我有两个数组：
var columns = ["Date", "Number", "Size", "Location", "Age"];
var rows = [
["2001", "5", "Big", "Sydney", "25"],
["2005", "2", "Med", "Melbourne", "50"],
["2012", "20", "Huge", "Brisbane", "80"]
];
我正在尝试将它们组合成一个 javascript 对象，用于 rows 数组中的每个项目。之后，我想将每个对象推入一个新数组中。
喜欢：
var newarray = [];
//'thing' should be the same structure for each row item
var thing = {
"Date": "2001",
"Number": "5",
"Size": "Big",
"Location": "Sydney",
"Age": "25"
}
newarray.push(thing);
当我知道列名时我可以这样做，但是当列名未知时我需要能够以这种方式存储数据 - 即基于列数组的索引。
我之前试过这样：
for (var y = 0; y < rows.length; y++) {
for (var i = 0; i < columns.length; i++) {
thing[columns[i]] = rows[i][i];
}
newarray.push(thing)
}
上面的代码仅一次又一次地存储第一项（根据 rows.length）。
我不明白如何将列名与行组合起来创建对象数组。 “行”包含数组这一事实尤其令人困惑..
您也可以以更加以数据为中心的方式来做到这一点：
const columns = ["Date", "Number", "Size", "Location", "Age"]
const rows = [
["2001", "5", "Big", "Sydney", "25"],
["2005", "2", "Med", "Melbourne", "50"],
["2012", "20", "Huge", "Brisbane", "80"]
]
const result = rows.map(row =>
row.reduce(
(result, field, index) => ({ ...result, [columns[index]]: field }),
{}
)
)
console.log(result)2022 年更新：使用现在更常见的语法（短函数）
这是原始片段：
var columns = ["Date", "Number", "Size", "Location", "Age"];
var rows = [
["2001", "5", "Big", "Sydney", "25"],
["2005", "2", "Med", "Melbourne", "50"],
["2012", "20", "Huge", "Brisbane", "80"]
];
var result = rows.map(function(row) {
return row.reduce(function(result, field, index) {
result[columns[index]] = field;
return result;
}, {});
});
console.log(result)这样你就不必处理临时数组了。
如果您的代码也可以在 ancient 浏览器上运行，我建议您查看 lodash 以使用
mapreduce在外循环每次迭代开始时创建一个新的
thingthing[columns[i]]newarray[i][i][y][i]var newarray = [],
thing;
for(var y = 0; y < rows.length; y++){
thing = {};
for(var i = 0; i < columns.length; i++){
thing[columns[i]] = rows[y][i];
}
newarray.push(thing)
}
“‘行’包含数组这一事实尤其令人困惑..”
对于您的示例数据，
rows.lengthrows[0]["2001", "5", "Big", "Sydney", "25"]
rows[0][3]以此为例，您应该能够看到
rows[y][i]yi_.object(list, [values])其实施方式如下：
_.object = function(list, values) {
if (list == null) return {};
var result = {};
for (var i = 0, length = list.length; i < length; i++) {
if (values) {
result[list[i]] = values[i];
} else {
result[list[i][0]] = list[i][1];
}
}
return result;
};
使用函数就像 nnnnnn 所说的那样，但有一个小的修正
var columns = ["Date", "Number", "Size", "Location", "Age"];
var rows = [
["2001", "5", "Big", "Sydney", "25"],
["2005", "2", "Med", "Melbourne", "50"],
["2012", "20", "Huge", "Brisbane", "80"]
];
var result = rows.map(function(row) {
return row.reduce(function(result, field, index) {
result[columns[index]] = field;
return result
}, {});
});
您需要指定正确的行。您正在使用该色谱柱。将
rows[i][i]rows[y][i]thingvar thing = {}; var newarray = [];
for (var y = 0; y < rows.length; y++) {
var thing = {};
for (var i = 0; i < columns.length; i++) {
thing[columns[i]] = rows[y][i];
}
newarray.push(thing);
}
您可以按照 ES6 代码使用此功能：
const assemblyData = ( columns, rows) => {
return rows.map((row) => {
return row.reduce((res, field, index) => {
res[columns[index]] = field;
return res
}, {});
});
}
// short version
const assemblyDataMin = (columns, rows) =>
