On 15/10/14 04:09, Kate Ignatius wrote:
In the sense - it does not work. it works when there are 50 samples
in the file, but it does not work when there is one.
The usual headings are: sample1.at sample1.dp
sample1.fg sample2.at sample2.dp sample2.fg.... and so on to a max of
sample50.at sample50.dp sample50.fg
using this greps out all the .at columns perfectly:
df[,grep(".at",colnames(df))]
When I come across a file when there is one sample:
sample1.at sample1.dp sample1.fg
Using this:
df[,grep(".at",colnames(df))]
returns nothing.
Oh - AT/at was just an example... thats not my problem...
You are being (deliberately?) obtuse.
It's *all* your problem. You have to be precise when working with
computers and when providing examples. Don't build examples with
confusing red herrings.
Your assertion that "df[,grep(".at",colnames(df))] returns nothing" is
simple ***INCORRECT***. It works just fine. See the (tidy, completely
reproducible) example in the attached file "kate.txt".
Note that, with a single ".at" column in your data frame, what is
returned is ***NOT*** a data frame but rather a vector. If you want a
(one-column) data frame you need to use "drop=FALSE" in your
subscripting call.
You need to study up on R and learn how it works (read the Introduction
to R) and stop going off half-cocked.
cheers,
Rolf Turner
P.S. It is a ***bad*** idea to use "df" as the name of a data frame.
The string "df" is the name of a *function* in base R (it is the
probability density function for the F distribution). Although R is
clever enough to distinguish functions from data objects in *most*
circumstances, at the very least confusion could arise.
R. T.
--
Rolf Turner
Technical Editor ANZJS
#
# Check it out.
#
# Data frame with one ".at" column.
d1 <- as.data.frame(matrix(1,ncol=3,nrow=10))
n1 <- c("sample1.at","sample1.dp","sample1.g")
names(d1) <- n1
# Data frame with many ".at" columns.
d2 <- as.data.frame(matrix(1,ncol=50,nrow=10))
set.seed(42)
n2 <- paste("sample",1:50,sample(c(".at",".dp",".fg"),50,TRUE),sep="")
names(d2) <- n2
# Extract the ".at" columns.
print(d1[,grep(".at",colnames(d1))])
print(d2[,grep(".at",colnames(d2))])
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