使用JavaScript級用戶的響應（比較兩個陣列）

Question

我工作的一個腳本來分級，通過比較兩個陣列的用戶響應。 （這是一個測驗，看他們如何很好地知道信息。）我已經有一些我需要的代碼，比如讓用戶響應小寫並分割它。我需要的是找出差異/錯誤的數量。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)); 

在這個特殊的例子，運行我正在尋找的函數將返回用戶做出5個錯誤（他們省略「快」，補充「上」，「和」 ，「真的」，並且將「狗」改爲「貓」）。如你所見，這兩個陣列的長度可能不同。

有誰知道如何處理呢？我想這可能是一個循環，如：

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

謝謝你看這個！我見過約翰Resig的差異溶液（http://ejohn.org/projects/javascript-diff-algorithm/）以及其他類似的東西，甚至幾陣比較，但似乎沒有什麼，因爲我發現返回的所有分歧，而我想看看有多少差異存在是那些工作。再次感謝您的關注，並請讓我知道任何問題。

更新：非常感謝Magnar的答案！它工作完美。

Answer 1

你所追求的是兩個陣列的The Levenshtein Distance。

它是計算一個序列轉換成另一種的添加，缺失和取代必要數量的算法。

的Wikipedia page I linked具有僞代碼實現。我已經做了線換行轉換爲JavaScript爲您提供：

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]; 
} 

這將記錄5到控制檯。正如你將會看到的那樣，數組之間的距離是正確的。該學生沒有添加lazy。所以它是1個刪除，3個添加和1個替換。

Answer 2

我不知道如果我完全理解你想要什麼，但我認爲這是解決方案。

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));