如何在Python中计算cohen的d？

Question

我需要计算cohen的d来确定实验的效果大小.我可以使用的声音库中是否有任何实现？如果没有,那么什么是好的实施？

Answer 1

在两组具有相同大小的特殊情况下,上述实现是正确的.基于维基百科和Robert Coe文章中的公式的更一般的解决方案是下面显示的第二种方法.请注意,分母是汇总标准差,通常只有在两组人口标准差相等的情况下才适用:

from numpy import std, mean, sqrt

#correct if the population S.D. is expected to be equal for the two groups.
def cohen_d(x,y):
    nx = len(x)
    ny = len(y)
    dof = nx + ny - 2
    return (mean(x) - mean(y)) / sqrt(((nx-1)*std(x, ddof=1) ** 2 + (ny-1)*std(y, ddof=1) ** 2) / dof)

#dummy data
x = [2,4,7,3,7,35,8,9]
y = [i*2 for i in x]
# extra element so that two group sizes are not equal.
x.append(10)

#correct only if nx=ny
d = (mean(x) - mean(y)) / sqrt((std(x, ddof=1) ** 2 + std(y, ddof=1) ** 2) / 2.0)
print ("d by the 1st method = " + str(d))
if (len(x) != len(y)):
    print("The first method is incorrect because nx is not equal to ny.")

#correct for more general case including nx !=ny
print ("d by the more general 2nd method = " + str(cohen_d(x,y)))

输出将是:

d by 1st method = -0.559662109472第一种方法不正确,因为nx不等于ny.d由更一般的第二种方法= -0.572015604666

Answer 2

从Python3.4开始,您可以使用该statistics模块计算扩展和平均指标.有了这个,Cohen的d可以很容易地计算出来:

from statistics import mean, stdev
from math import sqrt

# test conditions
c0 = [2, 4, 7, 3, 7, 35, 8, 9]
c1 = [i * 2 for i in c0]

cohens_d = (mean(c0) - mean(c1)) / (sqrt((stdev(c0) ** 2 + stdev(c1) ** 2) / 2))

print(cohens_d)

输出:

-0.5567679522645598

所以我们观察到中等效果.

Answer 3

在 Python 2.7 中，您可以使用numpy一些警告，正如我在改编 Python 3.4 中 Bengt 的答案时发现的那样。

1. 确保除法始终返回浮点数：from __future__ import division
2. 将方差的除法参数指定ddof=1为std函数 ，即numpy.std(c0, ddof=1)。numpy 的标准差默认行为是除以n，而 withddof=1它会除以n-1。

代码

from __future__ import division #Ensure division returns float
from numpy import mean, std # version >= 1.7.1
from math import sqrt
import sys
        

def cohen_d(x,y):
        return (mean(x) - mean(y)) / sqrt((std(x, ddof=1) ** 2 + std(y, ddof=1) ** 2) / 2.0)

if __name__ == "__main__":                
        # test conditions
        c0 = [2, 4, 7, 3, 7, 35, 8, 9]
        c1 = [i * 2 for i in c0]
        print(cohen_d(c0,c1))

输出将是：

-0.556767952265