dat1 is wrong also. It should read:
dat1 <-read.table(text="ID, x, y, z
A, 10, 34, 12
B, 25, 42, 18
C, 14, 20, 8
",sep=",",header=TRUE,stringsAsFactors=F)
Is this a homework problem? This list has a no homework policy.
Cheers,
Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along and
sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Sat, Sep 7, 2019 at 12:24 PM Val <valkremk at gmail.com> wrote:
> Hi all
>
> Correction for my previous posting.
> dat2 should be read as
> dat2 <-read.table(text="ID, weight
> A, 0.25
> B, 0.42
> C, 0.65 ",sep=",",header=TRUE,stringsAsFactors=F)
>
> On Sat, Sep 7, 2019 at 1:46 PM Val <valkremk at gmail.com> wrote:
> >
> > Hi All,
> >
> > I have two data frames with thousand rows and several columns. My
> > samples of the data frames are shown below
> >
> > dat1 <-read.table(text="ID, x, y, z
> > ID , x, y, z
> > A, 10, 34, 12
> > B, 25, 42, 18
> > C, 14, 20, 8
",sep=",",header=TRUE,stringsAsFactors=F)
> >
> > dat2 <-read.table(text="ID, x, y, z
> > ID, weight
> > A, 0.25
> > B, 0.42
> > C, 0.65 ",sep=",",header=TRUE,stringsAsFactors=F)
> >
> > My goal is to create an index value for each ID by mutliplying the
> > first row of dat1 by the second column of dat2.
> >
> > (10*0.25 ) + (34*0.42) + (12*0.65)= 24.58
> > (25*0.25 ) + (42*0.42) + (18*0.65)= 35.59
> > (14*0.25 ) + (20*0.42) + ( 8*0.65)= 19.03
> >
> > The desired out put is
> > dat3
> > ID, Index
> > A 24.58
> > B 35.59
> > C 19.03
> >
> > How do I do it in an efficent way?
> >
> > Thank you,
>
> ______________________________________________
> 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]]
Val has been posting to this list for almost a decade [1] so seems unlikely to be a student... but in all this time has yet to figure out how to post in plain text to avoid corruption of code on this plain text mailing list. The ability to generate small examples has improved, though execution still seems hazy. Why is there an ID column in dat2 at all? Try dat3 <- dat1[ 1,, drop=FALSE ] dat3$Index <- as.matrix( dat1[ -1 ] ) %*% dat2$weight [1] https://stat.ethz.ch/pipermail/r-help/2010-March/233533.html On September 7, 2019 12:38:12 PM PDT, Bert Gunter <bgunter.4567 at gmail.com> wrote:>dat1 is wrong also. It should read: > >dat1 <-read.table(text="ID, x, y, z > A, 10, 34, 12 > B, 25, 42, 18 > C, 14, 20, 8 ",sep=",",header=TRUE,stringsAsFactors=F) > >Is this a homework problem? This list has a no homework policy. > >Cheers, >Bert > >Bert Gunter > >"The trouble with having an open mind is that people keep coming along >and >sticking things into it." >-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) > > >On Sat, Sep 7, 2019 at 12:24 PM Val <valkremk at gmail.com> wrote: > >> Hi all >> >> Correction for my previous posting. >> dat2 should be read as >> dat2 <-read.table(text="ID, weight >> A, 0.25 >> B, 0.42 >> C, 0.65 ",sep=",",header=TRUE,stringsAsFactors=F) >> >> On Sat, Sep 7, 2019 at 1:46 PM Val <valkremk at gmail.com> wrote: >> > >> > Hi All, >> > >> > I have two data frames with thousand rows and several columns. >My >> > samples of the data frames are shown below >> > >> > dat1 <-read.table(text="ID, x, y, z >> > ID , x, y, z >> > A, 10, 34, 12 >> > B, 25, 42, 18 >> > C, 14, 20, 8 ",sep=",",header=TRUE,stringsAsFactors=F) >> > >> > dat2 <-read.table(text="ID, x, y, z >> > ID, weight >> > A, 0.25 >> > B, 0.42 >> > C, 0.65 ",sep=",",header=TRUE,stringsAsFactors=F) >> > >> > My goal is to create an index value for each ID by mutliplying >the >> > first row of dat1 by the second column of dat2. >> > >> > (10*0.25 ) + (34*0.42) + (12*0.65)= 24.58 >> > (25*0.25 ) + (42*0.42) + (18*0.65)= 35.59 >> > (14*0.25 ) + (20*0.42) + ( 8*0.65)= 19.03 >> > >> > The desired out put is >> > dat3 >> > ID, Index >> > A 24.58 >> > B 35.59 >> > C 19.03 >> > >> > How do I do it in an efficent way? >> > >> > Thank you, >> >> ______________________________________________ >> 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]] > >______________________________________________ >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.-- Sent from my phone. Please excuse my brevity.
