在Python中使用P值进行F检验

时间:2015-01-26 07:09:00

标签: python r statistics

R允许我们计算两个群体之间的F检验:

> d1 = c(2.5579227634, 1.7774243136, 2.0025207896, 1.9518876366, 0.0, 4.1984191803, 5.6170403364, 0.0)
> d2 = c(16.93800333, 23.2837045311, 1.2674791828, 1.0889208427, 1.0447584137, 0.8971380534, 0.0, 0.0)
> var.test(d1,d2)

    F test to compare two variances

data:  d1 and d2
F = 0.0439, num df = 7, denom df = 7, p-value = 0.000523
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
 0.008789447 0.219288957
sample estimates:
ratio of variances 
        0.04390249 

请注意,它也会报告P值。

另一个例子,R给出了这个:

> x1 = c(0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 68.7169110318)
> x2 = c(0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 2.1863361211)
> var.test(x1,x2)
#p-value = 1.223e-09

Python中的等价物是什么? 我检查了这个documentation,但似乎没有给出我想要的东西。

此代码提供不同的P值(尤其是示例2):

import statistics as stats
import scipy.stats as ss
def Ftest_pvalue(d1,d2):
    """docstring for Ftest_pvalue"""
    df1 = len(d1) - 1
    df2 = len(d2) - 1
    F = stats.variance(d1) / stats.variance(d2)
    single_tailed_pval = ss.f.cdf(F,df1,df2)
    double_tailed_pval = single_tailed_pval * 2
    return double_tailed_pval

Python给出了这个:

In [45]: d1 = [2.5579227634, 1.7774243136, 2.0025207896, 1.9518876366, 0.0, 4.1984191803, 5.6170403364, 0.0]
In [20]: d2 = [16.93800333, 23.2837045311, 1.2674791828, 1.0889208427, 1.0447584137, 0.8971380534, 0.0, 0.0]
In [64]: x1 = [0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 68.7169110318]
In [65]: x2 = [0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 2.1863361211]

In [69]: Ftest_pvalue(d1,d2)
Out[69]: 0.00052297887612346176

In [70]: Ftest_pvalue(x1,x2)
Out[70]: 1.9999999987772916

2 个答案:

答案 0 :(得分:2)

rpy2实施:

import rpy2.robjects as robjects
def Ftest_pvalue_rpy2(d1,d2):
    """docstring for Ftest_pvalue_rpy2"""
    rd1 = (robjects.FloatVector(d1))
    rd2 = (robjects.FloatVector(d2))
    rvtest = robjects.r['var.test']
    return rvtest(rd1,rd2)[2][0]

结果如下:

In [4]: x1 = [0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 68.7169110318]
In [5]: x2 = [0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 2.1863361211]
In [6]: Ftest_pvalue_rpy2(x1,x2)
Out[6]: 1.2227086010341282e-09

答案 1 :(得分:1)

我应该提到xalglib是一个包含统计方法的包,允许这样做: http://www.alglib.net/ http://www.alglib.net/hypothesistesting/variancetests.php 虽然它不如基于scipy的原始方法灵活。

我应该提到正确的双尾计算程序(在variancetests.c中)可以找到:

stat = ae_minreal(xvar / yvar,yvar / xvar,_state);     * bothtails = 1-(fdistribution(df1,df2,1 / stat,_state)-fdistribution(df1,df2,stat,_state))

虽然@Amit Kumar Gupta在他的评论中描述的是假的(如果你只是加倍1和单边p值之间的差异,你可以达到1以上的值)