我试图对我的数据框的所有列(一次两个)进行t检验,并仅提取p值.以下是我的想法:
for (i in c(5:525) ) {
t_test_p.value =sapply( Data[5:525], function(x) t.test(Data[,i],x, na.rm=TRUE)$p.value)
}
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我的问题是:1.有没有办法在没有循环的情况下做到这一点?2.如何捕获t检验的结果.
koh*_*ske 16
我建议将您的数据框转换为长格式并使用pairwise.t.test适当的p.adjust:
> library(reshape2)
>
> df <- data.frame(a=runif(100),
+ b=runif(100),
+ c=runif(100)+0.5,
+ d=runif(100)+0.5,
+ e=runif(100)+1,
+ f=runif(100)+1)
>
> d <- melt(df)
Using as id variables
>
> pairwise.t.test(d$value, d$variable, p.adjust = "none")
Pairwise comparisons using t tests with pooled SD
data: d$value and d$variable
a b c d e
b 0.86 - - - -
c <2e-16 <2e-16 - - -
d <2e-16 <2e-16 0.73 - -
e <2e-16 <2e-16 <2e-16 <2e-16 -
f <2e-16 <2e-16 <2e-16 <2e-16 0.63
P value adjustment method: none
> pairwise.t.test(d$value, d$variable, p.adjust = "bon")
Pairwise comparisons using t tests with pooled SD
data: d$value and d$variable
a b c d e
b 1 - - - -
c <2e-16 <2e-16 - - -
d <2e-16 <2e-16 1 - -
e <2e-16 <2e-16 <2e-16 <2e-16 -
f <2e-16 <2e-16 <2e-16 <2e-16 1
P value adjustment method: bonferroni
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MYa*_*208 15
试试这个吧
X <- rnorm(n=50, mean = 10, sd = 5)
Y <- rnorm(n=50, mean = 15, sd = 6)
Z <- rnorm(n=50, mean = 20, sd = 5)
Data <- data.frame(X, Y, Z)
library(plyr)
combos <- combn(ncol(Data),2)
adply(combos, 2, function(x) {
test <- t.test(Data[, x[1]], Data[, x[2]])
out <- data.frame("var1" = colnames(Data)[x[1]]
, "var2" = colnames(Data[x[2]])
, "t.value" = sprintf("%.3f", test$statistic)
, "df"= test$parameter
, "p.value" = sprintf("%.3f", test$p.value)
)
return(out)
})
X1 var1 var2 t.value df p.value
1 1 X Y -5.598 92.74744 0.000
2 2 X Z -9.361 90.12561 0.000
3 3 Y Z -3.601 97.62511 0.000
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这是另一种解决方案,使用outer.
outer(
1:ncol(Data), 1:ncol(Data),
Vectorize(
function (i,j) t.test(Data[,i], Data[,j])$p.value
)
)
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