Re: [R] how to use AND in grepl

[Tom Wright]

Please try to read my earlier comments.
In the absence of a proper example with expected output I think what you
are trying to achieve is:

# create a sample dataframe
df <- data.frame(Command=c("_localize_PD", "_localize_tre_t2",
"_abdomen_t1_seq", "knee_pd_t1_localize", "pd_local_abdomen_t2"))

# identify which rows in the dataframe set match the patterns
# note, the vectors PD, T2 and PDT2 are booleans indicating if a match was
made
PD <- grepl("pd", df$Command)
T2 <- grepl('t2', df$Command)
PDT2 <- grepl("(.*t2.*pd.*)|(.*pd.*t2.*)", df$Command)

# create the new column to hold the new names
df$new_name <- NA

df[PD,'new_name'] <- 'pd'
df[T2,'new_name'] <- 't2'
df[PDT2,'new_name'] <- 'pdt2'


# note 1: the order of these command is important, if the last command is
run first all matches will be overwritten by the single matches for 't2'
and 'pd'.
# note 2: There is no match for row 1 as "PD" != "pd", as suggested by John
McKown the ignore.case parameter for grepl can be used to change this
behaviour.

On Mon, May 2, 2016 at 11:01 AM,  wrote:

> I just changed all the names in Command to lowercase, then this
> str_extract works fine for "pd" and "t2", but not for "PDT2". Do you have
> any idea how I can bring PDT2  also in str_extract?
>
>
> On Monday, May 2, 2016 9:16 AM, Tom Wright  wrote:
>
>
>
> The first thing I notice here is that your first two subset statements are
> searching in an object named Command, not the column df$Command. I'm not at
> all sure what you are trying to achieve with the str_extract process but it
> is looking for the exact string 'PDT2' the vectors / dataframe formed in
> your previous commands are not being used at all.
> Moving forward I think you need to pay attention to case "PD" != "pd".
> Also the set PDT2 is going to be a subset of both  sets PD and t2, I don't
> think this is what you are after.
>
> On Mon, May 2, 2016, 8:49 AM   wrote:
>
> Yes it works, but let me explain what I am going to do. I extract all the
> names I want and then create a new column out of them for my plot. This is
> he whole thing I do:
> >  PD=subset(df,grepl("pd",Command)) //extract names in Command with only
> "pd"
> >  t2=subset(df,grepl("t2",Command)) //extract names with only "t2"
> >  PDT2=subset(df,grepl("(.*t2.*pd.*)|(.*pd.*t2.*)",df$Command) // extract
> names which contain both "pd" and "t2"
> >  v1=c('PD','t2','PDT2')// I create a vector with these conditions
> >  str_extract(df$Command,paste(v1,collaps='|')) //returning patterns,
> using stringr library
> >
> >here I see no pattern named PDT2 but there are only PD and t2 patterns.
> >On Monday, May 2, 2016 8:18 AM, Tom Wright  wrote:
> >
> >
> >
> >Sorry for the missed braces earlier. I was typing on a phone, not the
> best place to conjugate regular expressions.
> >Using the example you provided:
> >
> >> df=data.frame(Command=c("_localize_PD", "_localize_tre_t2",
> "_abdomen_t1_seq", "knee_pd_t1_localize", "pd_local_abdomen_t2"))
> >
> >> grepl("(.*t2.*pd.*)|(.*pd.*t2.*)",df$Command)
> >[1] FALSE FALSE FALSE FALSE  TRUE
> >
> >> subset(df,grepl("(.*t2.*pd.*)|(.*pd.*t2.*)",df$Command))
> >  Command
> >5 pd_local_abdomen_t2
> >
> >
> >
> >On Mon, May 2, 2016 at 7:42 AM,  wrote:
> >
> >Thanks Peter, you were right, the exact grepl is
> grepl("(.*t2.*pd.*)|(.*pd.*t2.*)",df$Command), but it does not change
> anything in Command, when I check the size of it by
> sum(grepl("(.*t2.*pd.*)|(.*pd.*t2.*)",df$Command))  the result is 0, but I
> am sure that the size is not 0. It seems that this AND does not work.
> >>
> >>
> >>
> >>On Monday, May 2, 2016 5:05 AM, peter dalgaard  wrote:
> >>
> >>On 02 May 2016, at 12:43 , ch.elahe via R-help 
> wrote:
> >>
> >>> Thanks for your reply tom. After using
> Subset(df,grepl("(.*t2.*pd.*)|(.*pd.*t2.*)"),df$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 
> 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$Command)
> >>>
> >>> On Sat, Apr 30, 2016, 7:07 PM Tom Wright  wrote:
> >>>
> >>> subset(df,grepl("t2|pd",x$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 

Re: [R] how to use AND in grepl

[John McKown]

On Mon, May 2, 2016 at 1:01 PM, ch.elahe via R-help 
wrote:

> I just changed all the names in Command to lowercase, then this
> str_extract works fine for "pd" and "t2", but not for "PDT2". Do you have
> any idea how I can bring PDT2  also in str_extract?
>

Looking at ​?grepl, I see the option: ignore.case=TRUE​

 PDT2=subset(df,grepl("(.*t2.*pd.*)|(.*pd.*t2.*)",df$
Command,ignore.case=TRUE)

Perhaps this will do the trick.

-- 
The unfacts, did we have them, are too imprecisely few to warrant our
certitude.

Maranatha! <><
John McKown

[[alternative HTML version deleted]]

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Re: [R] how to use AND in grepl

[ch.elahe via R-help]

I just changed all the names in Command to lowercase, then this str_extract 
works fine for "pd" and "t2", but not for "PDT2". Do you have any idea how I 
can bring PDT2  also in str_extract?  


