查找一个名称数组中包含了多少次字符

Question

我正在尝试在JavaScript中进行以下练习：我需要返回字母“ l”的次数在一系列名称中使用。

这是我必须使用的数组的一个示例：

["earth (c-137)","abadango","citadel of ricks","worldender's lair","anatomy park","interdimensional cable","immortality field resort","post-apocalyptic earth","purge planet","venzenulon 7","bepis 9","cronenberg earth","nuptia 4","giant's town","bird world","st. gloopy noops hospital","earth (5-126)","mr. goldenfold's dream","gromflom prime","earth (replacement dimension)","testicle monster dimension","signus 5 expanse","earth (c-500a)","rick's battery microverse","the menagerie","earth (k-83)","hideout planet","unity's planet","dorian 5","earth (unknown dimension)","earth (j19ζ7)","roy: a life well lived","eric stoltz mask earth","earth (evil rick's target dimension)","planet squanch","glaagablaaga","resort planet","interdimensional customs","galactic federation prison","gazorpazorp"]

请注意，字母“ l”可能会在一个名称中重复。

我可以使用任何JS方法吗？我已经尝试了索引，但无法正常工作

Answer 1

const stringArr = [
  'earth (c-137)', 'abadango', 'citadel of ricks', 'worldender\'s lair', 'anatomy park', 'interdimensional cable',
  'immortality field resort', 'post-apocalyptic earth', 'purge planet', 'venzenulon 7', 'bepis 9', 'cronenberg earth',
  'nuptia 4', 'giant\'s town', 'bird world', 'st. gloopy noops hospital', 'earth (5-126)', 'mr. goldenfold\'s dream',
  'gromflom prime', 'earth (replacement dimension)', 'testicle monster dimension', 'signus 5 expanse', 'earth (c-500a)',
  'rick\'s battery microverse', 'the menagerie', 'earth (k-83)', 'hideout planet', 'unity\'s planet', 'dorian 5',
  'earth (unknown dimension)', 'earth (j19ζ7)', 'roy: a life well lived', 'eric stoltz mask earth',
  'earth (evil rick\'s target dimension)', 'planet squanch', 'glaagablaaga', 'resort planet',
  'interdimensional customs', 'galactic federation prison', 'gazorpazorp'
]

const result = stringArr
  .map(item => item.split('l').length - 1)
  .reduce((acc, value) => acc + value, 0)

console.log(result)

Answer 2

尝试以下操作：

myStr.split('<my_char>').length - 1. 

每次您看到所需的字符时，都会拆分字符串，并在定界线减去1分裂后返回字符串数，这正是您想要的。在将字符串数组转换为单个字符串后执行此操作。 mystr = mystrary.tostring（）;

Answer 3

这是一种使用阵列方法” noreferrer'> foreach和a ...代码>循环

const stringArr = ["earth (c-137)","abadango","citadel of ricks","worldender's lair","anatomy park","interdimensional cable","immortality field resort","post-apocalyptic earth","purge planet","venzenulon 7","bepis 9","cronenberg earth","nuptia 4","giant's town","bird world","st. gloopy noops hospital","earth (5-126)","mr. goldenfold's dream","gromflom prime","earth (replacement dimension)","testicle monster dimension","signus 5 expanse","earth (c-500a)","rick's battery microverse","the menagerie","earth (k-83)","hideout planet","unity's planet","dorian 5","earth (unknown dimension)","earth (j19ζ7)","roy: a life well lived","eric stoltz mask earth","earth (evil rick's target dimension)","planet squanch","glaagablaaga","resort planet","interdimensional customs","galactic federation prison","gazorpazorp"]


let count = 0;

stringArr.forEach((string) => {
  for (let letter of string) {
    if (letter.toLowerCase() === "l") count += 1;
  }
});

console.log(count);

Answer 4

您可以使用与您要匹配的角色匹配的简单正则正等式，而在数组项目上迭代时。

const arrayToSearch = ["earth (c-137)", "abadango", "citadel of ricks", "worldender's lair", "anatomy park", "interdimensional cable", "immortality field resort", "post-apocalyptic earth", "purge planet", "venzenulon 7", "bepis 9", "cronenberg earth", "nuptia 4", "giant's town", "bird world", "st. gloopy noops hospital", "earth (5-126)", "mr. goldenfold's dream", "gromflom prime", "earth (replacement dimension)", "testicle monster dimension", "signus 5 expanse", "earth (c-500a)", "rick's battery microverse", "the menagerie", "earth (k-83)", "hideout planet", "unity's planet", "dorian 5", "earth (unknown dimension)", "earth (j19ζ7)", "roy: a life well lived", "eric stoltz mask earth", "earth (evil rick's target dimension)", "planet squanch", "glaagablaaga", "resort planet", "interdimensional customs", "galactic federation prison", "gazorpazorp"];

const REGEX_MATCH = /l/ig;

function getCount(array, regex) {
  const matchCount = array.reduce((memo, item) => {
    const matches = item.match(REGEX_MATCH);

    if (matches && matches.length > 0) {
      memo += matches.length;
    }

    return memo;
  }, 0);

  return matchCount;
}

const count = getCount(arrayToSearch, REGEX_MATCH);

console.log(`count is ${count}`);

Answer 5

const data = ["earth (c-137)", "abadango", "citadel of ricks", "worldender's lair", "anatomy park", "interdimensional cable", "immortality field resort", "post-apocalyptic earth", "purge planet", "venzenulon 7", "bepis 9", "cronenberg earth", "nuptia 4", "giant's town", "bird world", "st. gloopy noops hospital", "earth (5-126)", "mr. goldenfold's dream", "gromflom prime", "earth (replacement dimension)", "testicle monster dimension", "signus 5 expanse", "earth (c-500a)", "rick's battery microverse", "the menagerie", "earth (k-83)", "hideout planet", "unity's planet", "dorian 5", "earth (unknown dimension)", "earth (j19ζ7)", "roy: a life well lived", "eric stoltz mask earth", "earth (evil rick's target dimension)", "planet squanch", "glaagablaaga", "resort planet", "interdimensional customs", "galactic federation prison", "gazorpazorp"]

const howMany = (data, char) => data.join('').match(new RegExp(char, 'g')).length

console.log(howMany(data, 'p'))