如何将波斯语（波斯语）段落转换为 Javascript 中的单词列表

Posted 2023-04-14

【中文标题】如何将波斯语（波斯语）段落转换为 Javascript 中的单词列表

【英文标题】：how to turn a Persian (Farsi) paragraph into its list of words in Javascript

【发布时间】：2018-02-01 11:59:53

【问题描述】：

我正在尝试从显示单词及其频率的段落中创建一个对象。

var pattern = /\w+/g,
//the farsi paragraph
    string = "من امروز در مورد مهر خروج مشمولین اطلاعات جدیدی از سفارت ایران در مالزی گرفتم",
    matchedWords = string.match( pattern );

/* The Array.prototype.reduce method assists us in producing a single value from an
   array. In this case, we're going to use it to output an object with results. */
var counts = matchedWords.reduce(function ( stats, word ) 

    /* `stats` is the object that we'll be building up over time.
       `word` is each individual entry in the `matchedWords` array */
    if ( stats.hasOwnProperty( word ) ) 
        /* `stats` already has an entry for the current `word`.
           As a result, let's increment the count for that `word`. */
        stats[ word ] = stats[ word ] + 1;
     else 
        /* `stats` does not yet have an entry for the current `word`.
           As a result, let's add a new entry, and set count to 1. */
        stats[ word ] = 1;
    

    /* Because we are building up `stats` over numerous iterations,
       we need to return it for the next pass to modify it. */
    return stats;

, )

var dict = []; // create an empty array
// this for loop makes a dictionary for you
for (i in counts)
dict.push('text':i, "size": counts[i]);


;

/* lets print and see if you can solve your problem */

console.log( dict);

代码最初是为一个英文段落制定的。但是我需要将它用于波斯语。 我知道它应该是别的东西，而不是 "/\w+/g" in:

var pattern = /\w+/g,

但我不知道是什么。

【问题讨论】：

为什么不使用string.split(' ')？

也许，您可以使用/[آ-ی]+/g，但最好将\pL 构造（任何Unicode 字母）与XRegExp 一起使用。

【参考方案1】：

在您的正则表达式中，将变量用于“除空格之外的任何字符”，即\S。 编辑：空格被认为是换行符、制表符和空格）

var pattern = /\S+/g,
//the farsi paragraph
    string = "من امروز در مورد مهر خروج مشمولین اطلاعات جدیدی از سفارت ایران در مالزی گرفتم",
    matchedWords = string.match( pattern );

/* The Array.prototype.reduce method assists us in producing a single value from an
   array. In this case, we're going to use it to output an object with results. */
var counts = matchedWords.reduce(function ( stats, word ) 

    /* `stats` is the object that we'll be building up over time.
       `word` is each individual entry in the `matchedWords` array */
    if ( stats.hasOwnProperty( word ) ) 
        /* `stats` already has an entry for the current `word`.
           As a result, let's increment the count for that `word`. */
        stats[ word ] = stats[ word ] + 1;
     else 
        /* `stats` does not yet have an entry for the current `word`.
           As a result, let's add a new entry, and set count to 1. */
        stats[ word ] = 1;
    

    /* Because we are building up `stats` over numerous iterations,
       we need to return it for the next pass to modify it. */
    return stats;

, )

var dict = []; // create an empty array
// this for loop makes a dictionary for you
for (i in counts)
dict.push('text':i, "size": counts[i]);


;

/* lets print and see if you can solve your problem */

console.log( dict);

【讨论】：

\S 匹配任何非空格，OP 需要匹配字母。

标题和问题描述说的是单词，而不是字母。

是的，单词是由字母组成的，不是标点符号或符号字符。

@WiktorStribiżew 完全同意。无论如何，我不知道波斯语，但如果 OP 用标点符号和符号字符测试了他的第一个正则表达式 (\w)反对 (\w) 不包含除字母之外的任何内容。 OP 的一点澄清将使我们更准确地帮助他。

@Dez：我认为 Wiktor 的观点是，在像 Foo, bar? 这样的字符串中，OP 的版本将成功提取单词 Foo 和 bar，而您的版本将错误地提取 Foo,和bar?。

