R help - Jul 2009 - suggestion for paired t-tests

There's a funny inconsistency in how t.test handles paired=T or paired=F. If
x
and y parameters are lists, paired=F works, but paired=T doesn't.
> lg=read.csv("my.csv")
> a = subset(lg, condition=="a")["score"]
> b = subset(lg, condition=="b")["score"]
> t.test(a,b)
> t.test(a,b, paired=TRUE)
Error in `[.data.frame`(y, yok) : undefined columns selected

But this works
> a=a[,1]
> b=b[,1]
> t.test(a,b, paired=TRUE)
...

Hi Jack,

Maybe this helps.

#  make some data

set.seed(123)
condition <- factor(rep(c("a","b"), each = 5))
score <- rnorm(10);
lg <- data.frame(condition, score)

# Carry out commands

a <- subset(lg,condition=="a")["score"]
b <- subset(lg,condition=="b")["score"]

t.test(a,b,paired=TRUE)

#Error in `[.data.frame`(y, yok) : undefined columns selected

# a and b are still data frames

#So this works. We must refer to score, which is being compared.

t.test(a$score,b$score,paired=TRUE)


a=a[,1]
b=b[,1]

# This now works because a and b are vectors
# so we don't need $ to access score

t.test(a,b, paired=TRUE)

Regards,

Juliet

jacktanner wrote:
> 
> There's a funny inconsistency in how t.test handles paired=T or
paired=F.
> If x
> and y parameters are lists, paired=F works, but paired=T doesn't.
> 
>> lg=read.csv("my.csv")
>> a = subset(lg, condition=="a")["score"]
>> b = subset(lg, condition=="b")["score"]
>> t.test(a,b)
>> t.test(a,b, paired=TRUE)
> Error in `[.data.frame`(y, yok) : undefined columns selected
> 
> But this works
>> a=a[,1]
>> b=b[,1]
>> t.test(a,b, paired=TRUE)
> ...
> 
> 
It's sort of an accident that this works for the unpaired case.
You can follow what happens via debug(stats:::t.test.default) ...
there is some code

     if (paired) 
            xok <- yok <- complete.cases(x, y)
        else {
            yok <- !is.na(y)
            xok <- !is.na(x)
        }

if paired is FALSE, !is.na(y) and !is.na(x) happen to convert x and y
into matrices, whence they can be used for the rest of the computations.
If paired is TRUE, x and y remain data frames.

  Bottom line: a data frame with a single column in it really isn't the
same as a vector ...
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