Bert Gunter
2015-Jul-09 17:12 UTC
[R] detecting any element in a vector of strings, appearing anywhere in any of several character variables in a dataframe
Jeff: Well, it would be much better (no loops!) except, I think, for one issue: "red" would match "barred" and I don't think that this is what is wanted: the matches should be on whole "words" not just string patterns. So you would need to fix up the matching pattern to make this work, but it may be a little tricky, as arbitrary whitespace characters, e.g. " " or "\n" etc. could be in the strings to be matched separating the words or ending the "sentence." I'm sure it can be done, but I'll leave it to you or others to figure it out. Of course, if my diagnosis is wrong or silly, please point this out. Cheers, Bert Bert Gunter "Data is not information. Information is not knowledge. And knowledge is certainly not wisdom." -- Clifford Stoll On Thu, Jul 9, 2015 at 9:34 AM, Jeff Newmiller <jdnewmil at dcn.davis.ca.us> wrote:> I think grep is better suited to this: > > zz$v5 <- grepl( paste0( alarm.words, collapse="|" ), do.call( paste, zz[ , 2:3 ] ) ) ) > --------------------------------------------------------------------------- > Jeff Newmiller The ..... ..... Go Live... > DCN:<jdnewmil at dcn.davis.ca.us> Basics: ##.#. ##.#. Live Go... > Live: OO#.. Dead: OO#.. Playing > Research Engineer (Solar/Batteries O.O#. #.O#. with > /Software/Embedded Controllers) .OO#. .OO#. rocks...1k > --------------------------------------------------------------------------- > Sent from my phone. Please excuse my brevity. > > On July 9, 2015 8:51:10 AM PDT, Bert Gunter <bgunter.4567 at gmail.com> wrote: >>Here's a way to do it that uses %in% (i.e. match() ) and uses only a >>single, not a double, loop. It should be more efficient. >> >>> sapply(strsplit(do.call(paste,zz[,2:3]),"[[:space:]]+"), >>+ function(x)any(x %in% alarm.words)) >> >> [1] FALSE FALSE TRUE FALSE FALSE TRUE FALSE FALSE TRUE TRUE >> >>The idea is to paste the strings in each row (do.call allows an >>arbitrary number of columns) into a single string and then use >>strsplit to break the string into individual "words" on whitespace. >>Then the matching is vectorized with the any( %in% ... ) call. >> >>Cheers, >>Bert >>Bert Gunter >> >>"Data is not information. Information is not knowledge. And knowledge >>is certainly not wisdom." >> -- Clifford Stoll >> >> >>On Thu, Jul 9, 2015 at 6:05 AM, John Fox <jfox at mcmaster.ca> wrote: >>> Dear Chris, >>> >>> If I understand correctly what you want, how about the following? >>> >>>> rows <- apply(zz[, 2:3], 1, function(x) any(sapply(alarm.words, >>grepl, x=x))) >>>> zz[rows, ] >>> >>> v1 v2 v3 v4 >>> 3 -1.022329 green turtle ronald weasley 2 >>> 6 0.336599 waffle the hamster red sparks 1 >>> 9 -1.631874 yellow giraffe with a long neck gandalf the white 1 >>> 10 1.130622 black bear gandalf the grey 2 >>> >>> I hope this helps, >>> John >>> >>> ------------------------------------------------ >>> John Fox, Professor >>> McMaster University >>> Hamilton, Ontario, Canada >>> http://socserv.mcmaster.ca/jfox/ >>> >>> >>> On Wed, 08 Jul 2015 22:23:37 -0400 >>> "Christopher W. Ryan" <cryan at binghamton.edu> wrote: >>>> Running R 3.1.1 on windows 7 >>>> >>>> I want to identify as a case any record in a dataframe that contains >>any >>>> of several keywords in any of several variables. >>>> >>>> Example: >>>> >>>> # create a dataframe with 4 variables and 10 records >>>> v2 <- c("white bird", "blue bird", "green turtle", "quick brown >>fox", >>>> "big black dog", "waffle the hamster", "benny likes food a lot", >>"hello >>>> world", "yellow giraffe with a long neck", "black bear") >>>> v3 <- c("harry potter", "hermione grainger", "ronald weasley", >>"ginny >>>> weasley", "dudley dursley", "red sparks", "blue sparks", "white >>dress >>>> robes", "gandalf the white", "gandalf the grey") >>>> zz <- data.frame(v1=rnorm(10), v2=v2, v3=v3, v4=rpois(10, lambda=2), >>>> stringsAsFactors=FALSE) >>>> str(zz) >>>> zz >>>> >>>> # here are the keywords >>>> alarm.words <- c("red", "green", "turtle", "gandalf") >>>> >>>> # For each row/record, I want to test whether the string in v2 or >>the >>>> string in v3 contains any of the strings in alarm.words. And then if >>so, >>>> set zz$v5=TRUE for that record. >>>> >>>> # I'm thinking the str_detect function in the stringr package ought >>to >>>> be able to help, perhaps with some use of apply over the rows, but I >>>> obviously misunderstand something about how str_detect works >>>> >>>> library(stringr) >>>> >>>> str_detect(zz[,2:3], alarm.words) # error: the target of the >>search >>>> # must be a vector, not >>multiple >>>> # columns >>>> >>>> str_detect(zz[1:4,2:3], alarm.words) # same error >>>> >>>> str_detect(zz[,2], alarm.words) # error, length of alarm.words >>>> # is less than the number of >>>> # rows I am using for the >>>> # comparison >>>> >>>> str_detect(zz[1:4,2], alarm.words) # works as hoped when >>>> length(alarm.words) # confining nrows >>>> # to the length of alarm.words >>>> >>>> str_detect(zz, alarm.words) # obviously not right >>>> >>>> # maybe I need apply() ? >>>> my.f <- function(x){str_detect(x, alarm.words)} >>>> >>>> apply(zz[,2], 1, my.f) # again, a mismatch in lengths >>>> # between alarm.words and that >>>> # in which I am searching for >>>> # matching strings >>>> >>>> apply(zz, 2, my.f) # now I'm getting somewhere >>>> apply(zz[1:4,], 2, my.f) # but still only works with 4 >>>> # rows of the dataframe >>>> >>>> >>>> # perhaps %in% could do the job? >>>> >>>> Appreciate any advice. >>>> >>>> --Chris Ryan >>>> >>>> ______________________________________________ >>>> 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. >> >>______________________________________________ >>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. >
Jeff Newmiller
2015-Jul-09 17:30 UTC
[R] detecting any element in a vector of strings, appearing anywhere in any of several character variables in a dataframe
Just add a word break marker before and after:
zz$v5 <- grepl( paste0( "\\b(", paste0( alarm.words,
collapse="|" ), ")\\b" ), do.call( paste, zz[ , 2:3 ] ) ) )
---------------------------------------------------------------------------
Jeff Newmiller The ..... ..... Go Live...
DCN:<jdnewmil at dcn.davis.ca.us> Basics: ##.#. ##.#. Live
Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/Batteries O.O#. #.O#. with
/Software/Embedded Controllers) .OO#. .OO#. rocks...1k
---------------------------------------------------------------------------
Sent from my phone. Please excuse my brevity.
On July 9, 2015 10:12:23 AM PDT, Bert Gunter <bgunter.4567 at gmail.com>
wrote:>Jeff:
>
>Well, it would be much better (no loops!) except, I think, for one
>issue: "red" would match "barred" and I don't think
that this is what
>is wanted: the matches should be on whole "words" not just string
>patterns.
>
>So you would need to fix up the matching pattern to make this work,
>but it may be a little tricky, as arbitrary whitespace characters,
>e.g. " " or "\n" etc. could be in the strings to be
matched separating
>the words or ending the "sentence." I'm sure it can be done,
but I'll
>leave it to you or others to figure it out.
