我有一个javascript对象，我想通过id找到一个用户并更新他们的余额

我想这会达到你的目的，

var user = leaderboard.filter(function(user){return userId == msg.userId})[0];
if(user){
   user.balance = msg.userBalance;
} else {
   leaderboard.push({userId: msg.userId, balance: msg.userBalance});
   leaderboard.sort(function(a,b){return a.balance-b.balance});
   leaderboard.shift();     
}

我在这里做的是，我使用过滤器来获得与目标用户id匹配的用户，，如果没有找到，我将它添加到数组中，然后对数组进行升序排序，然后删除第一个元素，，这将是最低的

这可能不是最有效的方法，但是已经很晚了，而且我很累。

var lowest = Number.POSITIVE_INFINITY;             // set the lowest value to something really high
$.each(leaderboard, function(idx,item){            // get the lowest value
    if(item.balance < lowest){
        lowest = item.balance;
    }
});
var i = $.grep(leaderboard, function(item){        // get the item based on userId, if it already exists
    return item.userId == msg.userId;
})[0];
if(!i){                                                     // if it doesnt exist...
    if(msg.userBalance > lowest){                           // if the new items balance is greater than the lowest one
        leaderboard.push(msg);                              // add it
        leaderboard = $.grep(leaderboard, function(item){   // and remove the item with the lowest value from the array
            return item.balance != lowest;
        });
    }
} else {                                           // if it does exists
    i.balance = msg.userBalance;                   // update the balance
}

这是一把能用的小提琴。

让它按照你开始的方式工作是可能的——我从来都不是$的粉丝。地图，没有特别的原因。此解决方案只是可以实现的众多方法之一，并且可能帮助您了解所需的逻辑。

解决方案一次遍历数组。但是我们需要从头遍历一次来初始化子对象的值。

    leaderboard.max = -Infinity; // Set max balance
    leaderboard.min = Infinity; // Set min balance
    leaderboard.forEach( function( item, index, self ) {
        if (item.balance>self.max) self.max = item.balance;
        if (item.balance<self.min) self.min = item.balance;    
    });
    leaderboard.updateBalance = function ( userId, newBalance ) {
        var updated = false;
        var minIndex1 = -Infinity;
        var minIndex2 = Infinity;
        this.forEach( function( item, index, self ) {
            // Find 2 index with minimal balance
            if (item.balance === self.min) minIndex1 = index;
            if (item.balance > self.min && item.balance < minIndex2) minIndex2 = index;
            // If find user - update balance and min and max
            if (item.userId === userId) {
                updated = true; // Flag - user balance updated
                self[index].balance = newBalance;
                if (newBalance > self.max) self.max = newBalance;
                if (newBalance < self.min) self.min = newBalance;            
            }
        });
        // No user find? Push
        if (!updated && newBalance > leaderboard.max) {
            this.min = this[minIndex2].balance;
            if (this.length===5) this.splice(minIndex1, 1); // Remove user with minimal balance
            this.max = newBalance;
            this.push( { userId: userId, balance: newBalance } );
        }
    };

您可能会考虑将对象列表重组为散列表。它具有一个好的对象的所有条件，因为您的对象只是键值对。

那么你就可以更有力、更简单地根据键来参考你的平衡，以实现你所追求的。

// Initialise hash table
var leaderboard = {
    {10293, 1023}
    {20394, 1806},
    {45333, 2064},
    {57456, 2453},
    {24575, 2703}
};
// Initialise person and msg somewhere
// If the person exists already, update their balance
if(leaderboard[person.userId] != null) {
    leaderboard[person.userId] = msg.userBalance;
    return; // Return early, we've done our work
}
// If the person's balance is greater than any of the others in the hash, replace them
for(var i in leaderboard) {
    if(leaderboard[i] < msg.userBalance) {
        delete leaderboard[i];
        leaderboard[person.userId] = msg.userBalance
    }
}