我执行了wilcox测试,现在我想将p.value提取到列表或矩阵中。
DF <- data.frame(A1=sample(1:9, 10, T),
A2=sample(1:9, 10, T),
A3=sample(1:9, 10, T),
B1=sample(1:9, 10, T),
B2=sample(1:9, 10, T),
B3=sample(1:9, 10, T))
sampA <- DF[,grep('A', names(DF))] # Sample with columns A
sampB <- DF[,grep('B', names(DF))] # Sample with columns B
lapply(1:nrow(DF), function(i){
wilcox.test(as.numeric(sampA[i,]), as.numeric(sampB[i,]), exact=FALSE )
})
我对每行的wilcox测试结果如下: 我想知道如何在列表或矩阵中获取p.value以导出到excel文件? [[1]]
Wilcoxon rank sum test with continuity correction
data: as.numeric(sampA[i, ]) and as.numeric(sampB[i, ])
W = 3, p-value = 0.6579
alternative hypothesis: true location shift is not equal to 0
[[2]]
Wilcoxon rank sum test with continuity correction
data: as.numeric(sampA[i, ]) and as.numeric(sampB[i, ])
W = 0, p-value = 0.0722
alternative hypothesis: true location shift is not equal to 0
[[3]]
Wilcoxon rank sum test with continuity correction
data: as.numeric(sampA[i, ]) and as.numeric(sampB[i, ])
W = 6, p-value = 0.6579
alternative hypothesis: true location shift is not equal to 0
答案 0 :(得分:7)
只需添加$p.value
即可从wilcox.test
对象中提取您的p.value:
lapply(1:nrow(DF), function(i){
wilcox.test(as.numeric(sampA[i,]), as.numeric(sampB[i,]), exact=FALSE )$p.value
})
给出了:
[[1]]
[1] 1
[[2]]
[1] 1
[[3]]
[1] 0.8247781
[[4]]
[1] 0.8247781
使用sapply
代替lapply
,您将获得一个向量而不是列表,这可能更容易操作。
sapply(1:nrow(DF), function(i){
wilcox.test(as.numeric(sampA[i,]), as.numeric(sampB[i,]), exact=FALSE )$p.value
})
# [1] 1.0000000 1.0000000 0.8247781 0.8247781 0.0765225 0.8247781 1.0000000
# [8] 0.8247781 0.2682859 0.0765225
答案 1 :(得分:2)
使用sapply
并引用p.value名称...
sapply(1:nrow(DF), function(i){wilcox.test(as.numeric(sampA[i,]), as.numeric(sampB[i,]), exact=FALSE )$p.value})
#[1] 0.8247781 0.0765225 0.8247781 1.0000000 0.2682859 0.6625206 1.0000000
#[8] 0.1211833 0.5065552 0.8247781
这将返回一个向量而不是一个列表(我通常会发现向量更容易处理,例如因为你不能使用mylist[[1:5]]
索引列表,但你可以做myvector[1:5]
但是这是个人偏好