Most frequently used words in a text

4 kyu

題目描述

Description

Write a function that, given a string of text (possibly with punctuation and line-breaks), returns an array of the top-3 most occurring words, in descending order of the number of occurrences.

Assumptions:

• A word is a string of letters (A to Z) optionally containing one or more apostrophes (') in ASCII.
• Apostrophes can appear at the start, middle or end of a word ('abc, abc', 'abc', ab'c are all valid)
• Any other characters (e.g. #, , / , . ...) are not part of a word and should be treated as whitespace.
• Matches should be case-insensitive, and the words in the result should be lowercased.
• Ties may be broken arbitrarily.
• If a text contains fewer than three unique words, then either the top-2 or top-1 words should be returned, or an empty array if a text contains no words.

Examples:

"In a village of La Mancha, the name of which I have no desire to call to
mind, there lived not long since one of those gentlemen that keep a lance
in the lance-rack, an old buckler, a lean hack, and a greyhound for
coursing. An olla of rather more beef than mutton, a salad on most
nights, scraps on Saturdays, lentils on Fridays, and a pigeon or so extra
on Sundays, made away with three-quarters of his income."

--> ["a", "of", "on"]


"e e e e DDD ddd DdD: ddd ddd aa aA Aa, bb cc cC e e e"

--> ["e", "ddd", "aa"]


"  //wont won't won't"

--> ["won't", "wont"]

Bonus points (not really, but just for fun):

1. Avoid creating an array whose memory footprint is roughly as big as the input text.
2. Avoid sorting the entire array of unique words.

解答

Solution

function topThreeWords(text) {
  const counter = {};

  const words = text.toLowerCase().match(/\b[a-z']+\b/g);

  if (!words) return [];

  for (const word of words) {
    if (counter[word]) {
      counter[word]++;
    } else {
      counter[word] = 1;
    }
  }

  return Object.entries(counter)
    .sort(([, a], [, b]) => b - a)
    .slice(0, Math.min(3, Object.keys(counter).length))
    .map(([key]) => key);
}

解題思路

使用物件 Counter 來計算單詞出現次數，再用正則表達式篩選出符合題意的格式，重點為：轉小寫、忽略標點符號（但 won't 的標點不能被忽略掉），最後使用Object.entries(counter)轉換物件為陣列並依次數由大到小排序，並根據單詞總數決定要返回多少個單詞（使用 slice(0, Math.min(3, Object.keys(counter).length)) 動態調整返回數量。 ）。

心得

稍微進階的題目，能練習 Counter 與正則表達式與物件和陣列的操作，超值大禮包耶!