[R] how to use AND in grepl

[chalabi.elahe at yahoo.de]

Thanks for your reply tom. After using  Subset(df,grepl("(.*t2.*pd.*)|(.*pd.*t2.*)"),df$Command)  I get this error: Argument "x" is missing, with no default. Actually I don't know how to fix this. Do you have any idea?
Thanks,
Elahe 


On Saturday, April 30, 2016 7:35 PM, Tom Wright <tom at maladmin.com> wrote:



Actually not sure my previous answer does what you wanted. Using your approach:
 t2pd=subset(df,grepl("t2",df$Command) & grepl("pd",df$Command))
Should work.
I think the regex pattern you are looking for is:
 Subset(df,grepl("(.* t2.*pd.* )|(.* pd.* t2.*)",df$Command)

On Sat, Apr 30, 2016, 7:07 PM Tom Wright <tom at maladmin.com> wrote:

subset(df,grepl("t2|pd",x$Command))
>
>
>
>
>On Sat, Apr 30, 2016 at 2:38 PM, ch.elahe via R-help <r-help at r-project.org> wrote:
>
>Hi all,
>>
>>I have one factor variable in my df and I want to extract the names from it which contain both "t2" and "pd":
>>
>>  'data.frame': 36919 obs. of 162 variables
>>   $TE                :int 38,41,11,52,48,75,.....
>>   $TR                :int 100,210,548,546,.....
>>   $Command          :factor W/2229 levels "_localize_PD","_localize_tre_t2","_abdomen_t1_seq","knee_pd_t1_localize","pd_local_abdomen_t2"...
>>
>>I have tried this but I did not get result:
>>
>>  t2pd=subset(df,grepl("t2",Command) & grepl("pd",Command))
>>
>>
>>does anyone know how to apply AND in grepl?
>>
>>Thanks
>>Elahe
>>
>>______________________________________________
>>R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
>>https://stat.ethz.ch/mailman/listinfo/r-help
>>PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>>and provide commented, minimal, self-contained, reproducible code.
>>.