On Apr 14, 2012, at 4:44 PM, Junyu Lee wrote:
Hello everyone,
I posted a question this morning, when I got replies, I realized
that the
data I posted was messy. So I am going to re-post
It's still a mess.
Here is my data frame:
Learn to use dput:
dfrm <-
structure(list(group = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 2L,
2L, 3L, 3L, 3L, 3L), .Label = c("Control", "Test1", "Test2"), class =
"factor"),
gene1 = c(28.9776, 28.9499, 29.5468, 29.5246, 29.1864, 29.2048,
34.9563, 34.9464, 36.9566, 37.1309, 36.1017, 36.0883), gene2 =
c(9.9355,
10.0997, 14.2995, 13.9561, 9.7718, 10.0388, 11.9509, 11.8909,
14.5316, 14.5188, 29.5468, 29.5246)), .Names = c("group",
"gene1", "gene2"), class = "data.frame", row.names = c(NA, -12L))
I need to calculated t.test:
for gene1: test1 vs control (29.1864, 29.2048, 34.9563,
34.9464) vs (28.9776, 28.9499, 29.5468, 29.5246)
and test2 vs control (36.9566, 37.1309, 36.1017, 36.0883)
vs (28.9776, 28.9499, 29.5468, 29.5246) .
Then I'd like to perform the same on gene 2.
You appear resistant to the notion that you may not have firm grasp
on the statistical issues. Such hubris is at the root of quite a bit
of published material that geneticists are cranking out that is not
reproducible. This should be a more stiatically principled first step
before ad hoc tests.
lapply( dfrm[ , c("gene1", "gene2"}], function(x) summary(lm(x ~
dfrm[["group"]]) ) )
p.s. your offlist comment was amusing, but hardly accurate.
I'm sorry for the mess
Then don't create more of them. Read the Posting Guide and heed its
advice.
and thank you all in advance.
Junyu
[[alternative HTML version deleted]]
HTML posting is a sign that you are not reading the Positn Guide.
--
David Winsemius, MD, MPH
West Hartford, CT
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