>
>Of course, if my diagnosis is wrong or silly, please point this out.
>
>Cheers,
>Bert
>
>
>Bert Gunter
>
>"Data is not information. Information is not knowledge. And knowledge
>is certainly not wisdom."
> -- Clifford Stoll
>
>
>On Thu, Jul 9, 2015 at 9:34 AM, Jeff Newmiller
><jdnewmil at dcn.davis.ca.us> wrote:
>> I think grep is better suited to this:
>>
>> zz$v5 <- grepl( paste0( alarm.words, collapse="|" ),
do.call( paste,
>zz[ , 2:3 ] ) ) )
>>
>---------------------------------------------------------------------------
>> Jeff Newmiller The ..... ..... Go
>Live...
>> DCN:<jdnewmil at dcn.davis.ca.us> Basics: ##.#.
##.#. Live
>Go...
>> Live: OO#.. Dead: OO#..
>Playing
>> Research Engineer (Solar/Batteries O.O#. #.O#. with
>> /Software/Embedded Controllers) .OO#. .OO#.
>rocks...1k
>>
>---------------------------------------------------------------------------
>> Sent from my phone. Please excuse my brevity.
>>
>> On July 9, 2015 8:51:10 AM PDT, Bert Gunter <bgunter.4567 at
gmail.com>
>wrote:
>>>Here's a way to do it that uses %in% (i.e. match() ) and uses
only a
>>>single, not a double, loop. It should be more efficient.
>>>
>>>>
sapply(strsplit(do.call(paste,zz[,2:3]),"[[:space:]]+"),
>>>+ function(x)any(x %in% alarm.words))
>>>
>>> [1] FALSE FALSE TRUE FALSE FALSE TRUE FALSE FALSE TRUE TRUE
>>>
>>>The idea is to paste the strings in each row (do.call allows an
>>>arbitrary number of columns) into a single string and then use
>>>strsplit to break the string into individual "words" on
whitespace.
>>>Then the matching is vectorized with the any( %in% ... ) call.
>>>
>>>Cheers,
>>>Bert
>>>Bert Gunter
>>>
>>>"Data is not information. Information is not knowledge. And
knowledge
>>>is certainly not wisdom."
>>> -- Clifford Stoll
>>>
>>>
>>>On Thu, Jul 9, 2015 at 6:05 AM, John Fox <jfox at mcmaster.ca>
wrote:
>>>> Dear Chris,
>>>>
>>>> If I understand correctly what you want, how about the
following?
>>>>
>>>>> rows <- apply(zz[, 2:3], 1, function(x)
any(sapply(alarm.words,
>>>grepl, x=x)))
>>>>> zz[rows, ]
>>>>
>>>> v1 v2 v3
v4
>>>> 3 -1.022329 green turtle ronald weasley
2
>>>> 6 0.336599 waffle the hamster red sparks
1
>>>> 9 -1.631874 yellow giraffe with a long neck gandalf the white
1
>>>> 10 1.130622 black bear gandalf the grey
2
>>>>
>>>> I hope this helps,
>>>> John
>>>>
>>>> ------------------------------------------------
>>>> John Fox, Professor
>>>> McMaster University
>>>> Hamilton, Ontario, Canada
>>>> http://socserv.mcmaster.ca/jfox/
>>>>
>>>>
>>>> On Wed, 08 Jul 2015 22:23:37 -0400
>>>> "Christopher W. Ryan" <cryan at
binghamton.edu> wrote:
>>>>> Running R 3.1.1 on windows 7
>>>>>
>>>>> I want to identify as a case any record in a dataframe that
>contains
>>>any
>>>>> of several keywords in any of several variables.
