All standard variants of t-test use sample variances in their formulas, and you cannot compute that from one observation as you are dividing with n-1, where n is sample size.
This would probably be the easiest modification, although I cannot test it as you did not provide sample data (you could dput
your data to your question):
t<- lapply(1:length(x), function(i){
if(length(x[[i]][[2]])>1){
t.test(dat$Value,x[[i]][[2]])
} else "Only one observation in subset" #or NA or something else
})
Another option would be to modify the indices which are used in lapply
:
ind<-which(sapply(x,function(i) length(i[[2]])>1))
t<- lapply(ind, function(i) t.test(dat$Value,x[[i]][[2]]))
Here's an example of the first case with artificial data:
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
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