Javascript 拆分数组

Question

我正在尝试编写一个自定义字符串拆分函数，它比我预期的要难。

基本上，我传入一个字符串和一个字符串将拆分的值的数组，它将返回一个子字符串数组，删除空的并包括它拆分的值。如果字符串可以在同一位置被两个不同的值分割，则较长的优先。

也就是说，

split("Go ye away, I want some peace && quiet. & Thanks.", ["Go ", ",", "&&", "&", "."]);

应该返回

["Go ", "ye away", ",", " I want some peace ", "&&", " quiet", ".", " ", "&", " Thanks", "."]

你能想出一个相当简单的算法吗？如果在 Javascript 中有内置的方法可以做到这一点（我认为没有），那就更好了。

Answer 1

类似this?

function mySplit(input, delimiters) {

    // Sort delimiters array by length to avoid ambiguity
    delimiters.sort(function(a, b) {
       if (a.length > b.length) { return -1; }
       return 0;
    }

    var result = [];

    // Examine input one character at a time
    for (var i = 0; i < input.length; i++) {
        for (var j = 0; j < delimiters.length; j++) {
            if (input.substr(i, delimiters[j].length) == delimiters[j]) {

                // Add first chunk of input to result
                if (i > 0) {
                    result.push(input.substr(0, i));
                }
                result.push(delimiters[j]);

                // Reset input and iteration
                input = input.substr(i + delimiters[j].length);
                i = 0;
                j = 0;
            }
        }
    }

    return result;
}

var input      = "Go ye away, I want some peace && quiet. & Thanks.";
var delimiters = ["Go ", ",", "&&", "&", "."];

console.log(mySplit(input, delimiters));
// Output: ["Go ", "ye away", ",", " I want some peace ",
//          "&&", " quiet", ".", " ", "&", " Thanks", "."]

Answer 2

要求的确切解决方案：

function megasplit(toSplit, splitters) {
    var splitters = splitters.sorted(function(a,b) {return b.length-a.length});
                                                          // sort by length; put here for readability, trivial to separate rest of function into helper function
    if (!splitters.length)
        return toSplit;
    else {
        var token = splitters[0];
        return toSplit
            .split(token)             // split on token
            .map(function(segment) {  // recurse on segments
                 return megasplit(segment, splitters.slice(1))
             })
            .intersperse(token)       // re-insert token
            .flatten()                // rejoin segments
            .filter(Boolean);
    }
}

演示：

> megasplit(
      "Go ye away, I want some peace && quiet. & Thanks.",
      ["Go ", ",", "&&", "&", "."]
  )
["Go ", "ye away", ",", " I want some peace ", "&", "&", " quiet", ".", " ", "&", " Thanks", "."]

机械（可重复使用！）：

Array.prototype.copy = function() {
    return this.slice()
}
Array.prototype.sorted = function() {
    var copy = this.copy();
    copy.sort.apply(copy, arguments);
    return copy;
}
Array.prototype.flatten = function() {
    return [].concat.apply([], this)
}
Array.prototype.mapFlatten = function() {
    return this.map.apply(this,arguments).flatten()
}
Array.prototype.intersperse = function(token) {
    // [1,2,3].intersperse('x') -> [1,'x',2,'x',3]
    return this.mapFlatten(function(x){return [token,x]}).slice(1)
}

---

注意事项：

• 这需要大量的研究才能优雅地完成：
  • (Deep) copying an array using jQuery
  • What is the most efficient way to concatenate N arrays in JavaScript?（创建了我自己的不那么难看的方法）
  • How can I split text on commas not within double quotes, while keeping the quotes?（垃圾答案，再次创建了我自己的方法）
• 由于规范要求令牌（尽管它们将留在字符串中）不应该被拆分（否则你会得到"&", "&"），这使情况变得更加复杂。这使得使用reduce 成为不可能和必要的递归。
• 我个人也不会忽略带有拆分的空字符串。我可以理解不想在令牌上递归拆分，但我个人会简化函数并使输出表现得像普通的 .split 和 ["", "Go ", "ye away", ",", " I want some peace ", "&&", " quiet", ".", " ", "&", " Thanks", ".", ""]
• 我应该指出，如果您愿意稍微放宽您的要求，这将从 15/20-liner 变为 1/3-liner：

---

1-liner，如果遵循规范拆分行为：

Array.prototype.mapFlatten = function() {
    ...
}
function megasplit(toSplit, splitters) {
    return splitters.sorted(...).reduce(function(strings, token) {
        return strings.mapFlatten(function(s){return s.split(token)});
    }, [toSplit]);
}

3-liner，如果上面的内容难以阅读：

Array.prototype.mapFlatten = function() {
    ...
}
function megasplit(toSplit, splitters) {
    var strings = [toSplit];
    splitters.sorted(...).forEach(function(token) {
        strings = strings.mapFlatten(function(s){return s.split(token)});
    });
    return strings;
}