如何通过任何字符串项的前 3 个字符对字符串类型的数组项进行分组和收集?

时间:2020-12-22 17:25:45

标签: javascript arrays algorithm grouping reduce

提供的是一个单词数组,其中一些单词最有可能具有共同的前 3 个字符(考虑不区分大小写)......类似于这个单词列表示例......

['Bob', 'Bobby', 'Robert', 'Robby']

由于 'Bob''Bobby' 以及 'Robert''Robby' 确实有共同点 'Bob' 分别为 'Rob',我正在寻找一种将返回两个分组单词数组的数组的方法,对于前一个示例是 ...

[['Bob', 'Bobby'], ['Robert', 'Robby']]

这样的方法会是什么样子?

2 个答案:

答案 0 :(得分:1)

将名称以开头为键收集到一个对象中(参见String.substring),然后获取该对象的值(参见Object.values()

function getArraysByNameFirstThreeChars(names) {  
  // declare the object for collecting similarely beginning names
  const resultObject = {};

  // iterate through the names
  for (const name of names) {
    // get 3 first characters of the name
    const strStart = name.substring(0, name.length >= 3 ? 3 : name.length);
    // if there's already a member in the object
    // for this beginning, push the name to it
    if (resultObject[strStart]) {
      resultObject[strStart].push(name);
    } else {
      // otherwise, create a member in the collection object
      // for this name-beginning, comprising of an array
      // with only current name in it
      resultObject[strStart] = [name];
    }
  }
  return Object.values(resultObject);
}

const names = ['Bob', 'Bobby', 'Robert', 'Robby'];
const whatYouWant = getArraysByNameFirstThreeChars(names);

console.log(whatYouWant);
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答案 1 :(得分:0)

Array.prototype.reduce 经常证明自己是瑞士军刀

一个可配置的 reducer 函数即可完成这项工作。另外提供的起始值是收集器对象,它具有 groups 对象,该对象稍后保存所有收集和分组的单词列表。也可以配置 length 属性。它表示任何单词的起始序列的字符数,所有单词将根据该序列进行分组和收集。省略此属性在内部默认为 3 的数值。

reducer 专门收集唯一的词。因此,将过滤要处理的数组中的重复项。

对于预期为单词数组的最终结果,只需将返回值的 groups 键传递给 Object.values ...

// Added clarification.
// In this case they share the first 3 letters, bob and rob.

function groupAndCollectWordsBySimilarStartSequences(collector, word) {
  const { groups, length = 3 } = collector;
  const index = collector.index || (collector.index = {});

  // process non duplicate words of minimum length only.
  if ((word.length >= length) && !(word in index)) {

    const groupKey = word.slice(0, length).toLowerCase();
    const groupList = groups[groupKey] || (groups[groupKey] = []);

    groupList.push(word);
    index[word] = word;
  }
  return collector;
}

console.log(
  'grouped index :',

  ['Bob', 'Bobby', 'Robert', 'Robby', 'Bob', 'Robby'].reduce(
    groupAndCollectWordsBySimilarStartSequences, {
      groups: {},
      length: 2,
    }
  ).groups
);
console.log(
  'array of arrays :',

  Object.values(['Bob', 'Bobby', 'Robert', 'Robby'].reduce(
    groupAndCollectWordsBySimilarStartSequences, {
      groups: {},
      length: 2,
    }
  ).groups)
);
console.log('\n');

console.log(
  'grouped index :',

  ['Bob', 'Bobby', 'Robert', 'Robby', 'Bob', 'Robby'].reduce(
    groupAndCollectWordsBySimilarStartSequences, {
      groups: {},
    }
  ).groups
);
console.log(
  'array of arrays :',

  Object.values(['Bob', 'Bobby', 'Robert', 'Robby'].reduce(
    groupAndCollectWordsBySimilarStartSequences, {
      groups: {},
    }
  ).groups)
);
console.log('\n');

console.log(
  'grouped index :',

  ['Bob', 'Bobby', 'Robert', 'Robby', 'Bob', 'Robby'].reduce(
    groupAndCollectWordsBySimilarStartSequences, {
      groups: {},
      length: 4,
    }
  ).groups
);
console.log(
  'array of arrays :',

  Object.values(['Bob', 'Bobby', 'Robert', 'Robby'].reduce(
    groupAndCollectWordsBySimilarStartSequences, {
      groups: {},
      length: 4,
    }
  ).groups)
);
console.log('\n');
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