t 检验的所有标准变体都在其公式中使用样本方差,并且您无法从一次观察中计算出它,因为您正在除以 n-1,其中 n 是样本大小。
这可能是最简单的修改,尽管我无法测试它,因为您没有提供示例数据(您可以dput
将数据用于您的问题):
t<- lapply(1:length(x), function(i){
if(length(x[[i]][[2]])>1){
t.test(dat$Value,x[[i]][[2]])
} else "Only one observation in subset" #or NA or something else
})
另一种选择是修改以下中使用的索引lapply
:
ind<-which(sapply(x,function(i) length(i[[2]])>1))
t<- lapply(ind, function(i) t.test(dat$Value,x[[i]][[2]]))
这是第一个使用人工数据的案例的示例:
x<-list(a=cbind(1:5,rnorm(5)),b=cbind(1,rnorm(1)),c=cbind(1:3,rnorm(3)))
y<-rnorm(20)
t<- lapply(1:length(x), function(i){
if(length(x[[i]][,2])>1){ #note the indexing x[[i]][,2]
t.test(y,x[[i]][,2])
} else "Only one observation in subset"
})
t
[[1]]
Welch Two Sample t-test
data: y and x[[i]][, 2]
t = -0.4695, df = 16.019, p-value = 0.645
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-1.2143180 0.7739393
sample estimates:
mean of x mean of y
0.1863028 0.4064921
[[2]]
[1] "Only one observation in subset"
[[3]]
Welch Two Sample t-test
data: y and x[[i]][, 2]
t = -0.6213, df = 3.081, p-value = 0.5774
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-3.013287 2.016666
sample estimates:
mean of x mean of y
0.1863028 0.6846135
Welch Two Sample t-test
data: y and x[[i]][, 2]
t = 5.2969, df = 10.261, p-value = 0.0003202
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
3.068071 7.496963
sample estimates:
mean of x mean of y
5.5000000 0.2174829