[R] Removing rows in a data frame containing string in rownames

[Jeff Newmiller]

Not clear what you did. Is this an example of FAQ 7.16?
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Sent from my phone. Please excuse my brevity.

On November 16, 2014 7:47:44 PM PST, Steven Yen <syen04 at gmail.com> wrote:
>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 ignored. Hmm...
>
>Steven
>
>At 10:22 PM 11/16/2014, William Dunlap wrote:
>>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 <http://tibco.com>tibco.com
>>
>>On Sun, Nov 16, 2014 at 6:31 PM, Steven Yen 
>><<mailto:syen04 at gmail.com>syen04 at gmail.com> wrote:
>>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�  �  �  �  �  3.502�  3.930 0.891 0.373 *
>>p.children�  �  � 3.693�  4.521 0.817 0.414 *
>>p.occu�  �  �  �  � 0.740�  1.116 0.663 0.508
>>p.rating�  �  �  -7.897�  1.331 5.933 0.000
>>c.(Intercept)�  1.861�  0.965 1.929 0.054
>>c.sex�  �  �  �  �  0.221�  0.249 0.889 0.374 *
>>c.children�  �  � 0.234�  0.289 0.810 0.418 *
>>c.occu�  �  �  �  � 0.052�  0.079 0.663 0.508
>>c.rating�  �  �  -0.556�  0.102 5.451 0.000
>>u.(Intercept)�  1.943�  1.017 1.910 0.057
>>u.sex�  �  �  �  �  0.221�  0.248 0.888 0.375 *
>>u.children�  �  � 0.229�  0.276 0.827 0.409 *
>>u.occu�  �  �  �  � 0.054�  0.082 0.663 0.508
>>u.rating�  �  �  -0.581�  0.109 5.331 0.000
>>
>> > out<-subset(out,!(names(out) %in% c("sex","rating")))
>> > out
>>�  �  �  �  �  �  �  �  � est�  �  � se�  �  � t�  �  � p disc
>>p.(Intercept) 26.430 13.605 1.943 0.053
>>p.sex�  �  �  �  �  3.502�  3.930 0.891 0.373 *
>>p.children�  �  � 3.693�  4.521 0.817 0.414 *
>>p.occu�  �  �  �  � 0.740�  1.116 0.663 0.508
>>p.rating�  �  �  -7.897�  1.331 5.933 0.000
>>c.(Intercept)�  1.861�  0.965 1.929 0.054
>>c.sex�  �  �  �  �  0.221�  0.249 0.889 0.374 *
>>c.children�  �  � 0.234�  0.289 0.810 0.418 *
>>c.occu�  �  �  �  � 0.052�  0.079 0.663 0.508
>>c.rating�  �  �  -0.556�  0.102 5.451 0.000
>>u.(Intercept)�  1.943�  1.017 1.910 0.057
>>u.sex�  �  �  �  �  0.221�  0.248 0.888 0.375 *
>>u.children�  �  � 0.229�  0.276 0.827 0.409 *
>>u.occu�  �  �  �  � 0.054�  0.082 0.663 0.508
>>u.rating�  �  �  -0.581�  0.109 5.331 0.000
>>
>>______________________________________________
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>>PLEASE do read the posting guide 
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>>and provide commented, minimal, self-contained, reproducible code.
>>
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