Python：在python中做几何平均值的简单方法？
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Python : easy way to do geometric mean in python?
提问by Marcin Wojnarski
I wonder is there any easy way to do geometric mean using python but without using python package. If there is not, is there any simple package to do geometric mean?
我想知道有没有什么简单的方法可以使用 python 但不使用 python 包来做几何平均。如果没有，是否有任何简单的包来做几何平均？
回答by Willem Van Onsem
The formula of the gemetric mean is:
几何均值的公式为：
So you can easily write an algorithm like:
因此，您可以轻松编写如下算法：
import numpy as np
def geo_mean(iterable):
a = np.array(iterable)
return a.prod()**(1.0/len(a))You do not have to use numpy for that, but it tends to perform operations on arrays faster than Python (since there is less "overhead" with casting).
您不必为此使用 numpy，但它倾向于比 Python 更快地对数组执行操作（因为转换的“开销”较少）。
In case the chances of overfloware high, you can map the numbers to a logdomain first, calculate the sum of these logs, then multiply by 1/n and finally calculate the exponent, like:
如果溢出的可能性很高，您可以先将数字映射到对数域，计算这些对数的总和，然后乘以 1/n，最后计算指数，例如：
import numpy as np
def geo_mean_overflow(iterable):
a = np.log(iterable)
return np.exp(a.sum()/len(a))回答by Marcin Wojnarski
In case someone is looking here for a library implementation, there is gmean()in scipy, possibly faster and numerically more stable than a custom implementation:
如果有人在这里寻找库实现，scipy 中有gmean()，可能比自定义实现更快且数值更稳定：
>>> from scipy.stats.mstats import gmean
>>> gmean([1.0, 0.00001, 10000000000.])
46.415888336127786
回答by Xavier Guihot
Starting Python 3.8, the standard library comes with the geometric_meanfunction as part of the statisticsmodule:
开始Python 3.8，标准库附带该geometric_mean功能作为statistics模块的一部分：
from statistics import geometric_mean
geometric_mean([1.0, 0.00001, 10000000000.]) // 46.415888336127786
回答by rmmh
Here's an overflow-resistant version in pure Python, basically the same as the accepted answer.
这是纯Python中的防溢出版本，与接受的答案基本相同。
import math
def geomean(xs):
return math.exp(math.fsum(math.log(x) for x in xs) / len(xs))
回答by Liam
just do this:
只需这样做：
numbers = [1, 3, 5, 7, 10]
print reduce(lambda x, y: x*y, numbers)**(1.0/len(numbers))
回答by user12295593
Geometric mean
几何平均数
import pandas as pd
geomean=Variable.product()**(1/len(Variable))
print(geomean)
Geometric mean with Scipy
Scipy 的几何平均值
from scipy import stats
print(stats.gmean(Variable))
回答by gil.fernandes
You can also calculate the geometrical mean with numpy:
您还可以使用 numpy 计算几何平均值：
import numpy as np
np.exp(np.mean(np.log([1, 2, 3])))
result:
结果：
1.8171205928321397
