Not clear what you did. Is this an example of FAQ 7.16?
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Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#. ##.#. Live Go...
Liv
Thank you Bill and Dennis. grepl worked great.
However, for reason I am not figuring out, the
code worked as I included the procedure
(subroutine) with a source command, viz.,
source("z:\\R\\mylib\\me.R")
Compiling the routine into a library/package, as
I always do, then the command got ig
Try grepl() to do pattern matching in strings. ("%in%" checks for
equality.) E.g., using your original 'out' do
out[ !grepl("sex|rating", rownames(out), ]
to get all but the rows whose names contain the character sequences
"sex" or "rating".
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Su
I like to remove from a data frame rows with labels containing
certain string, e.g., "sex" and "rating". Below is a list of the data
frame and my failed attempt to the rows. Any clues? Thanks.
> out
est se t p disc
p.(Intercept) 26.430 13.605 1.943 0.053
p.sex
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