On 02 May 2016, at 12:43 , ch.elahe via R-help <r-help@r-project.org> wrote:
> Thanks for your reply tom. After using
> Subset(df,grepl("(.*t2.*pd.*)|(.*pd.*t2.*)"),df$Command) I get this error:
> Argument "x" is missing, with no default. Actually I don't know how to fix
> this. Do you have any idea?
Tom's code was missing a ")" but not where you put one. He probably also didn't
intend to capitalize "subset".
-pd
> Thanks,
> Elahe
>
>
> On Saturday, April 30, 2016 7:35 PM, Tom Wright <t...@maladmin.com> wrote:
>
>
>
> Actually not sure my previous answer does what you wanted. Using your
> approach:
> t2pd=subset(df,grepl("t2",df$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 <t...@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@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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>
> ______________________________________________
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> 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....@cbs.dk Priv: pda...@gmail.com
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