Question

我一直试图想出这个练习的答案，但是我不能看到它的错误。我将分享练习的描述，我的代码，最后是我的输出。

描述：

现在我们将在一个名为array_lengths

的新数组中保存每个数组的最长长度

打开控制台并重新加载索引页以查看代码的结果。

编写一个charFreq（）函数，它接受一个字符串并构建其中包含的字符的频率列表。

作为条件，您必须使用对象freqCounter。

将频率列表表示为Javascript对象。尝试使用像charFreq（“abbabcbdbabdbdbabababcbcbab”）这样的东西。

CODE：

function charFreq(string) {

  var array_lengths = [];

  // compute frequencies of each value
  for (var i = 0; i < string.length; i++) {
    value = string[i];
    if (value in array_lengths) {
      array_lengths[value] ++;
    } else {
      array_lengths[value] = 1;
    }
  }


  // make array from the frequency object to de-duplicate
  var freqCounter = [];
  for (value in array_lengths) {
    freqCounter.push(value);
  }

  // sort the uniques array in descending order by frequency
  function compareFrequency(a, b) {
    return array_lengths[b] - array_lengths[a];
  }

  return freqCounter.sort(compareFrequency);
}

//OUTPUT: 

// if I had it right I should get "true" from every of the following values of the function but the point is that I'm getting all false...where's the error?

counter = charFreq("abbabcbdbabdbdbabababcbcbab");

console.log(counter['a'] === 7);

console.log(counter.b === 14);

console.log(counter['c'] === 3);

Answer 1

两个问题

使用对象
```
var array_lengths = {}; // object
```
返回该对象而不进行排序和其他内容
```
return array_lengths;
```

function charFreq( string ) {
    var array_lengths = {}; // object

    // compute frequencies of each value
    for(var i = 0; i < string.length; i++) {
        value = string[i];
        if(value in array_lengths) {
            array_lengths[value]++;
        } else {
            array_lengths[value] = 1;
        }
    }
    return array_lengths;
}

//OUTPUT:
// if I had it right I should get "true" from every of the following values of the function but the point is that I'm getting all false...where's the error?

counter = charFreq("abbabcbdbabdbdbabababcbcbab");
console.log(counter);
console.log( counter['a'] === 7);
console.log( counter.b === 14);
console.log( counter['c'] === 3);

Answer 2

分解。你需要迭代字符串的每个部分，对吗？将其拆分为数组并为对象中的每个元素创建一个计数器：

var counter = {};
var string = "abbabcbdbabdbdbabababcbcbab";
var stringArray = string.split('');
stringArray.forEach(function(character){
  if(!counter.hasOwnProperty(character)){
    counter[character] = 0;
  }
  counter[character]++;
});
console.log(counter);

Answer 3

您可以使用ES6执行此操作 - 使用Set通过将字符串转换为数组并将整个内容传递到let path = NSBundle.mainBundle().pathForResource("movielist2", ofType: "txt") let filemgr = NSFileManager.defaultManager() if filemgr.fileExistsAtPath(path!){ do{ let fullText = try String(contentsOfFile: path!, encoding: NSUTF8StringEncoding) let readings = fullText.componentsSeparatedByString("@") as [String] for i in 1..<readings.count { let movieData = readings[i].componentsSeparatedByString("\t") dictMovies["MovieTitle1"] = "\(movieData[0])" dictMovies["MovieYear"] = "\(movieData[1])" arrayMovies.addObject(dictMovies) } } catch let error as NSError{ print("Error: \(error)") } }

来获取数组中的唯一项

new Set

Answer 4

您正在使用数组作为对象我留下了var freqCounter = Object.keys(array_lengths);，因为它是字符串中的唯一字符。但是，它在您描述的代码中没有为您做任何事情。

＆＃13;

function charFreq(string) {

  var array_lengths = {};
  var value;
  // compute frequencies of each value
  for (var i = 0; i < string.length; i++) {
    value = string[i];
    if (value in array_lengths) {
      array_lengths[value] ++;
    } else {
      array_lengths[value] = 1;
    }
  }


  // make array from the frequency object to de-duplicate
  var freqCounter = Object.keys(array_lengths);

  // sort the uniques array in descending order by frequency
  function compareFrequency(a, b) {
    return array_lengths[b] - array_lengths[a];
  }

  return array_lengths;
}

//OUTPUT: 

// if I had it right I should get "true" from every of the following values of the function but the point is that I'm getting all false...where's the error?

counter = charFreq("abbabcbdbabdbdbabababcbcbab");
console.log(counter);
console.log(counter['a'] === 7);

console.log(counter.b === 14);

console.log(counter['c'] === 3);

＆＃13;

Javascript计算字符串的频率字母

4 个答案: