英文:
Give space to array of signs in the string
问题
考虑到每个标点符号始终只出现在字符串的开头或结尾，我们希望在每个标点符号之前或之后添加一个空格，使其成为一个单词：
所需的结果已经在注释中给出，但是我的函数存在问题，如你所见：
modify('you always have to wake up?'); // you always have to wake up ?
modify('...you always have to wake up'); // ... you always have to wake up
modify('...you always have to wake up?!'); // ... you always have to wake up ?!
function modify(string) {
const sgins = ['?', '!', '...', '?!', '!?', '!!', '.'];
for (let a = 0; a < sgins.length; a++) {
const split = string.split(sgins[a]);
string = string.replaceAll(sgins[a], ` ${sgins[a]} `).trim();
}
console.log(string);
}
你会如何解决这个问题？
英文:
considering each sign always in the start or end of the string only, we want to give each sgin a space before or after so as a word:
The desired result is commented, My function has an issue as you see:
<!-- begin snippet: js hide: false console: true babel: false -->
<!-- language: lang-js -->
modify('you always have to wake up?'); // you always have to wake up ?
modify('...you always have to wake up'); // ... you always have to wake up
modify('...you always have to wake up?!'); // ... you always have to wake up ?!
function modify(string) {
const sgins = ['?', '!', '...', '?!', '!?', '!!', '.'];
for (let a = 0; a < sgins.length; a++) {
const split = string.split(sgins[a]);
string = string.replaceAll(sgins[a], ` ${sgins[a]} `).trim();
}
console.log(string);
}
<!-- end snippet -->
How would you do this?
答案1
得分: 1
你可以使用简单的正则表达式来实现：
string.replace(/[$?!]+/g, '$& ');
这个正则表达式的意思是将连续的$?!字符替换为这些字符加上一个空格，放在字符串的开头和结尾。
如果你需要精确的前缀，你可以构建一个正则表达式：
function modify(string) {
const signs = ['?', '!', '...', '?!', '!?', '!!', '.'];
const options = signs.sort((a, b) => b.length - a.length).map(prefix => [...prefix].map(c => '\\' + c).join('')).join('|');
string = string.replace(new RegExp('^(' + options + ')'), '$1 ').replace(new RegExp('(' + options + ')$'), ' $1');
console.log(string);
}
这个函数会根据给定的前缀列表构建一个正则表达式，并将字符串中符合前缀的部分替换为这些前缀加上一个空格，放在字符串的开头和结尾。
英文:
You could use a simple regex:
string.replace(/$[?!.]+/, '$& ');
Means replace all continuous ?!. characters with the characters plus a space in the beginning of the string. The same for the end.
<!-- begin snippet: js hide: false console: true babel: false -->
<!-- language: lang-js -->
modify('you always have to wake up?'); // you always have to wake up ?
modify('...you always have to wake up'); // ... you always have to wake up
modify('...you always have to wake up?!'); // ... you always have to wake up ?!
function modify(string) {
string = string.replace(/^[?!.]+/, '$& ').replace(/[?!.]+$/, ' $&');
console.log(string);
}
<!-- end snippet -->
If you need exact prefixes you can build the regexp:
<!-- begin snippet: js hide: false console: true babel: false -->
<!-- language: lang-js -->
modify('you always have to wake up?'); // you always have to wake up ?
modify('...you always have to wake up'); // ... you always have to wake up
modify('...you always have to wake up?!'); // ... you always have to wake up ?!
function modify(string) {
const sgins = ['?', '!', '...', '?!', '!?', '!!', '.'];
const options = sgins.sort((a, b) => b.length - a.length).map(prefix => [...prefix].map(c => '\\' + c).join('')).join('|');
string = string.replace(new RegExp('^(' + options + ')'), '$1 ').replace(new RegExp('(' + options + ')$'), ' $1');
console.log(string);
}
<!-- end snippet -->
答案2
得分: 1
首先，老实说，我不会使用split()。我会这样写，但这不是最优化的代码。这只适用于某些情况，我不得不替换一些数组的符号才能正常工作。如果你想要完美的代码，那么你需要为许多不同的情况编写程序，所以代码会稍微长一些。例如，如果已经有空格了怎么办？等等。
function modify(string) {
let didEnd = false;
let didStart = false;
const sgins = ['?!', '!?', '!!', '!', '?', '...', '.'];
for (let a = 0; a < sgins.length; a++) {
if (string.startsWith(sgins[a]) && !didStart) {
string = string.replace(sgins[a], `${sgins[a]} `);
didStart = true;
}
if (string.endsWith(sgins[a]) && !didEnd) {
string = string.replace(sgins[a], ` ${sgins[a]}`);
didEnd = true;
}
}
console.log(string);
}
modify('you always have to wake up?');
modify('...you always have to wake up');
modify('...you always have to wake up?');
希望能帮到你！
英文:
First off I would not use split() to be honest with You. I would write this like that, but this is not optimal code. This is not for all cases, I had to replace some signs of Your array to work. If You want to make it perfect, then you have to program this for a lot of different cases so code would be a little bigger. Cause for example what if there already is space? ETC.
<!-- begin snippet: js hide: false console: true babel: false -->
<!-- language: lang-js -->
function modify(string) {
let didEnd = false;
let didStart = false;
const sgins = ['?!', '!?', '!!', '!', '?', '...', '.'];
for (let a = 0; a < sgins.length; a++) {
if (string.startsWith(sgins[a]) && !didStart) {
string = string.replace(sgins[a], `${sgins[a]} `);
didStart = true;
}
if (string.endsWith(sgins[a]) && !didEnd) {
string = string.replace(sgins[a], ` ${sgins[a]}`);
didEnd = true;
}
}
console.log(string);
}
modify('you always have to wake up?');
modify('...you always have to wake up');
modify('...you always have to wake up?!');
<!-- end snippet -->
答案3
得分: 0
作为替代方案，您可以使用具有全局匹配 /g 和 2 个捕获组的模式。
在回调函数中，检查第 1 组或第 2 组，以添加或追加一个空格。
const regex = /^([?!.]+)|([?!.]+)$/g;
modify('you always have to wake up?'); // you always have to wake up ?
modify('...you always have to wake up'); // ... you always have to wake up
modify('...you always have to wake up?!'); // ... you always have to wake up ?!
function modify(string) {
const regex = /^([?!.]+)|([?!.]+)$/g;
console.log(string.replace(regex, (_, g1, g2) => g1 ? g1 + " " : " " + g2));
}
希望对您有所帮助！
英文:
As an alternative, you could use a pattern with a global match /g with 2 capture groups.
In the callback check for group 1 or group 2 to prepend or append a space.
const regex = /^([?!.]+)|([?!.]+)$/g;
<!-- begin snippet: js hide: false console: true babel: false -->
<!-- language: lang-js -->
modify('you always have to wake up?'); // you always have to wake up ?
modify('...you always have to wake up'); // ... you always have to wake up
modify('...you always have to wake up?!'); // ... you always have to wake up ?!
function modify(string) {
const regex = /^([?!.]+)|([?!.]+)$/g;
console.log(string.replace(regex, (_, g1, g2) => g1 ? g1 + " " : " " + g2));
}
<!-- end snippet -->
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