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,..... > >>> $Command :factor W/2229 levels > "_localize_PD","_localize_tre_t2","_abdomen_t1_seq","knee_pd_t1_localize","pd_local_abdomen_t2"... > >>> > >>> I have tried this but I did not get result: > >>> > >>> t2pd=subset(df,grepl("t2",Command) & grepl("pd",Command)) > >>> > >>> > >>> does anyone know how to apply AND in grepl? > >>> > >>> Thanks > >>> Elahe > >>> > >>> ______________________________________________ > >>> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see > >>> https://stat.ethz.ch/mailman/listinfo/r-help > >>> PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > >>> and provide commented, minimal, self-contained, reproducible code. > >>> . > > > > ______________________________________________ > > R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see > > https://stat.ethz.ch/mailman/listinfo/r-help > > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > > and provide commented, minimal, self-contained, reproducible code. > > -- > Peter Dalgaard, Professor, > Center for Statistics, Copenhagen Business School > Solbjerg Plads 3, 2000 Frederiksberg, Denmark > Phone: (+45)38153501 > Office: A 4.23 > Email: pd.mes at cbs.dk Priv: PDalgd at gmail.com >[[alternative HTML version deleted]]
Yes it works, but let me explain what I am going to do. I extract all the names
I want and then create a new column out of them for my plot. This is he whole
thing I do:
PD=subset(df,grepl("pd",Command)) //extract names in Command with
only "pd"
t2=subset(df,grepl("t2",Command)) //extract names with only
"t2"
PDT2=subset(df,grepl("(.*t2.*pd.*)|(.*pd.*t2.*)",df$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,.....
>>>> $Command :factor W/2229 levels
"_localize_PD","_localize_tre_t2","_abdomen_t1_seq","knee_pd_t1_localize","pd_local_abdomen_t2"...
>>>>
>>>> I have tried this but I did not get result:
>>>>
>>>> t2pd=subset(df,grepl("t2",Command) &
grepl("pd",Command))
>>>>
>>>>
>>>> does anyone know how to apply AND in grepl?
>>>>
>>>> Thanks
>>>> Elahe
>>>>
>>>> ______________________________________________
>>>> R-help at r-project.org mailing list -- To UNSUBSCRIBE and
more, see
>>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>>> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
>>>> and provide commented, minimal, self-contained, reproducible
code.
>>>> .
>>
>> ______________________________________________
>> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>
>--
>Peter Dalgaard, Professor,
>Center for Statistics, Copenhagen Business School
>Solbjerg Plads 3, 2000 Frederiksberg, Denmark
>Phone: (+45)38153501
>Office: A 4.23
>Email: pd.mes at cbs.dk Priv: PDalgd at gmail.com
>
The first thing I notice here is that your first two subset statements are searching in an object named Command, not the column df$Command. I'm not at all sure what you are trying to achieve with the str_extract process but it is looking for the exact string 'PDT2' the vectors / dataframe formed in your previous commands are not being used at all. Moving forward I think you need to pay attention to case "PD" != "pd". Also the set PDT2 is going to be a subset of both sets PD and t2, I don't think this is what you are after. On Mon, May 2, 2016, 8:49 AM <chalabi.elahe at yahoo.de> wrote:> 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 > >>>> > >>>> ______________________________________________ > >>>> 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 > > >[[alternative HTML version deleted]]