rows.map(row =>
row.reduce((res, field, index) => {
res[columns[index]] = field;
return res;
}, {})
);
Object.fromEntries()const columns = ["Date", "Number", "Size", "Location", "Age"];
const rows = [["2001", "5", "Big", "Sydney", "25"],["2005", "2", "Med", "Melbourne", "50"],["2012", "20", "Huge", "Brisbane", "80"]];
const res = rows.map(row => Object.fromEntries(
columns.map((key, i) => [key, row[i]]))
);
console.log(res);在上面，我正在为
[[key1, value1], [key2, value2], ...].map()构建一个
row 形式的数组，其中 rows ，并将其传递到 Object.fromEntries(){key1: value1, key2: value2}您可以将 Array.prototype.map() 与 Object.fromEntries()
结合使用代码：
const columns = ["Date", "Number", "Size", "Location", "Age"];
const rows = [["2001", "5", "Big", "Sydney", "25"],["2005", "2", "Med", "Melbourne", "50"],["2012", "20", "Huge", "Brisbane", "80"]]
const result = rows.map(r => Object.fromEntries(r.map((c, i) => [columns[i], c])))
console.log(result)您只需重置
thing for(var y = 0; y < rows.length; y++){
for(var i = 0; i < columns.length; i++){
thing[columns[i]] = rows[y][i];
}
newarray.push(thing);
thing = {};
}
查看更新：小提琴
我认为我的解决方案可以解决您的问题
代码：
var keys = columns,
values = rows,
finalarray = [];
values.forEach((data,index) =>{
var objJSON = new Object();
for (i = 0; i < keys.length; i++) {
objJSON[keys[i]] = data[i];
}
finalarray.push(objJSON);
});
答案：
console.log(finalarray)
[{
"Date" : "2001",
"Number" : "5",
"Size":"Big",
"Location":"Sydney",
"Age":"25"
},
...
}]
const columns = ["Date", "Number", "Size", "Location", "Age"]
const rows = [
["2001", "5", "Big", "Sydney", "25"],
["2005", "2", "Med", "Melbourne", "50"]
]
const a = Array.from(rows, r => Object.fromEntries(
Array.from(r, (v, i) => [columns[i], v])
));
console.log(a)您还可以使用 columns 数组生成 Row 类（使用 defineProperty），然后使用该 row 类将行数组包装在 rows 数组中，而不是循环遍历整个表。然后，生成的 Row 对象数组将与原始 rows 数组绑定，这样一个数组中的更改就会反映在另一个数组中。根据您的用例，这可能有用也可能有问题。下面的代码片段给出了一个工作示例。
var columns = ["Date", "Number", "Size", "Location", "Age"];
var rows = [
["2001", "5", "Big", "Sydney", "25"],
["2005", "2", "Med", "Melbourne", "50"],
["2012", "20", "Huge", "Brisbane", "80"]
];
var table = create_table(columns, rows);
//
for (var i = 0; i<table.length; i++) {
document.writeln("The thing in " + table[i].Location + " is " + table[i].Size + "<br/>");
}
// The rows in table are still tied to those in the rows variable
table[0].Size = "Small";
document.writeln(JSON.stringify(rows[0]));
function create_table(columns, rows) {
// Create a Row class with the given columns
for (var i=0; i<columns.length; i++) {
var c = columns[i];
Object.defineProperty(Row.prototype, c, {
enumerable: true,
get: getter(i),
set: setter(i),
});
}
// Wrap the rows in row objects and return the resulting array
var r = new Array(rows.length);
for (var i=0; i<rows.length; i++) {
r[i] = new Row(rows[i]);
}
return r;
function Row(row) { this._ = row; }
// Generators for the getter and setter functions, with variable i stored in a closure.
function getter(i) {
return function(){
return this._[i];
};
}
function setter(i) {
return function(val){
return this._[i] = val;
};
}
}