search for: bressan

Given this data frame (a simplified, essential reproducible example) A<-c(8,7,10,1,5) A_flag<-c(10,0,1,0,2) B<-c(5,6,2,1,0) B_flag<-c(12,9,0,5,0) mydf<-data.frame(A, A_flag, B, B_flag) # this is my initial df mydf I want to get to this final situation i<-which(mydf$A_flag==0) mydf$A[i]<-NA ii<-which(mydf$B_flag==0) mydf$B[ii]<-NA

...ct I was hoping to manage everything in a "one-go": i.e. without the need to create the "intermediate" variable called "speed_vehicles" and with the use of the function weighted.mean() any hints for a different approach much appreciated thanks Da: "Massimo Bressan" <massimo.bressan at arpa.veneto.it> A: "r-help" <r-help at r-project.org> Inviato: Gioved?, 9 novembre 2017 12:20:52 Oggetto: weighted average grouped by variables hi all I have this dataframe (created as a reproducible example) mydf<-structure(list(date_time...

...ll, I was hoping for a compact solution because I need to deal with MANY columns (more than 40) in data frame with the same basic structure as the simplified example I posted thanks m ----- Messaggio originale ----- Da: "Bert Gunter" <bgunter.4567 at gmail.com> A: "Massimo Bressan" <massimo.bressan at arpa.veneto.it> Cc: "r-help" <r-help at r-project.org> Inviato: Mercoled?, 22 novembre 2017 17:32:33 Oggetto: Re: [R] assign NA to rows by test on multiple columns of a data frame Do you mean like this: mydf <- within(mydf, { is.na(A)<-...

...well, I don't think this is exactly the expected result (see my post) to be noted that the columns affected should be "A" and "B" thanks for the help max ----- Messaggio originale ----- Da: "Rui Barradas" <ruipbarradas at sapo.pt> A: "Massimo Bressan" <massimo.bressan at arpa.veneto.it>, "r-help" <r-help at r-project.org> Inviato: Mercoled?, 22 novembre 2017 11:49:08 Oggetto: Re: [R] assign NA to rows by test on multiple columns of a data frame Hello, Try the following. icol <- which(grepl("flag", nam...

...naar het Herman Teirlinckgebouw op de site Thurn & Taxis. Vanaf dan ben je welkom op het nieuwe adres: Havenlaan 88 bus 73, 1000 Brussel. /////////////////////////////////////////////////////////////////////////////////////////// <https://www.inbo.be> 2017-11-09 14:16 GMT+01:00 Massimo Bressan <massimo.bressan at arpa.veneto.it>: > Hello > > an update about my question: I worked out the following solution (with the > package "dplyr") > > library(dplyr) > > mydf%>% > mutate(speed_vehicles=n_vehicles*mydf$speed) %>% > group_by(date_time,...

hi all I have this dataframe (created as a reproducible example) mydf<-structure(list(date_time = structure(c(1508238000, 1508238000, 1508238000, 1508238000, 1508238000, 1508238000, 1508238000), class = c("POSIXct", "POSIXt"), tzone = ""), direction = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 2L), .Label = c("A", "B"), class =

...f[icol] <- lapply(mydf[icol], function(x){ is.na(x) <- x == 0 x }) mydf # A A_flag B B_flag #1 8 10 5 12 #2 7 NA 6 9 #3 10 1 2 NA #4 1 NA 1 5 #5 5 2 0 NA Hope this helps, Rui Barradas On 11/22/2017 10:34 AM, Massimo Bressan wrote: > > > Given this data frame (a simplified, essential reproducible example) > > > > > A<-c(8,7,10,1,5) > > A_flag<-c(10,0,1,0,2) > > B<-c(5,6,2,1,0) > > B_flag<-c(12,9,0,5,0) > > > > > mydf<-data.frame(A, A_f...

...0 1 5 5 5 2 NA 0 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 Wed, Nov 22, 2017 at 2:34 AM, Massimo Bressan < massimo.bressan at arpa.veneto.it> wrote: > > > Given this data frame (a simplified, essential reproducible example) > > > > > A<-c(8,7,10,1,5) > > A_flag<-c(10,0,1,0,2) > > B<-c(5,6,2,1,0) > > B_flag<-c(12,9,0,5,0) > > > >...