Thank you Jeff and all. I wish to go back to my student life. ID is not necessary in dat2, sorry for that. On Sat, Sep 7, 2019 at 5:10 PM Jeff Newmiller <jdnewmil at dcn.davis.ca.us> wrote:> > Val has been posting to this list for almost a decade [1] so seems unlikely to be a student... but in all this time has yet to figure out how to post in plain text to avoid corruption of code on this plain text mailing list. The ability to generate small examples has improved, though execution still seems hazy. Why is there an ID column in dat2 at all? > > Try > > dat3 <- dat1[ 1,, drop=FALSE ] > dat3$Index <- as.matrix( dat1[ -1 ] ) %*% dat2$weight > > [1] https://stat.ethz.ch/pipermail/r-help/2010-March/233533.html > > On September 7, 2019 12:38:12 PM PDT, Bert Gunter <bgunter.4567 at gmail.com> wrote: > >dat1 is wrong also. It should read: > > > >dat1 <-read.table(text="ID, x, y, z > > A, 10, 34, 12 > > B, 25, 42, 18 > > C, 14, 20, 8 ",sep=",",header=TRUE,stringsAsFactors=F) > > > >Is this a homework problem? This list has a no homework policy. > > > >Cheers, > >Bert > > > >Bert Gunter > > > >"The trouble with having an open mind is that people keep coming along > >and > >sticking things into it." > >-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) > > > > > >On Sat, Sep 7, 2019 at 12:24 PM Val <valkremk at gmail.com> wrote: > > > >> Hi all > >> > >> Correction for my previous posting. > >> dat2 should be read as > >> dat2 <-read.table(text="ID, weight > >> A, 0.25 > >> B, 0.42 > >> C, 0.65 ",sep=",",header=TRUE,stringsAsFactors=F) > >> > >> On Sat, Sep 7, 2019 at 1:46 PM Val <valkremk at gmail.com> wrote: > >> > > >> > Hi All, > >> > > >> > I have two data frames with thousand rows and several columns. > >My > >> > samples of the data frames are shown below > >> > > >> > dat1 <-read.table(text="ID, x, y, z > >> > ID , x, y, z > >> > A, 10, 34, 12 > >> > B, 25, 42, 18 > >> > C, 14, 20, 8 ",sep=",",header=TRUE,stringsAsFactors=F) > >> > > >> > dat2 <-read.table(text="ID, x, y, z > >> > ID, weight > >> > A, 0.25 > >> > B, 0.42 > >> > C, 0.65 ",sep=",",header=TRUE,stringsAsFactors=F) > >> > > >> > My goal is to create an index value for each ID by mutliplying > >the > >> > first row of dat1 by the second column of dat2. > >> > > >> > (10*0.25 ) + (34*0.42) + (12*0.65)= 24.58 > >> > (25*0.25 ) + (42*0.42) + (18*0.65)= 35.59 > >> > (14*0.25 ) + (20*0.42) + ( 8*0.65)= 19.03 > >> > > >> > The desired out put is > >> > dat3 > >> > ID, Index > >> > A 24.58 > >> > B 35.59 > >> > C 19.03 > >> > > >> > How do I do it in an efficent way? > >> > > >> > Thank you, > >> > >> ______________________________________________ > >> 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]] > > > >______________________________________________ > >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. > > -- > Sent from my phone. Please excuse my brevity.
Hello, Bert: Quoting Jeff, "Val has been posting to this list for almost a decade". I didn't know about the "decade" part (didn't look it up) but that's why I answered the question. Val: It's not that hard to run the code you post *before* posting it. Please. Rui Barradas ?s 23:10 de 07/09/19, Jeff Newmiller escreveu:> Val has been posting to this list for almost a decade [1] so seems unlikely to be a student... but in all this time has yet to figure out how to post in plain text to avoid corruption of code on this plain text mailing list. The ability to generate small examples has improved, though execution still seems hazy. Why is there an ID column in dat2 at all? > > Try > > dat3 <- dat1[ 1,, drop=FALSE ] > dat3$Index <- as.matrix( dat1[ -1 ] ) %*% dat2$weight > > [1] https://stat.ethz.ch/pipermail/r-help/2010-March/233533.html > > On September 7, 2019 12:38:12 PM PDT, Bert Gunter <bgunter.4567 at gmail.com> wrote: >> dat1 is wrong also. It should read: >> >> dat1 <-read.table(text="ID, x, y, z >> A, 10, 34, 12 >> B, 25, 42, 18 >> C, 14, 20, 8 ",sep=",",header=TRUE,stringsAsFactors=F) >> >> Is this a homework problem? This list has a no homework policy. >> >> Cheers, >> Bert >> >> Bert Gunter >> >> "The trouble with having an open mind is that people keep coming along >> and >> sticking things into it." >> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) >> >> >> On Sat, Sep 7, 2019 at 12:24 PM Val <valkremk at gmail.com> wrote: >> >>> Hi all >>> >>> Correction for my previous posting. >>> dat2 should be read as >>> dat2 <-read.table(text="ID, weight >>> A, 0.25 >>> B, 0.42 >>> C, 0.65 ",sep=",",header=TRUE,stringsAsFactors=F) >>> >>> On Sat, Sep 7, 2019 at 1:46 PM Val <valkremk at gmail.com> wrote: >>>> >>>> Hi All, >>>> >>>> I have two data frames with thousand rows and several columns. >> My >>>> samples of the data frames are shown below >>>> >>>> dat1 <-read.table(text="ID, x, y, z >>>> ID , x, y, z >>>> A, 10, 34, 12 >>>> B, 25, 42, 18 >>>> C, 14, 20, 8 ",sep=",",header=TRUE,stringsAsFactors=F) >>>> >>>> dat2 <-read.table(text="ID, x, y, z >>>> ID, weight >>>> A, 0.25 >>>> B, 0.42 >>>> C, 0.65 ",sep=",",header=TRUE,stringsAsFactors=F) >>>> >>>> My goal is to create an index value for each ID by mutliplying >> the >>>> first row of dat1 by the second column of dat2. >>>> >>>> (10*0.25 ) + (34*0.42) + (12*0.65)= 24.58 >>>> (25*0.25 ) + (42*0.42) + (18*0.65)= 35.59 >>>> (14*0.25 ) + (20*0.42) + ( 8*0.65)= 19.03 >>>> >>>> The desired out put is >>>> dat3 >>>> ID, Index >>>> A 24.58 >>>> B 35.59 >>>> C 19.03 >>>> >>>> How do I do it in an efficent way? >>>> >>>> Thank you, >>> >>> ______________________________________________ >>> 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]] >> >> ______________________________________________ >> 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. >