何时使用几何平均数与算术平均数？
所以我想这在技术上不是一个代码问题，但我确信其他人和我自己在编写代码时都会遇到这个问题，所以希望它仍然是一个发布在 SO 上的好问题。
谷歌为我提供了很多关于何时使用财务数据等方面的详细解释。
但我的特定背景不适合，我想知道这里是否有人有一些见解。我需要收集大量个人用户对某个特定项目的“好”程度的投票。即，一定数量的用户每个给特定项目的分数在 0 到 10 之间，我想报告“典型”分数是多少。将几何和/或算术平均值报告为典型响应的直观原因是什么？
或者，就此而言，我最好报告中位数吗？
我想“最好”的方法可能涉及到一些心理学......
无论如何，你已经知道了。
谢谢！
So I guess this isn't technically a code question, but it's something that I'm sure will come up for other folks as well as myself while writing code, so hopefully it's still a good one to post on SO.
The Google has directed me to plenty of nice lengthy explanations of when to use one or the other as regards financial numbers, and things like that.
But my particular context doesn't fit in, and I'm wondering if anyone here has some insight. I need to take a whole bunch of individual users' votes on how "good" a particular item is. I.e., some number of users each give a particular item a score between 0 and 10, and I want to report on what the 'typical' score is. What would be the intuitive reasons to report the geometric and/or arithmetic mean as the typical response?
Or, for that matter, would I be better off reporting the median instead?
I imagine there's some psychology involved in what the "best" method might be...
Anyway, there you have it.
Thanks!
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一般来说，算术平均值就足够了。与几何平均值（涉及求 n 次方根）相比，它的计算强度要小得多。
至于所涉及的心理学，几何平均值永远不会大于算术平均值，因此如果您通常希望获得更高的分数，算术是最佳选择。
当数据集相对较小并且出现大量异常值的可能性相对较高时，中位数最有用。根据这些投票的精度，中位数有时可能会有点随意。
如果您确实想要最准确的答案，您可以计算算术几何平均值。然而，这涉及重复计算算术平均值和几何平均值，因此相比而言，计算量非常大。
Generally speaking, the arithmetic mean will suffice. It is much less computationally intensive than the geometric mean (which involves taking an n-th root).
As for the psychology involved, the geometric mean is never greater than the arithmetic mean, so arithmetic is the best choice if you'd prefer higher scores in general.
The median is most useful when the data set is relatively small and the chance of a massive outlier relatively high. Depending on how much precision these votes can take, the median can sometimes end up being a bit arbitrary.
If you really want the most accurate answer possible, you could go for calculating the arithmetic-geometric mean. However, this involves calculating both arithmetic and geometric means repeatedly, so it is very computationally intensive in comparison.
你想要算术平均值。因为你没有测量平均值的平均变化或其他东西。
you want the arithmetic mean. since you aren't measuring the average change in average or something.
算术平均值是正确的。
您的比例是人为的：
，但对于其他比例，您需要考虑 正确的使用方法。
其他一些例子
在数钱时，有人认为财富具有对数效用。因此，比尔·盖茨的财富与市中心的流浪汉之间的财富中位数将是一个中等成功的商人。 （算术平均值会让你震惊拉里佩奇。）
在测量声级时，分贝已经标准化了效果。所以你可以取分贝的算术平均值。
但如果您以瓦特为单位测量体积，则使用二次均值 (RMS)。
Arithmetic mean is correct.
Your scale is artificial:
But for other scales, you would need to consider the correct mean to use.
Some other examples
In counting money, it has been argued that wealth has logarithmic utility. So the median between Bill Gates' wealth and a bum in the inner city would be a moderately successful business person. (Arithmetic average would hive you Larry Page.)
In measuring sound level, decibels already normalizes the effect. So you can take arithmetic average of decibels.
But if you are measuring volume in watts, then use quadratic means (RMS).
答案取决于上下文和您的目的。有人提到百分比变化是使用几何平均值的好时机。我在计算天线和频率时使用几何平均值，因为百分比变化比频率范围的平均值或中间或天线的平均尺寸更重要。如果您的数字差异很大，特别是如果大多数数字相似，但一两个是“传单”（远离其他数字的范围），则几何平均值将“平滑”结果（不会让不同的数字对结果产生变化）超过他们应该的）。此方法用于计算子弹组大小（“传单”可能是人为错误，而不是设备错误，因此在这种情况下平均值“不公平”）。与几何平均数类似的另一种变化是均方根方法。你取数字的平方根，取平均值，然后对你的答案进行平方（这提供了更多的平滑），这通常用于电气计算，大多数电表都是以“RMS”（均方根）计算的，而不是。希望这会有所帮助，这是一个解释得很好的网站。
The answer depends on the context and your purpose. Percent changes were mentioned as a good time to use geometric mean. I use geometric mean when calculating antennas and frequencies since the percentage change is more important than the average or middle of the frequency range or average size of the antenna is concerned. If you have wildly varying numbers, especially if most are similar but one or two are "flyers" (far from the range of the others) the geometric mean will "smooth" the results (not let the different ones exert a change in the results more than they should). This method is used to calculate bullet group sizes (the "flyer" was probably human error, not the equipment, so the average is ""unfair" in that case). Another variation similar to geometric mean is the root mean square method. First you take the square root of the numbers, take THAT mean, and then square your answer (this provides even more smoothing). This is often used in electrical calculations and most electical meters are calculated in "RMS" (root mean square), not average readings. Hope this helps a little. Here is a web site that explains it pretty well. standardwisdom.com