我可以计算这些数字的平均值吗?

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我想知道是否可以计算一些数字的平均值:

I was wondering if it's possible to calculate the average of some numbers if I have this:

int currentCount = 12; float currentScore = 6.1123 (this is a range of 1 <-> 10).

现在，如果我得到另一个分数(比如4.5)，我可以重新计算平均值了吗?

Now, if I receive another score (let's say 4.5), can I recalculate the average so it would be something like:

int currentCount now equals 13 float currentScore now equals ?????

或者这是不可能的，我仍然需要记住分数列表吗?

or is this impossible and I still need to remember the list of scores?

推荐答案

以下公式可让您根据需要从存储的平均值和计数中跟踪平均值.

The following formulas allow you to track averages just from stored average and count, as you requested.

currentScore = (currentScore * currentCount + newValue) / (currentCount + 1) currentCount = currentCount + 1

这取决于您的平均值是当前的总和除以计数的事实.因此，您只需将计数乘以平均值即可得到总和，将新值相加并除以(count + 1)，然后增加计数.

This relies on the fact that your average is currently your sum divided by the count. So you simply multiply count by average to get the sum, add your new value and divide by (count+1), then increase count.

因此，假设您有数据{7,9,11,1,12}，并且唯一要保留的是平均值和计数.随着每个数字的添加，您将得到:

So, let's say you have the data {7,9,11,1,12} and the only thing you're keeping is the average and count. As each number is added, you get:

+--------+-------+----------------------+----------------------+ | Number | Count | Actual average | Calculated average | +--------+-------+----------------------+----------------------+ | 7 | 1 | (7)/1 = 7 | (0 * 0 + 7) / 1 = 7 | | 9 | 2 | (7+9)/2 = 8 | (7 * 1 + 9) / 2 = 8 | | 11 | 3 | (7+9+11)/3 = 9 | (8 * 2 + 11) / 3 = 9 | | 1 | 4 | (7+9+11+1)/4 = 7 | (9 * 3 + 1) / 4 = 7 | | 12 | 5 | (7+9+11+1+12)/5 = 8 | (7 * 4 + 12) / 5 = 8 | +--------+-------+----------------------+----------------------+