【参考方案2】：

要匹配任何字母，您需要使用 XRegExp 包和 \pL Unicode 属性类：

var pattern = new XRegExp("[_\\pL\\pN]+", "g");
var s = "من امروز در مورد مهر خروج مشمولین اطلاعات جدیدی از سفارت ایران در مالزی گرفتم";
var matchedWords = s.match( pattern );
var counts = matchedWords.reduce(function ( stats, word ) 
 if ( stats.hasOwnProperty( word ) ) 
    stats[ word ] = stats[ word ] + 1;
  else 
    stats[ word ] = 1;
 
 return stats;
, )

var dict = [];
for (i in counts)
	dict.push('text':i, "size": counts[i]);
 
console.log(dict);

<script src="https://cdnjs.cloudflare.com/ajax/libs/xregexp/3.2.0/xregexp-all.min.js"></script>

[_\\pL\\pN]+ 模式匹配一​​个或多个下划线（_，我包含它是因为您的原始正则表达式中的\w 也匹配_）、Unicode 字母（\pL）和数字（\pN） .

要只计算由字母组成的单词，只需使用

var pattern = new XRegExp("\\pL+", "g");

【参考方案3】：

在您的情况下，为什么不直接使用 split 和 reduce 呢？示例：

const p = 'من امروز در مورد مهر خروج مشمولین اطلاعات جدیدی از سفارت ایران در مالزی گرفتم';

const counted = p.split( ' ' ).reduce( ( collected, item ) => 
  collected[ item ] = ( collected[ item ] || 0 ) + 1;
  return collected;
,  /* initial empty object */  );
const dict = Object.keys( counted ).map( key => 
  return 
    text: key,
    size: counted[ key ],
  ;
 );

console.log( 'در:', counted[ 'در' ] );
console.log( dict );

它更简单，性能更好。您甚至可以省略 const dict... 部分。

【参考方案4】：

您可以使用 JS 等效词和量词 \w+ 这将匹配大约 119,000 个 Unicode 9 字字符。 这包括所有非字母、非数字、其他单词字符 像下划线一样，其中大约有 1,100 个。

注意 - 它运行得非常快，但是我会将这个正则表达式设为全局并且 编译一次以备后用。

此外，这是从 ICU 数据库生成的，它提供了完整的 在 U+000000 到 U+10FFFF 之间的单词 \w 的样本，来自这个正则表达式 在RegexFormat 应用程序中使用UCD Interface 生成。

