如何从JavaScript中的数组中删除拼接列表? [英] How to remove list of splices from array in JavaScript?

问题描述

我有索引列表:

var remove_list = [ [ 7, 12 ], [ 12, 14 ] ];

以及Scheme Lisp中S-Expression的标记列表.

and list of tokens of S-Expression from Scheme lisp.

var tokens = [
  '(',  'let', '(',   '(',
  'x',  '10',  ')',   '#;',
  '(',  'foo', 'bar', ')',
  '#;', 'xxx', ')',   '(',
  '*',  'x',   'x',   ')',
  ')'
];

从令牌中删除所有项目的最佳方法是什么.(它假定按照R7RS中的指定删除内联命令#;(foo bar) ad #; xxx ).

what is the best way to remove all items from tokens. (it suppose to remove inline commands #;(foo bar) ad #;xxx as specified in R7RS).

我可以使用这个

for (let remove of remove_list) {
    tokens.splice(...remove);
}

因为在第一次调用拼接索引后将发生变化.从数组中删除所有指定范围的最简单方法是什么?

because after first call to splice indexes will change. What is simplest way to remove all specified ranges from array?

在上下文中，我具有应删除内联注释的功能，因此我从将索引从数组中删除的方式拆分了计算方法，这是一种好方法吗?

And for context I have this function that should remove inline comments, I've splitted calculating the indexes from removing them from array, is this good approach?

function strip_s_comments(tokens) {
    var s_count = 0;
    var s_start = null;
    var remove_list = [];
    for (let i = 0; i < tokens.length; ++i) {
        const token = tokens[i];
        if (token === '#;') {
            if (['(', '['].includes(tokens[i + 1])) {
                s_count = 1;
                s_start = i;
            } else {
                remove_list.push([i, i + 2]);
            }
            i += 1;
            continue;
        }
        if (s_start !== null) {
            if ([')', ']'].includes(token)) {
                s_count--;
            } else if (['(', '['].includes(token)) {
                s_count++;
            }
            if (s_count === 0) {
                remove_list.push([s_start, i + 1]);
                s_start = null;
            }
        }
    }
    for (let remove of remove_list) {
        tokens.splice(...remove);
    }
    return tokens;
}

推荐答案

您有两个选择:

1. 以相反的顺序工作，或者

1. Work in reverse order, or

跟踪删除的内容，并在以后的索引中加以考虑

Keep track of how much you've removed and take that into account with later indexes

这里是#1的示例:

const reverse = remove_list.sort(([a], [b]) => b - a);
for (const [begin, end] of reverse) {
    tokens.splice(begin, end - begin);
}

var remove_list = [ [ 7, 12 ], [ 12, 14 ] ];

var tokens = [
  '(',  'let', '(',   '(',
  'x',  '10',  ')',   '#;',
  '(',  'foo', 'bar', ')',
  '#;', 'xxx', ')',   '(',
  '*',  'x',   'x',   ')',
  ')'
];

const reverse = remove_list.sort(([a], [b]) => b - a);
for (const [begin, end] of reverse) {
    tokens.splice(begin, end - begin);
}

console.log(tokens);

这里是#2的示例:

let removed = 0;
for (const [begin, end] of remove_list) {
    removed += tokens.splice(begin - removed, end - begin).length;
}

var remove_list = [ [ 7, 12 ], [ 12, 14 ] ];

var tokens = [
  '(',  'let', '(',   '(',
  'x',  '10',  ')',   '#;',
  '(',  'foo', 'bar', ')',
  '#;', 'xxx', ')',   '(',
  '*',  'x',   'x',   ')',
  ')'
];

let removed = 0;
for (const [begin, end] of remove_list) {
    removed += tokens.splice(begin - removed, end - begin).length;
}

console.log(tokens);

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