R 弗里德曼&x27;s测试手动计算
我从弗里德曼的测试中得到了一个负值。数据如下: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
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?