j]) !== 'undefined') l.push(i[j]);
  for (k in n) m.push({ name: k, data: m[k] });
  return m
}

Answer 10

我的解决方案获取所有具有数值（超过 PO 的条款）的键并完全链接为一行答案。

var data = [{ "Fruits ": "Apples", "Fresh": "12", "Rotten": "5", "Total": "17" }, { "Fruits ": "Oranges", "Fresh": "34", "Rotten": "6", "Total": "40" }, { "Fruits ": "Strawberries", "Fresh": "67", "Rotten": "8", "Total": "75" }, { "Fruits ": "Bananas", "Fresh": "23", "Rotten": "5", "Total": "28" }];

var result = Object.keys(data[0])
  .filter(k => !isNaN(data[0][k]))
  .map(k => ({
    name: k,
    data: data.map(d => d[k])
  }));

console.log(result);

如何在数组中迭代两次

How to iterate two times inside an array

javascript

underscore.js

angularjs