我的实验有36个比较,3个间隔,21个数据点每隔室。数据集有2268个数据集。总共。第一行如下所示:
Comparison Value Compartment
A -0.715126087 1
A -0.481391603 2
A 0.374693449 1
A null 2
A 0.857450232 1
A null 2
A -1.608637992 3
A -1.670859336 3
A -1.731618976 3
...
AJ -5.84850106 3我对每项比较进行了一次方差分析,并检验了假设,但在某些比较中有违规之处。我想运行一个非参数Brown-Forsythe F*测试,然后进行成对比较。我的尝试:
if(!require(onewaytests)){install.packages("onewaytests")} # for Brown-Forsythe Test
if(!require(tidyverse)){install.packages("tidyverse")}
if(!require(broom)){install.packages("broom")}
fm3 <- my_data %>%
group_by(Comparison) %>%
do(multitst = paircomp(bf.test(Value ~ Compartment, data = .)))
fm3 %>% tidy(multitst)问题: fm3不是创建的,因为它一碰到第一个没有统计意义的比较就会停止运行。这是我得到的错误:
误差在paircomp.owt中(bf.test(值~室,:成对比较)不能进行,因为差异没有统计学意义(α=0.0 5)。
希望:在这样的表中有一个输出:
Comparison term comparison estimate conf.low conf.high adj.p.value
1 A Compartment 1-2 3.360531 2.9495551 3.7715075 2.023126e-11
2 A Compartment 3-2 3.537098 3.1293925 3.9448041 2.023126e-11
3 A Compartment 3-1 0.176567 -0.2382312 0.5913652 5.779274e-01
...
108 AJ Compartment 3-1 4.50731363 2.85507469 6.1595525749 1.954819e-07有人能帮忙吗?提前感谢!
发布于 2020-06-09 00:49:06
您可以包含一个ifelse()语句,在比较没有意义时填充NA,当比较没有显着性时填充结果。
Not familiar with Brown-Forsythe test, but my go-to in such cases is a randomization test. I guess the decision should be based upon how much information is being lost in using this particular non-parametric test.https://stackoverflow.com/questions/62223978
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