创建多维数组，然后将每个多维数组推入另一个数组

Question

有可能创建一个多维数组，然后在每个多维数组上推送另一个数组吗？ 假设变量

arr = ["apple", "orange",  "Avocados", "Tomato", "Tangerine"]

我想要的输出是：

[
  ["a", ["apple", "avocados"] ], 
  [ "o", ["orange"] ], 
  ["T", ["Tomato", "Tangering"]]
]

关于在输出之前创建第一个从头到新的arr多维的示例，我们像这样[[ "a"],["o"],["T"]]创建，我想要的输出在上方（框码）

然后再次检查该第一个初始值是否与第一个初始数组相同，例如，将其推入那些2d数组，但是如果其他数组的长度不相同，则应创建一个函数，稍后再使用

Answer 1

我认为您的输出太复杂而无法产生。您可以像这样简化输出格式：

{
   "a" : ["apple", "avocados"], 
   "o": ["orange"], 
   "t": ["Tomato", "Tangering"] 
 }

您可以使用Lodash轻松生成这样的输出：

_.groupBy(["apple", "orange",  "Avocados", "Tomato", "Tangerine"], function (item) {
  return item[0].toLowerCase()
});

或迭代数组：

var arr = ["apple", "orange",  "Avocados", "Tomato", "Tangerine"];
var output = [];
for (var i = 0 ; i < arr.length ; i++){
  if (output[arr[i][0].toLowerCase()] == undefined)
    output[arr[i][0].toLowerCase()] = [];
  output[arr[i][0].toLowerCase()].push(arr[i]);
}

Answer 2

您可以通过查找具有该值的数组来对数据进行分组，或者为结果集创建一个新数组。

var array = ["apple", "orange", "avocados", "tomato", "tangerine"],
    result = array.reduce((r, s) => {
        var temp = r.find(([c]) => c === s[0]);
        if (temp) {
            temp[1].push(s);
        } else {
            r.push([s[0], [s]]);
        }
        return r;
    }, []);
    
console.log(result);

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更好的结构是使用Map并将所有字符串收集到相同的起始字母。

var array = ["apple", "orange", "avocados", "tomato", "tangerine"],
    result = Array.from(
        array.reduce((map, s) => map.set(s[0], [...(map.get(s[0]) || []), s]), new Map)
    );
    
console.log(result);

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Answer 3

您可以将数组简化为对象

var arr = ["apple", "orange",  "Avocados", "Tomato", "Tangerine"];

var output = arr.reduce(function(res, item) {
  if(Object.keys(res).indexOf(item.charAt(0).toLowerCase()) == -1) 
  {
    res[item.charAt(0).toLowerCase()] = [];
  }
  res[item.charAt(0).toLowerCase()].push(item);
  return res;
},{});

console.log(output);

Answer 4

还有另一种使用普通JavaScript和一些功能编程的替代方法：

const input = ["apple", "orange", "Avocados", "Tomato", "Tangerine"]

// (1) An unique list of lower-cased initials of all given input words
// * Set gives us the unique behavior
const initials = [...new Set (
      input.map (([initial]) => initial.toLowerCase ())
)]

const byInitial = ([initial]) => ([initial_]) => 
      initial_.toLowerCase () == initial

// (2) This groups each world by their initial
// Iterates each lowered initial and filters the input by each initial 
// to build the groups!
const grouped = initials.reduce ((o, initial) =>
    [
      ...o, // <--- this accumulates the new result with the previous one
      [
        initial, 
        input.filter (byInitial(initial))
      ]
    ], [])
    
console.log (grouped)

另一种方法，现在使用zip：

const input = ["apple", "orange", "Avocados", "Tomato", "Tangerine"]

// (1) An unique list of lower-cased initials of all given input words
// * Set gives us the unique behavior
const initials = [...new Set (
      input.map (([initial]) => initial.toLowerCase ())
)]

const byInitial = ([initial]) => ([initial_]) => 
      initial_.toLowerCase () == initial

// (2) This builds groups each world by their initial
const groups = initials.map (initial => 
      input.filter (byInitial (initial))
)

const zip = (xs, ys) => 
      xs.map((x, i) => [x, ys[i]])
      
// zip builds the pairs, where each one is the initial and the group
const grouped = zip (initials, groups)
    
console.log (grouped)

Answer 5

使用reduce，创建一个对象，其中每个项目的第一个char为key，并创建一个以该字符开头的所有项目的数组，形式为value。然后调用Object.entries获得所需的输出：

const arr = ["apple", "orange",  "avocados", "tomato", "tangerine"],
group = Object.entries(arr.reduce((a, i) => ((a[i[0]] = a[i[0]] || []).push(i), a), {}));

console.log(group);