Hello, Using base R only, the following seems to do what you want. with(mydf, ave(speed, date_time, type, FUN = weighted.mean, w = n_vehicles)) Hope this helps, Rui Barradas Em 09-11-2017 13:16, Massimo Bressan escreveu: > Hello > > an update about my question: I worked out the following solution (with the package "dplyr") > > library(dplyr) > > mydf%>% > mutate(speed_vehicles=n_vehicles*mydf$speed) %>% > group_by(date_time,type) %>% > summarise( > sum_n...

...function(speed, n_vehicles) sum(mydf$speed*mydf$n_vehicles)/sum(mydf$n_vehicles)) gives you averages aggregate(mydf$n_vehicles, list(mydf$type), sum)$x gives you sums Cheers Petr > -----Original Message----- > From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of Massimo > Bressan > Sent: Thursday, November 9, 2017 2:17 PM > To: r-help <r-help at r-project.org> > Subject: Re: [R] weighted average grouped by variables > > Hello > > an update about my question: I worked out the following solution (with the > package "dplyr") > > li...

hi thierry thanks for your reply yes, you are right, your solution is more straightforward best Da: "Thierry Onkelinx" <thierry.onkelinx at inbo.be> A: "Massimo Bressan" <massimo.bressan at arpa.veneto.it> Cc: "r-help" <r-help at r-project.org> Inviato: Gioved?, 9 novembre 2017 15:17:31 Oggetto: Re: [R] weighted average grouped by variables Dear Massimo, It seems straightforward to use weighted.mean() in a dplyr context library(...

...viously now the same as the dplyr method. Berend Hasselman > aggregate(mydf$n_vehicles, list(mydf$type), sum)$x > gives you sums > > Cheers > Petr > >> -----Original Message----- >> From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of Massimo >> Bressan >> Sent: Thursday, November 9, 2017 2:17 PM >> To: r-help <r-help at r-project.org> >> Subject: Re: [R] weighted average grouped by variables >> >> Hello >> >> an update about my question: I worked out the following solution (with the >> packa...

Hi *Massimo,* *Try this.* *a <- mydf==0mydf[a] <- NAHTHEK* On Wed, Nov 22, 2017 at 5:34 AM, Massimo Bressan < massimo.bressan at arpa.veneto.it> wrote: > > > Given this data frame (a simplified, essential reproducible example) > > > > > A<-c(8,7,10,1,5) > > A_flag<-c(10,0,1,0,2) > > B<-c(5,6,2,1,0) > > B_flag<-c(12,9,0,5,0) > > > >...

...A and B.. But you could modify it to work for specific columns i believe. EK On Wed, Nov 22, 2017 at 8:37 AM, Ek Esawi <esawiek at gmail.com> wrote: > Hi *Massimo,* > > *Try this.* > > *a <- mydf==0mydf[a] <- NAHTHEK* > > On Wed, Nov 22, 2017 at 5:34 AM, Massimo Bressan < > massimo.bressan at arpa.veneto.it> wrote: > >> >> >> Given this data frame (a simplified, essential reproducible example) >> >> >> >> >> A<-c(8,7,10,1,5) >> >> A_flag<-c(10,0,1,0,2) >> >> B<-c(5,6,2,1,0...

? stato filtrato un testo allegato il cui set di caratteri non era indicato... Nome: non disponibile URL: <https://stat.ethz.ch/pipermail/r-help/attachments/20100308/a80f5468/attachment.pl>

...suffix = ou=people #ldap filter = ($(uid=%u)(objectclass=sambaSAMAccount)) ldap idmap suffix = ou=Idmap idmap backend = ldap:ldaps://host.domain.com add machine script = /var/lib/samba/sbin/smbldap-useradd.pl -w '%u' thanks -- Simone Cittadini ================== COMVERT S.R.L. via F.lli Bressan, 21 20126 Milano - ITALY Tel +39.02.27006796(aspetta un beep)103 simonec@comvert.com http://www.comvert.com

Guys. What do you think about Dell hardware and Asterisk? Whos using it, comments, any special specs recommended or models?

I am trying to create a frequency distribution and I am a bit confused. Here are the commands I have entered: > data <- read.csv(file="40609_sortedfinal.csv",head=TRUE,sep=",") > NumberOfActionsByStatus = data$STATUS > NumberOfActionsByUser = data$ETS_LOGIN > NumberOfBidOffer = data$BID_OFFER > NumberOfActionsByUser.freq = table(NumberOfActionsByUser) >