how to use AND in grepl
Yes it works, but let me explain what I am going to do. I extract all the names I want and then create a new column out of them for my plot. This is he whole thing I do:
PD=subset(df,grepl("pd",Command)) //extract names in Command with only "pd"
t2=subset(df,grepl("t2",Command)) //extract names with only "t2"
PDT2=subset(df,grepl("(.*t2.*pd.*)|(.*pd.*t2.*)",df$Command) // extract names which contain both "pd" and "t2"
v1=c('PD','t2','PDT2')// I create a vector with these conditions
str_extract(df$Command,paste(v1,collaps='|')) //returning patterns, using stringr library
here I see no pattern named PDT2 but there are only PD and t2 patterns.
On Monday, May 2, 2016 8:18 AM, Tom Wright <tom at maladmin.com> wrote:
Sorry for the missed braces earlier. I was typing on a phone, not the best place to conjugate regular expressions. Using the example you provided:
df=data.frame(Command=c("_localize_PD", "_localize_tre_t2", "_abdomen_t1_seq", "knee_pd_t1_localize", "pd_local_abdomen_t2"))
grepl("(.*t2.*pd.*)|(.*pd.*t2.*)",df$Command)
[1] FALSE FALSE FALSE FALSE TRUE
subset(df,grepl("(.*t2.*pd.*)|(.*pd.*t2.*)",df$Command))
Command 5 pd_local_abdomen_t2
On Mon, May 2, 2016 at 7:42 AM, <chalabi.elahe at yahoo.de> wrote:
Thanks Peter, you were right, the exact grepl is grepl("(.*t2.*pd.*)|(.*pd.*t2.*)",df$Command), but it does not change anything in Command, when I check the size of it by sum(grepl("(.*t2.*pd.*)|(.*pd.*t2.*)",df$Command)) the result is 0, but I am sure that the size is not 0. It seems that this AND does not work.
On Monday, May 2, 2016 5:05 AM, peter dalgaard <pdalgd at gmail.com> wrote: On 02 May 2016, at 12:43 , ch.elahe via R-help <r-help at r-project.org> wrote:
Thanks for your reply tom. After using Subset(df,grepl("(.*t2.*pd.*)|(.*pd.*t2.*)"),df$Command) I get this error: Argument "x" is missing, with no default. Actually I don't know how to fix this. Do you have any idea?
Tom's code was missing a ")" but not where you put one. He probably also didn't intend to capitalize "subset". -pd
Thanks,
Elahe
On Saturday, April 30, 2016 7:35 PM, Tom Wright <tom at maladmin.com> wrote:
Actually not sure my previous answer does what you wanted. Using your approach:
t2pd=subset(df,grepl("t2",df$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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______________________________________________ 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.
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