similar to: weighted average grouped by variables

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_times_speed=sum(speed_vehicles), n_vehicles=sum(n_vehicles), vel=sum(speed_vehicles)/sum(n_vehicles) ) In fact I was hoping to manage everything in a

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%>% >

Hi Thanks for working example. you could use split/ lapply approach, however it is probably not much better than dplyr method. sapply(split(mydf, mydf$type), function(speed, n_vehicles) sum(mydf$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

> On 9 Nov 2017, at 14:58, PIKAL Petr <petr.pikal at precheza.cz> wrote: > > Hi > > Thanks for working example. > > you could use split/ lapply approach, however it is probably not much better than dplyr method. > > sapply(split(mydf, mydf$type), function(speed, n_vehicles) sum(mydf$speed*mydf$n_vehicles)/sum(mydf$n_vehicles)) > gives you averages > The

Dear Massimo, It seems straightforward to use weighted.mean() in a dplyr context library(dplyr) mydf %>% group_by(date_time, type) %>% summarise(vel = weighted.mean(speed, n_vehicles)) Best regards, ir. Thierry Onkelinx Statisticus / Statistician Vlaamse Overheid / Government of Flanders INSTITUUT VOOR NATUUR- EN BOSONDERZOEK / RESEARCH INSTITUTE FOR NATURE AND FOREST Team

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

yes, it works, even if I do not really get how and why it's working the combination of logical results (could you provide some insights for that?) moreover, and most of all, 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:

Hello, Try the following. icol <- which(grepl("flag", names(mydf))) mydf[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

Do you mean like this: mydf <- within(mydf, { is.na(A)<- !A_flag is.na(B)<- !B_flag } ) > mydf A A_flag B B_flag 1 8 10 5 12 2 NA 0 6 9 3 10 1 NA 0 4 NA 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

...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

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) >

OPS, Sorry i did not read the post carfully. Mine will not work if you have zeros on columns 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

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

Hi All. I have a mysql statement that I would really really like to call from my Ruby program which goes like this: SELECT a, b, DAYOFWEEK(date_time) as DOW, HOUR(date_time) at hr, AVG(x/y) FROM records; This is possible by creating a 3-dimentional array of a, b, date_time containing x/y, and then finding averages and putting it into a 4-dimensional array of a, b, dow,

Hi, I have a irregularly spaced time series dataset, which reads in from a .csv. I need to convert this to a regularly spaced time series by filling in missing rows of data with NAs. So my data, called NtuMot, looks like this (I've removed some of the additional rows for simplicity).... ELEID date_time height slope 1 2009-06-24 00:00:00

Dear All, I seem to have stumbled across an AGI problem; I have written an AGI Script (bottom of this email); The script does the following; Makes a CDR entry when called Records the call Updates the CDR Finds a corresponding DNIS from the SMDR table (captured via a serial port logger) Matches up the record and updates the CDR. The script works perfectly in my test lab and has been doing so

Look at the bottom of the message for my question #here is a little function that I wrote USGS <- function(input="discharge", days=7){ library(chron) library(gsubfn) #021973269 is the Waynesboro Gauge on the Savannah River Proper (SRS) #02102908 is the Flat Creek Gauge (ftbrfcms) #02133500 is the Drowning Creek (ftbrbmcm) #02341800 is the Upatoi Creek Near Columbus (ftbn) #02342500 is

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Hi, I'm new to R and new to this forum. I'm struggling with trying to extract certain rows of data from my data.frame. The data.frame has eleven columns. Among those columns are "FISH_ID" and "DATE_TIME". FISH_ID is a factor. For each of my 21 unique FISH_IDs (levels) I have a few to a few thousand rows, each row with a unique DATE_TIME value. I would like to obtain,

note: whole function is below- I am sure I am doing something silly. when I use it like USGS(input="precipitation") it is choking on the precip.1 <- subset(DF, precipitation!="NA") b <- ddply(precip.1$precipitation, .(precip.1$gauge_name), cumsum) DF.precip <- precip.1 DF.precip$precipitation <- b$.data part, but runs fine outside of the function: days=7