这是 XRegExp 做不到的。

演示：

https://regex101.com/r/sjLmMC/1

(?:[\u0030-\u0039\u0041-\u005A\u005F\u0061-\u007A\u00AA\u00B5\u00BA\u00C0-\u00D6\u00D8-\u00F6\u00F8-\u02C1\u02C6-\u02D1\u02E0-\u02E4\u02EC\u02EE\u0300-\u0374\u0376-\u0377\u037A-\u037D\u037F\u0386\u0388-\u038A\u038C\u038E-\u03A1\u03A3-\u03F5\u03F7-\u0481\u0483-\u0487\u048A-\u052F\u0531-\u0556\u0559\u0561-\u0587\u0591-\u05BD\u05BF\u05C1-\u05C2\u05C4-\u05C5\u05C7\u05D0-\u05EA\u05F0-\u05F2\u0610-\u061A\u0620-\u0669\u066E-\u06D3\u06D5-\u06DC\u06DF-\u06E8\u06EA-\u06FC\u06FF\u0710-\u074A\u074D-\u07B1\u07C0-\u07F5\u07FA\u0800-\u082D\u0840-\u085B\u08A0-\u08B4\u08B6-\u08BD\u08D4-\u08E1\u08E3-\u0902\u0904-\u093A\u093C-\u093D\u0941-\u0948\u094D\u0950-\u0963\u0966-\u096F\u0971-\u0981\u0985-\u098C\u098F-\u0990\u0993-\u09A8\u09AA-\u09B0\u09B2\u09B6-\u09B9\u09BC-\u09BD\u09C1-\u09C4\u09CD-\u09CE\u09DC-\u09DD\u09DF-\u09E3\u09E6-\u09F1\u0A01-\u0A02\u0A05-\u0A0A\u0A0F-\u0A10\u0A13-\u0A28\u0A2A-\u0A30\u0A32-\u0A33\u0A35-\u0A36\u0A38-\u0A39\u0A3C\u0A41-\u0A42\u0A47-\u0A48\u0A4B-\u0A4D\u0A51\u0A59-\u0A5C\u0A5E\u0A66-\u0A75\u0A81-\u0A82\u0A85-\u0A8D\u0A8F-\u0A91\u0A93-\u0AA8\u0AAA-\u0AB0\u0AB2-\u0AB3\u0AB5-\u0AB9\u0ABC-\u0ABD\u0AC1-\u0AC5\u0AC7-\u0AC8\u0ACD\u0AD0\u0AE0-\u0AE3\u0AE6-\u0AEF\u0AF9\u0B01\u0B05-\u0B0C\u0B0F-\u0B10\u0B13-\u0B28\u0B2A-\u0B30\u0B32-\u0B33\u0B35-\u0B39\u0B3C-\u0B3D\u0B3F\u0B41-\u0B44\u0B4D\u0B56\u0B5C-\u0B5D\u0B5F-\u0B63\u0B66-\u0B6F\u0B71\u0B82-\u0B83\u0B85-\u0B8A\u0B8E-\u0B90\u0B92-\u0B95\u0B99-\u0B9A\u0B9C\u0B9E-\u0B9F\u0BA3-\u0BA4\u0BA8-\u0BAA\u0BAE-\u0BB9\u0BC0\u0BCD\u0BD0\u0BE6-\u0BEF\u0C00\u0C05-\u0C0C\u0C0E-\u0C10\u0C12-\u0C28\u0C2A-\u0C39\u0C3D-\u0C40\u0C46-\u0C48\u0C4A-\u0C4D\u0C55-\u0C56\u0C58-\u0C5A\u0C60-\u0C63\u0C66-\u0C6F\u0C80-\u0C81\u0C85-\u0C8C\u0C8E-\u0C90\u0C92-\u0CA8\u0CAA-\u0CB3\u0CB5-\u0CB9\u0CBC-\u0CBD\u0CBF\u0CC6\u0CCC-\u0CCD\u0CDE\u0CE0-\u0CE3\u0CE6-\u0CEF\u0CF1-\u0CF2\u0D01\u0D05-\u0D0C\u0D0E-\u0D10\u0D12-\u0D3A\u0D3D\u0D41-\u0D44\u0D4D-\u0D4E\u0D54-\u0D56\u0D5F-\u0D63\u0D66-\u0D6F\u0D7A-\u0D7F\u0D85-\u0D96\u0D9A-\u0DB1\u0DB3-\u0DBB\u0DBD\u0DC0-\u0DC6\u0DCA\u0DD2-\u0DD4\u0DD6\u0DE6-\u0DEF\u0E01-\u0E3A\u0E40-\u0E4E\u0E50-\u0E59\u0E81-\u0E82\u0E84\u0E87-\u0E88\u0E8A\u0E8D\u0E94-\u0E97\u0E99-\u0E9F\u0EA1-\u0EA3\u0EA5\u0EA7\u0EAA-\u0EAB\u0EAD-\u0EB9\u0EBB-\u0EBD\u0EC0-\u0EC4\u0EC6\u0EC8-\u0ECD\u0ED0-\u0ED9\u0EDC-\u0EDF\u0F00\u0F18-\u0F19\u0F20-\u0F29\u0F35\u0F37\u0F39\u0F40-\u0F47\u0F49-\u0F6C\u0F71-\u0F7E\u0F80-\u0F84\u0F86-\u0F97\u0F99-\u0FBC\u0FC6\u1000-\u102A\u102D-\u1030\u1032-\u1037\u1039-\u103A\u103D-\u1049\u1050-\u1055\u1058