>>>>>
>>>>> Example:
>>>>>
>>>>> # create a dataframe with 4 variables and 10 records
>>>>> v2 <- c("white bird", "blue bird",
"green turtle", "quick brown
>>>fox",
>>>>> "big black dog", "waffle the hamster",
"benny likes food a lot",
>>>"hello
>>>>> world", "yellow giraffe with a long neck",
"black bear")
>>>>> v3 <- c("harry potter", "hermione
grainger", "ronald weasley",
>>>"ginny
>>>>> weasley", "dudley dursley", "red
sparks", "blue sparks", "white
>>>dress
>>>>> robes", "gandalf the white", "gandalf
the grey")
>>>>> zz <- data.frame(v1=rnorm(10), v2=v2, v3=v3,
v4=rpois(10,
>lambda=2),
>>>>> stringsAsFactors=FALSE)
>>>>> str(zz)
>>>>> zz
>>>>>
>>>>> # here are the keywords
>>>>> alarm.words <- c("red", "green",
"turtle", "gandalf")
>>>>>
>>>>> # For each row/record, I want to test whether the string in
v2 or
>>>the
>>>>> string in v3 contains any of the strings in alarm.words.
And then
>if
>>>so,
>>>>> set zz$v5=TRUE for that record.
>>>>>
>>>>> # I'm thinking the str_detect function in the stringr
package
>ought
>>>to
>>>>> be able to help, perhaps with some use of apply over the
rows, but
>I
>>>>> obviously misunderstand something about how str_detect
works
>>>>>
>>>>> library(stringr)
>>>>>
>>>>> str_detect(zz[,2:3], alarm.words) # error: the target of
the
>>>search
>>>>> # must be a vector,
not
>>>multiple
>>>>> # columns
>>>>>
>>>>> str_detect(zz[1:4,2:3], alarm.words) # same error
>>>>>
>>>>> str_detect(zz[,2], alarm.words) # error, length of
>alarm.words
>>>>> # is less than the
number of
>>>>> # rows I am using for
the
>>>>> # comparison
>>>>>
>>>>> str_detect(zz[1:4,2], alarm.words) # works as hoped when
>>>>> length(alarm.words) # confining nrows
>>>>> # to the length of
>alarm.words
>>>>>
>>>>> str_detect(zz, alarm.words) # obviously not right
>>>>>
>>>>> # maybe I need apply() ?
>>>>> my.f <- function(x){str_detect(x, alarm.words)}
>>>>>
>>>>> apply(zz[,2], 1, my.f) # again, a mismatch in lengths
>>>>> # between alarm.words and that
>>>>> # in which I am searching for
>>>>> # matching strings
>>>>>
>>>>> apply(zz, 2, my.f) # now I'm getting somewhere
>>>>> apply(zz[1:4,], 2, my.f) # but still only works with 4
>>>>> # rows of the dataframe
>>>>>
>>>>>
>>>>> # perhaps %in% could do the job?
>>>>>
>>>>> Appreciate any advice.
>>>>>
>>>>> --Chris Ryan
>>>>>
>>>>> ______________________________________________
>>>>> 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.
>>>
>>>______________________________________________
>>>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.