On Monday, May 2, 2016 9:16 AM, Tom Wright  wrote:



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




 On Sat, Apr 30, 2016 at 2:38 PM, ch.elahe via R-help 
  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@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@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 

Re: [R] how to use AND in grepl

[Tom Wright]

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

Re: [R] how to use AND in grepl

[ch.elahe via R-help]

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  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,  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  wrote:
>
>On 02 May 2016, at 12:43 , ch.elahe via R-help  wrote:
>
>> Thanks for your reply tom. After using  
>> Subset(df,grepl("(.*t2.*pd.*)|(.*pd.*t2.*)"),df$Command)  I get this error: 
>> Argument "x" is missing, with no default. Actually I don't know how to fix 
>> this. Do you have any idea?
>
>Tom's code was missing a ")" but not where you put one. He probably also 
>didn't intend to capitalize "subset".
>
>
>-pd
>
>> Thanks,
>> Elahe
>>
>>
>> On Saturday, April 30, 2016 7:35 PM, Tom Wright  wrote:
>>
>>
>>
>> Actually not sure my previous answer does what you wanted. Using your 
>> approach:
>> t2pd=subset(df,grepl("t2",df$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  wrote:
>>
>> subset(df,grepl("t2|pd",x$Command))
>>>
>>>
>>>
>>>
>>> On Sat, Apr 30, 2016 at 2:38 PM, ch.elahe via R-help  
>>> 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@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@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@cbs.dk  Priv: pda...@gmail.com
>

__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] how to use AND in grepl

[Tom Wright]

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

[[alternative HTML version deleted]]

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Re: [R] how to use AND in grepl

[ch.elahe via R-help]

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  wrote:

On 02 May 2016, at 12:43 , ch.elahe via R-help  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  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$Command)
> 
> On Sat, Apr 30, 2016, 7:07 PM Tom Wright  wrote:
> 
> subset(df,grepl("t2|pd",x$Command))
>> 
>> 
>> 
>> 
>> On Sat, Apr 30, 2016 at 2:38 PM, ch.elahe via R-help  
>> 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@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@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@cbs.dk  Priv: pda...@gmail.com

__
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and provide commented, minimal, self-contained, reproducible code.

Re: [R] how to use AND in grepl

[peter dalgaard]

On 02 May 2016, at 12:43 , ch.elahe via R-help  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  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$Command)
> 
> On Sat, Apr 30, 2016, 7:07 PM Tom Wright  wrote:
> 
> subset(df,grepl("t2|pd",x$Command))
>> 
>> 
>> 
>> 
>> On Sat, Apr 30, 2016 at 2:38 PM, ch.elahe via R-help  
>> 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@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@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@cbs.dk  Priv: pda...@gmail.com

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and provide commented, minimal, self-contained, reproducible code.

Re: [R] how to use AND in grepl

[ch.elahe via R-help]

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  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  wrote:

subset(df,grepl("t2|pd",x$Command))
>
>
>
>
>On Sat, Apr 30, 2016 at 2:38 PM, ch.elahe via R-help  
>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@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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Re: [R] how to use AND in grepl

[Tom Wright]

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  wrote:

> subset(df,grepl("t2|pd",x$Command))
>
>
> On Sat, Apr 30, 2016 at 2:38 PM, ch.elahe via R-help  > wrote:
>
>> Hi all,
>>
>> I have one factor variable in my df and I want to extract the names from
>> it which contain both "t2" and "pd":
>>
>>   'data.frame': 36919 obs. of 162 variables
>>$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@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]]

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Re: [R] how to use AND in grepl

[Tom Wright]

subset(df,grepl("t2|pd",x$Command))


On Sat, Apr 30, 2016 at 2:38 PM, ch.elahe via R-help 
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@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]]

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Re: [R] how to use AND in grepl

[William Dunlap via R-help]

Your code looks fine to me.  What did t2pd look like?

I tried reproducing the problem in R-3.2.4(Revised) and everything worked
(although the output of str() looked a bit different - perhaps you have an
old version of R)

> df <- data.frame(TE=1:10, TR=101:110,
Command=c("pd_local_abdomen_t2","knee_pd_t1_localize","PD_localize_tre_t2","t2_localize_PD")[rep(1:4,len=10)])
> str(df)
'data.frame':   10 obs. of  3 variables:
 $ TE : int  1 2 3 4 5 6 7 8 9 10
 $ TR : int  101 102 103 104 105 106 107 108 109 110
 $ Command: Factor w/ 4 levels "knee_pd_t1_localize",..: 2 1 3 4 2 1 3 4 2 1
> subset(df,grepl("t2",Command) & grepl("pd",Command))
  TE  TR Command
1  1 101 pd_local_abdomen_t2
5  5 105 pd_local_abdomen_t2
9  9 109 pd_local_abdomen_t2
> subset(df,grepl("t2",Command,ignore.case=TRUE) &
grepl("pd",Command,ignore.case=TRUE))
  TE  TR Command
1  1 101 pd_local_abdomen_t2
3  3 103  PD_localize_tre_t2
4  4 104  t2_localize_PD
5  5 105 pd_local_abdomen_t2
7  7 107  PD_localize_tre_t2
8  8 108  t2_localize_PD
9  9 109 pd_local_abdomen_t2


Bill Dunlap
TIBCO Software
wdunlap tibco.com

On Sat, Apr 30, 2016 at 2:38 PM, ch.elahe via R-help 
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@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]]

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[R] how to use AND in grepl

[ch.elahe via R-help]

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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