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,.....
>> $Command :factor W/2229 levels
>>
"_localize_PD","_localize_tre_t2","_abdomen_t1_seq","knee_pd_t1_localize","pd_local_abdomen_t2"...
>>
>> I have tried this but I did not get result:
>>
>> t2pd=subset(df,grepl("t2",Command) &
grepl("pd",Command))
>>
>>
>> does anyone know how to apply AND in grepl?
>>
>> Thanks
>> Elahe
>>
>> ______________________________________________
>> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
[[alternative HTML version deleted]]
Thanks for your reply tom. After using
Subset(df,grepl("(.*t2.*pd.*)|(.*pd.*t2.*)"),df$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,.....
>> $Command :factor W/2229 levels
"_localize_PD","_localize_tre_t2","_abdomen_t1_seq","knee_pd_t1_localize","pd_local_abdomen_t2"...
>>
>>I have tried this but I did not get result:
>>
>> t2pd=subset(df,grepl("t2",Command) &
grepl("pd",Command))
>>
>>
>>does anyone know how to apply AND in grepl?
>>
>>Thanks
>>Elahe
>>
>>______________________________________________
>>R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
>>https://stat.ethz.ch/mailman/listinfo/r-help
>>PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
>>and provide commented, minimal, self-contained, reproducible code.
>>.
On 02 May 2016, at 12:43 , ch.elahe via R-help <r-help at r-project.org> wrote:> Thanks for your reply tom. After using Subset(df,grepl("(.*t2.*pd.*)|(.*pd.*t2.*)"),df$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 >>> >>> ______________________________________________ >>> 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. >>> . > > ______________________________________________ > 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.-- Peter Dalgaard, Professor, Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Office: A 4.23 Email: pd.mes at cbs.dk Priv: PDalgd at gmail.com