how to use AND in grepl
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
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