-\u1061\u1065-\u1066\u106E-\u1082\u1085-\u1086\u108D-\u108E\u1090-\u1099\u109D\u10A0-\u10C5\u10C7\u10CD\u10D0-\u10FA\u10FC-\u1248\u124A-\u124D\u1250-\u1256\u1258\u125A-\u125D\u1260-\u1288\u128A-\u128D\u1290-\u12B0\u12B2-\u12B5\u12B8-\u12BE\u12C0\u12C2-\u12C5\u12C8-\u12D6\u12D8-\u1310\u1312-\u1315\u1318-\u135A\u135D-\u135F\u1380-\u138F\u13A0-\u13F5\u13F8-\u13FD\u1401-\u166C\u166F-\u167F\u1681-\u169A\u16A0-\u16EA\u16F1-\u16F8\u1700-\u170C\u170E-\u1714\u1720-\u1734\u1740-\u1753\u1760-\u176C\u176E-\u1770\u1772-\u1773\u1780-\u17B5\u17B7-\u17BD\u17C6\u17C9-\u17D3\u17D7\u17DC-\u17DD\u17E0-\u17E9\u180B-\u180D\u1810-\u1819\u1820-\u1877\u1880-\u18AA\u18B0-\u18F5\u1900-\u191E\u1920-\u1922\u1927-\u1928\u1932\u1939-\u193B\u1946-\u196D\u1970-\u1974\u1980-\u19AB\u19B0-\u19C9\u19D0-\u19D9\u1A00-\u1A18\u1A1B\u1A20-\u1A54\u1A56\u1A58-\u1A5E\u1A60\u1A62\u1A65-\u1A6C\u1A73-\u1A7C\u1A7F-\u1A89\u1A90-\u1A99\u1AA7\u1AB0-\u1ABD\u1B00-\u1B03\u1B05-\u1B34\u1B36-\u1B3A\u1B3C\u1B42\u1B45-\u1B4B\u1B50-\u1B59\u1B6B-\u1B73\u1B80-\u1B81\u1B83-\u1BA0\u1BA2-\u1BA5\u1BA8-\u1BA9\u1BAB-\u1BE6\u1BE8-\u1BE9\u1BED\u1BEF-\u1BF1\u1C00-\u1C23\u1C2C-\u1C33\u1C36-\u1C37\u1C40-\u1C49\u1C4D-\u1C7D\u1C80-\u1C88\u1CD0-\u1CD2\u1CD4-\u1CE0\u1CE2-\u1CF1\u1CF4-\u1CF6\u1CF8-\u1CF9\u1D00-\u1DF5\u1DFB-\u1F15\u1F18-\u1F1D\u1F20-\u1F45\u1F48-\u1F4D\u1F50-\u1F57\u1F59\u1F5B\u1F5D\u1F5F-\u1F7D\u1F80-\u1FB4\u1FB6-\u1FBC\u1FBE\u1FC2-\u1FC4\u1FC6-\u1FCC\u1FD0-\u1FD3\u1FD6-\u1FDB\u1FE0-\u1FEC\u1FF2-\u1FF4\u1FF6-\u1FFC\u2071\u207F\u2090-\u209C\u20D0-\u20DC\u20E1\u20E5-\u20F0\u2102\u2107\u210A-\u2113\u2115\u2119-\u211D\u2124\u2126\u2128\u212A-\u212D\u212F-\u2139\u213C-\u213F\u2145-\u2149\u214E\u2183-\u2184\u2C00-\u2C2E\u2C30-\u2C5E\u2C60-\u2CE4\u2CEB-\u2CF3\u2D00-\u2D25\u2D27\u2D2D\u2D30-\u2D67\u2D6F\u2D7F-\u2D96\u2DA0-\u2DA6\u2DA8-\u2DAE\u2DB0-\u2DB6\u2DB8-\u2DBE\u2DC0-\u2DC6\u2DC8-\u2DCE\u2DD0-\u2DD6\u2DD8-\u2DDE\u2DE0-\u2DFF\u2E2F\u3005-\u3006\u302A-\u302D\u3031-\u3035\u303B-\u303C\u3041-\u3096\u3099-\u309A\u309D-\u309F\u30A1-\u30FA\u30FC-\u30FF\u3105-\u312D\u3131-\u318E\u31A0-\u31BA\u31F0-\u31FF\u3400-\u4DB5\u4E00-\u9FD5\uA000-\uA48C\uA4D0-\uA4FD\uA500-\uA60C\uA610-\uA62B\uA640-\uA66F\uA674-\uA67D\uA67F-\uA6E5\uA6F0-\uA6F1\uA717-\uA71F\uA722-\uA788\uA78B-\uA7AE\uA7B0-\uA7B7\uA7F7-\uA822\uA825-\uA826\uA840-\uA873\uA882-\uA8B3\uA8C4-\uA8C5\uA8D0-\uA8D9\uA8E0-\uA8F7\uA8FB\uA8FD\uA900-\uA92D\uA930-\uA951\uA960-\uA97C\uA980-\uA982\uA984-\uA9B3\uA9B6-\uA9B9\uA9BC\uA9CF-\uA9D9\uA9E0-\uA9FE\uAA00-\uAA2E\uAA31-\uAA32\uAA35-\u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