Question

我正在编写一个脚本，通过比较两个数组来评估用户响应。（这是一个测验，看看他们对单词逐字逐句了解的信息。）我已经有了一些我需要的代码，比如将用户响应小写并拆分。我只需要找到差异/错误的数量。例如：

var correctanswer = ["The","quick","brown","fox","jumped","over","the","lazy","dog"];
var useranswer = ["The","brown","fox","jumped","up","and","over","the","really","lazy","cat"];
alert(counterrors(correctanswer, useranswer));

在这个特定示例中，运行我正在寻找的功能会返回用户犯了5个错误（省略“快速”，添加“up”， “和”，“真的”，改变“狗”到“猫”）。如您所见，这两个数组的长度可能不同。

有人知道如何处理这个问题吗？我当时认为这可能是一个循环：

for (x in correctanswer) {
    // compare correctanswer[x] to useranswer[x]... not sure how exactly. Seems tricky...
}

感谢您关注此事！我已经看过John Resig的diff解决方案（http://ejohn.org/projects/javascript-diff-algorithm/）和其他类似的东西，甚至是一些数组比较，但似乎没有任何工作，因为我发现的返回所有的差异，而我想知道有多少差异是。再次，感谢您的关注，请让我知道任何问题。

更新：非常感谢Magnar的回答！它运作得很好。

Answer 1

你所追求的是两个阵列的The Levenshtein Distance。

这是一种计算将一个序列转换为另一个序列所需的添加，删除和替换的数量。

Wikipedia page I linked有一个伪代码实现。我已经为您完成了JavaScript的逐行转换：

var correctanswer = ["The","quick","brown","fox","jumped","over","the","lazy","dog"];
var useranswer =    ["The","brown","fox","jumped","up","and","over","the","really","lazy","cat"];

console.log(calculate_levenshtein_distance(correctanswer, useranswer));

function calculate_levenshtein_distance(s, t) {
  var m = s.length + 1, n = t.length + 1;
  var i, j;

  // for all i and j, d[i,j] will hold the Levenshtein distance between
  // the first i words of s and the first j words of t;
  // note that d has (m+1)x(n+1) values
  var d = [];

  for (i = 0; i < m; i++) {
    d[i] = [i]; // the distance of any first array to an empty second array
  }
  for (j = 0; j < n; j++) {
    d[0][j] = j; // the distance of any second array to an empty first array
  }

  for (j = 1; j < n; j++) {
    for (i = 1; i < m; i++) {
      if (s[i - 1] === t[j - 1]) {
        d[i][j] = d[i-1][j-1];           // no operation required
      } else {
        d[i][j] = Math.min(
                    d[i - 1][j] + 1,     // a deletion
                    d[i][j - 1] + 1,     // an insertion
                    d[i - 1][j - 1] + 1  // a substitution
                  );
      }
    }
  }

  return d[m - 1][n - 1];
}

这会将5记录到控制台。正如您将看到的那样，阵列之间的距离正确。学生没有添加lazy。所以这是1次删除，3次添加和1次替换。

Answer 2

我不确定我是否完全明白你想要什么，但我认为这是解决方案。

function counterrors(a, b) {
    var max = Math.max(a.length, b.length);
    var min = Math.min(a.length, b.length);
    var count = 0;
    for (var i = 0; i < min; i+=1) {
        if (a[i] !== b[i]) {
            count += 1;
        }
    }
    return count + max - min;  // max - min for any extra things that don't match
}
var correctanswer = ["The", "quick", "brown", "fox", "jumped", "over", "the", "lazy", "dog"];
var useranswer = ["The", "brown", "fox", "jumped", "up", "and", "over", "the", "really", "lazy", "cat"];
alert(counterrors(correctanswer, useranswer));

使用JavaScript对用户响应进行评级（比较两个数组）

2 个答案: