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，每当 allPlayers[i] 出现在 roundWinners 中时，allPlayers 中的每个玩家都会递增 playerScores。

目前 imJbVrPfGG 只加注到 1 的计数，而应该是 3.

数组：

allPlayers = ["iMJbVrPfGG","PjCtgHizDF","iQtZ88qyeX","zhvd8mAdhl"]

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

代码:

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]++;
    }
}

Answer 1

您的 if 语句设置不正确。现在，当你增加回合索引时，你正在增加你的玩家索引......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，然后更新分数是这样的：

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]++;
    }
}

请记住，我假设 playerScores 数组保持与 allPlayers 数组完全同步。

Answer 2

由于每个获胜者可能有多场比赛，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 3

你需要两个循环。外循环将遍历每个玩家，内循环将搜索并计算所有匹配项。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]++;
        }
    }
}

计算元素是否出现在两个数组中

Counting If Element Appears in Both Arrays

javascript

arrays

counting