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R 弗里德曼&x27;s测试手动计算_R_Statistics_Calculation - Fatal编程技术网

R 弗里德曼&x27;s测试手动计算

R 弗里德曼&x27;s测试手动计算,r,statistics,calculation,R,Statistics,Calculation,我从弗里德曼的测试中得到了一个负值。数据如下: Full MIC ReliefF LCorrel InfoGain equinox 69.939 80.178 78.794 75.205 62.268 lucene 78.175 84.103 79.017 82.044 75.564 mylyn 75.531 78.006 77.161 47.711 81.575 pde 70.282 82.686 81.884 75.0

我从弗里德曼的测试中得到了一个负值。数据如下:

        Full    MIC     ReliefF LCorrel InfoGain
equinox 69.939  80.178  78.794  75.205  62.268
lucene  78.175  84.103  79.017  82.044  75.564
mylyn   75.531  78.006  77.161  47.711  81.575
pde     70.282  82.686  81.884  75.07   79.476
jdt     71.675  93.202  95.387  85.878  82.818
排名低于

        Full    MIC     ReliefF LCorrel InfoGain
equinox 2       5       4       3       1
lucene  2       5       3       4       1
mylyn   2       4       3       1       5
pde     1       5       4       2       3
jdt     1       4       5       3       2
Sum     8      23      19      13      12
弗里德曼F计算公式:

F = (5/[5*5*(5+1)] * [8*8 + 23*23 + 19*19 + 13*13 + 12*12] - [5*5*(5+1)]
我得到的值是
-107.7666667

我怎么解释呢?我看到的例子都有积极的结果


我知道R代码,但需要手动计算。

这就是我生成结果的方式,它可以工作

pacc_part
f1 <- friedman.test(pacc_part) 
print (f1) 
# Post-hoc tests are conducted only if omnimus Kruskal-Wallis test p-value 
is 0.05 or less. 
if ( f1$p.value < 0.05 ) 
{ 

n1 <- posthoc.friedman.nemenyi.test(pacc_part) 
} 
n1; 
# alternate representation of post-hoc test results 
summary(n1); 
pacc\u零件

f1试一试abs
可能吗?