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Hello, Or even simpler, without ifelse, colordata$response <- colordata$color == 'blue' + 0 Hope this helps, Rui Barradas ? Citando David Barron <dnbarron at gmail.com>: > ifelse is vectorised, so just use that without the loop. > > colordata$response <- ifelse(colordata$color == 'blue', 1, 0) > > David > > On 7 April 2016 at 12:41, Michael Artz <michaeleartz at gmail.com> wrote: >> Hi I'm not sure...

Thaks so much! And how would you incorporate lapply() here? On Thu, Apr 7, 2016 at 6:52 AM, David Barron <dnbarron at gmail.com> wrote: > ifelse is vectorised, so just use that without the loop. > > colordata$response <- ifelse(colordata$color == 'blue', 1, 0) > > David > > On 7 April 2016 at 12:41, Michael Artz <michaeleartz at gmail.com> wrote: > >> Hi I'...

ifelse is vectorised, so just use that without the loop. colordata$response <- ifelse(colordata$color == 'blue', 1, 0) David On 7 April 2016 at 12:41, Michael Artz <michaeleartz at gmail.com> wrote: > Hi I'm not sure how to ask this, but its a very easy question to answer for > an R person. > > What is an easy way to check for a column value and then assigne a

Hi I'm not sure how to ask this, but its a very easy question to answer for an R person. What is an easy way to check for a column value and then assigne a new column a value based on that old column value? For example, Im doing colordata <- data.frame(id = c(1,2,3,4,5), color = c("blue", "red", "green", "blue", "orange")) for (i in

...heza.cz<mailto:petr.pikal at precheza.cz>> wrote: Hi > -----Original Message----- > From: R-help [mailto:r-help-bounces at r-project.org<mailto:r-help-bounces at r-project.org>] On Behalf Of Michael > Artz > Sent: Thursday, April 7, 2016 1:57 PM > To: David Barron <dnbarron at gmail.com<mailto:dnbarron at gmail.com>> > Cc: r-help at r-project.org<mailto:r-help at r-project.org> > Subject: Re: [R] simple question on data frames assignment > > Thaks so much! And how would you incorporate lapply() here? Why do you want to use lapply? What is...

...tor' On Thu, Apr 7, 2016 at 8:43 AM, <ruipbarradas at sapo.pt> wrote: > Hello, > > Or even simpler, without ifelse, > > colordata$response <- colordata$color == 'blue' + 0 > > Hope this helps, > > Rui Barradas > > > Citando David Barron <dnbarron at gmail.com>: > > ifelse is vectorised, so just use that without the loop. > > colordata$response <- ifelse(colordata$color == 'blue', 1, 0) > > David > > On 7 April 2016 at 12:41, Michael Artz <michaeleartz at gmail.com> wrote: > > Hi I'm not...

== is also vectorised, and you're better off with TRUE and FALSE rather than 1 and 0, so I'd recommend: colordata$response <- colordata$color == 'blue' Hadley On Thu, Apr 7, 2016 at 6:52 AM, David Barron <dnbarron at gmail.com> wrote: > ifelse is vectorised, so just use that without the loop. > > colordata$response <- ifelse(colordata$color == 'blue', 1, 0) > > David > > On 7 April 2016 at 12:41, Michael Artz <michaeleartz at gmail.com> wrote: > >> Hi I'm...

...> Sent from my phone. Please excuse my brevity. > > On April 7, 2016 4:57:04 AM PDT, Michael Artz <michaeleartz at gmail.com> > wrote: >> >> Thaks so much! And how would you incorporate lapply() here? >> >> On Thu, Apr 7, 2016 at 6:52 AM, David Barron <dnbarron at gmail.com> wrote: >> >> ifelse is vectorised, so just use that without the loop. >>> >>> colordata$response <- ifelse(colordata$color == 'blue', 1, 0) >>> >>> David >>> >>> On 7 April 2016 at 12:41, Michael Artz...

Also, check whether R runs standalone, either as Rgui or in Terminal.app (just start Terminal and type "R" if you haven't tried it before.) -pd > On 26 Oct 2017, at 14:57 , David Barron <dnbarron at gmail.com> wrote: > > I've seen similar issues reported on the RStudio community site: > https://community.rstudio.com/. You might want to check in there, as I > think this may well be an RStudio issue rather than a problem with R itself. > > Dave > > On 26 Oct...

...ata2 data frame to have more columns. -- Sent from my phone. Please excuse my brevity. On April 7, 2016 4:57:04 AM PDT, Michael Artz <michaeleartz at gmail.com> wrote: >Thaks so much! And how would you incorporate lapply() here? > >On Thu, Apr 7, 2016 at 6:52 AM, David Barron <dnbarron at gmail.com> >wrote: > >> ifelse is vectorised, so just use that without the loop. >> >> colordata$response <- ifelse(colordata$color == 'blue', 1, 0) >> >> David >> >> On 7 April 2016 at 12:41, Michael Artz <michaeleartz at gmail.co...

I've seen similar issues reported on the RStudio community site: https://community.rstudio.com/. You might want to check in there, as I think this may well be an RStudio issue rather than a problem with R itself. Dave On 26 October 2017 at 12:11, Eric Berger <ericjberger at gmail.com> wrote: > How about going back to earlier versions if you don't need the latest ones? >

...excuse my brevity. >> >> On April 7, 2016 4:57:04 AM PDT, Michael Artz ><michaeleartz at gmail.com> >> wrote: >>> >>> Thaks so much! And how would you incorporate lapply() here? >>> >>> On Thu, Apr 7, 2016 at 6:52 AM, David Barron <dnbarron at gmail.com> >wrote: >>> >>> ifelse is vectorised, so just use that without the loop. >>>> >>>> colordata$response <- ifelse(colordata$color == 'blue', 1, 0) >>>> >>>> David >>>> >>>> On...

...wickham at gmail.com> wrote: > == is also vectorised, and you're better off with TRUE and FALSE > rather than 1 and 0, so I'd recommend: > > colordata$response <- colordata$color == 'blue' > > Hadley > > On Thu, Apr 7, 2016 at 6:52 AM, David Barron <dnbarron at gmail.com> wrote: > > ifelse is vectorised, so just use that without the loop. > > > > colordata$response <- ifelse(colordata$color == 'blue', 1, 0) > > > > David > > > > On 7 April 2016 at 12:41, Michael Artz <michaeleartz at gmail.com&...

...exp(2) = > 7.39 > ...etc > > > -- > View this message in context: http://www.nabble.com/axis-on-a-non-uniform-scale-tp24032916p24032916.html > Sent from the R help mailing list archive at Nabble.com. > > > > > > > > > > From: David Barron <dnbarron@googlemail.com> > Date: 15 June 2009 9:42:26 PM > To: poleteiep34@hotmail.com > Cc: r-help@r-project.org > Subject: Re: [R] NA as a result of using GLM > > > Because you have 16 variables but only 10 observations. > > David > > On Mon, Jun 15, 2009 at 12:09, pole...