>>
Bert Gunter
2015-Jul-09 17:52 UTC
[R] detecting any element in a vector of strings, appearing anywhere in any of several character variables in a dataframe
Yup, that does it. Let grep figure out what's a word rather than doing it manually. Forgot about "\b" Cheers, Bert Bert Gunter "Data is not information. Information is not knowledge. And knowledge is certainly not wisdom." -- Clifford Stoll On Thu, Jul 9, 2015 at 10:30 AM, Jeff Newmiller <jdnewmil at dcn.davis.ca.us> wrote:> Just add a word break marker before and after: > > zz$v5 <- grepl( paste0( "\\b(", paste0( alarm.words, collapse="|" ), ")\\b" ), do.call( paste, zz[ , 2:3 ] ) ) ) > --------------------------------------------------------------------------- > Jeff Newmiller The ..... ..... Go Live... > DCN:<jdnewmil at dcn.davis.ca.us> Basics: ##.#. ##.#. Live Go... > Live: OO#.. Dead: OO#.. Playing > Research Engineer (Solar/Batteries O.O#. #.O#. with > /Software/Embedded Controllers) .OO#. .OO#. rocks...1k > --------------------------------------------------------------------------- > Sent from my phone. Please excuse my brevity. > > On July 9, 2015 10:12:23 AM PDT, Bert Gunter <bgunter.4567 at gmail.com> wrote: >>Jeff: >> >>Well, it would be much better (no loops!) except, I think, for one >>issue: "red" would match "barred" and I don't think that this is what >>is wanted: the matches should be on whole "words" not just string >>patterns. >> >>So you would need to fix up the matching pattern to make this work, >>but it may be a little tricky, as arbitrary whitespace characters, >>e.g. " " or "\n" etc. could be in the strings to be matched separating >>the words or ending the "sentence." I'm sure it can be done, but I'll >>leave it to you or others to figure it out. >> >>Of course, if my diagnosis is wrong or silly, please point this out. >> >>Cheers, >>Bert >> >> >>Bert Gunter >> >>"Data is not information. Information is not knowledge. And knowledge >>is certainly not wisdom." >> -- Clifford Stoll >> >> >>On Thu, Jul 9, 2015 at 9:34 AM, Jeff Newmiller >><jdnewmil at dcn.davis.ca.us> wrote: >>> I think grep is better suited to this: >>> >>> zz$v5 <- grepl( paste0( alarm.words, collapse="|" ), do.call( paste, >>zz[ , 2:3 ] ) ) ) >>> >>--------------------------------------------------------------------------- >>> Jeff Newmiller The ..... ..... Go >>Live... >>> DCN:<jdnewmil at dcn.davis.ca.us> Basics: ##.#. ##.#. Live >>Go... >>> Live: OO#.. Dead: OO#.. >>Playing >>> Research Engineer (Solar/Batteries O.O#. #.O#. with >>> /Software/Embedded Controllers) .OO#. .OO#. >>rocks...1k >>> >>--------------------------------------------------------------------------- >>> Sent from my phone. Please excuse my brevity. >>> >>> On July 9, 2015 8:51:10 AM PDT, Bert Gunter <bgunter.4567 at gmail.com> >>wrote: >>>>Here's a way to do it that uses %in% (i.e. match() ) and uses only a >>>>single, not a double, loop. It should be more efficient. >>>> >>>>> sapply(strsplit(do.call(paste,zz[,2:3]),"[[:space:]]+"), >>>>+ function(x)any(x %in% alarm.words)) >>>> >>>> [1] FALSE FALSE TRUE FALSE FALSE TRUE FALSE FALSE TRUE TRUE >>>> >>>>The idea is to paste the strings in each row (do.call allows an >>>>arbitrary number of columns) into a single string and then use >>>>strsplit to break the string into individual "words" on whitespace. >>>>Then the matching is vectorized with the any( %in% ... ) call. >>>> >>>>Cheers, >>>>Bert >>>>Bert Gunter >>>> >>>>"Data is not information. Information is not knowledge. And knowledge >>>>is certainly not wisdom." >>>> -- Clifford Stoll >>>> >>>> >>>>On Thu, Jul 9, 2015 at 6:05 AM, John Fox <jfox at mcmaster.ca> wrote: >>>>> Dear Chris, >>>>> >>>>> If I understand correctly what you want, how about the following? >>>>> >>>>>> rows <- apply(zz[, 2:3], 1, function(x) any(sapply(alarm.words, >>>>grepl, x=x))) >>>>>> zz[rows, ] >>>>> >>>>> v1 v2 v3 v4 >>>>> 3 -1.022329 green turtle ronald weasley 2 >>>>> 6 0.336599 waffle the hamster red sparks 1 >>>>> 9 -1.631874 yellow giraffe with a long neck gandalf the white 1 >>>>> 10 1.130622 black bear gandalf the grey 2 >>>>> >>>>> I hope this helps, >>>>> John >>>>> >>>>> ------------------------------------------------ >>>>> John Fox, Professor >>>>> McMaster University >>>>> Hamilton, Ontario, Canada >>>>> http://socserv.mcmaster.ca/jfox/ >>>>> >>>>> >>>>> On Wed, 08 Jul 2015 22:23:37 -0400 >>>>> "Christopher W. Ryan" <cryan at binghamton.edu> wrote: >>>>>> Running R 3.1.1 on windows 7 >>>>>> >>>>>> I want to identify as a case any record in a dataframe that >>contains >>>>any >>>>>> of several keywords in any of several variables. >>>>>> >>>>>> Example: >>>>>> >>>>>> # create a dataframe with 4 variables and 10 records >>>>>> v2 <- c("white bird", "blue bird", "green turtle", "quick brown >>>>fox", >>>>>> "big black dog", "waffle the hamster", "benny likes food a lot", >>>>"hello >>>>>> world", "yellow giraffe with a long neck", "black bear") >>>>>> v3 <- c("harry potter", "hermione grainger", "ronald weasley", >>>>"ginny >>>>>> weasley", "dudley dursley", "red sparks", "blue sparks", "white >>>>dress >>>>>> robes", "gandalf the white", "gandalf the grey") >>>>>> zz <- data.frame(v1=rnorm(10), v2=v2, v3=v3, v4=rpois(10, >>lambda=2), >>>>>> stringsAsFactors=FALSE) >>>>>> str(zz) >>>>>> zz >>>>>> >>>>>> # here are the keywords >>>>>> alarm.words <- c("red", "green", "turtle", "gandalf") >>>>>> >>>>>> # For each row/record, I want to test whether the string in v2 or >>>>the >>>>>> string in v3 contains any of the strings in alarm.words. And then >>if >>>>so, >>>>>> set zz$v5=TRUE for that record. >>>>>> >>>>>> # I'm thinking the str_detect function in the stringr package >>ought >>>>to >>>>>> be able to help, perhaps with some use of apply over the rows, but >>I >>>>>> obviously misunderstand something about how str_detect works >>>>>> >>>>>> library(stringr) >>>>>> >>>>>> str_detect(zz[,2:3], alarm.words) # error: the target of the >>>>search >>>>>> # must be a vector, not >>>>multiple >>>>>> # columns >>>>>> >>>>>> str_detect(zz[1:4,2:3], alarm.words) # same error >>>>>> >>>>>> str_detect(zz[,2], alarm.words) # error, length of >>alarm.words >>>>>> # is less than the number of >>>>>> # rows I am using for the >>>>>> # comparison >>>>>> >>>>>> str_detect(zz[1:4,2], alarm.words) # works as hoped when >>>>>> length(alarm.words) # confining nrows >>>>>> # to the length of >>alarm.words >>>>>> >>>>>> str_detect(zz, alarm.words) # obviously not right >>>>>> >>>>>> # maybe I need apply() ? >>>>>> my.f <- function(x){str_detect(x, alarm.words)} >>>>>> >>>>>> apply(zz[,2], 1, my.f) # again, a mismatch in lengths >>>>>> # between alarm.words and that >>>>>> # in which I am searching for >>>>>> # matching strings >>>>>> >>>>>> apply(zz, 2, my.f) # now I'm getting somewhere >>>>>> apply(zz[1:4,], 2, my.f) # but still only works with 4 >>>>>> # rows of the dataframe >>>>>> >>>>>> >>>>>> # perhaps %in% could do the job? >>>>>> >>>>>> Appreciate any advice. >>>>>> >>>>>> --Chris Ryan >>>>>> >>>>>> ______________________________________________ >>>>>> 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. >>>> >>>>______________________________________________ >>>>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. >>> >