计算元素是否出现在两个数组中

Question

我正在尝试创建一个计数器，每playerScores allPlayers allPlayers[i]出现roundWinners，每个玩家的imJbVrPfGG都会增加{。}}。

目前1仅提升为3，而应为allPlayers = ["iMJbVrPfGG","PjCtgHizDF","iQtZ88qyeX","zhvd8mAdhl"]。

阵列

roundWinners = ["iMJbVrPfGG","iMJbVrPfGG","iMJbVrPfGG"]

for (i = 0; i < allPlayers.length; i++) { playersScores[i] = 0; } for (i = 0; i < roundWinners.length; i++) { if(roundWinners[i] == allPlayers[i]){ var index = allPlayers.indexOf(roundWinners[i]) playersScores[index]++; } }

代码：

{{1}}

Answer 1

由于每个获胜者可能有多个匹配项，因此您需要一个内部循环：

for (i = 0; i < allPlayers.length; i++) {
    playersScores[i] = 0;
    for (j = 0; j < roundWinners.length; j++) {
        if (roundWinners[j] == allPlayers[i]) playersScores[i]++;
    }
}

结果：[3, 0, 0, 0]

演示：http://jsfiddle.net/3pm0msh6/1/

Answer 2

您的if语句设置不正确。现在，当您递增圆形索引时，您将增加播放器索引...并将Spot 0与Spot 0，Spot 1与Spot 1进行比较。相反，您只需要查看数组中是否存在该值，并像这样更新分数：

for (i = 0; i < allPlayers.length; i++) {
    playersScores[i] = 0;
}
for (i = 0; i < roundWinners.length; i++) {
    var index = allPlayers.indexOf(roundWinners[i])
    if(index >= 0) {
        playersScores[index]++;
    }
}

请注意，我假设playerScores数组保持与allPlayers数组完全同步。

Answer 3

你需要两个循环。外部循环将遍历每个玩家，内部循环将搜索并计算所有匹配。例如：

for (i = 0; i < allPlayers.length; i++) {
    playersScores[i] = 0;

    for (var j = 0; j < roundWinners.length; j++) { 
        if(roundWinners[j] == allPlayers[i]){
            playersScores[i]++;
        }
    }
}