Here's how I would have done the data.table method. It's a bit faster
than the ave approach on my machine:
> # install.packages("data.table",repos="http://R-Forge.R-project.org";)
> library(data.table)
>
> f3 <- function(frame) {
+ frame <- as.data.table(frame)
+ frame[, lapply(.SD[,2:ncol(.SD), with = FALSE],
+ function(x) x / mean(x, na.rm = TRUE)),
+ by = "group"]
+ }
>
> system.time(new.frame2 <- f2(frame)) # ave
user system elapsed
0.50 0.08 1.24
> system.time(new.frame3 <- f3(frame)) # data.table
user system elapsed
0.25 0.01 0.30
- Tom
Tom Short
On Wed, Apr 7, 2010 at 12:46 PM, Dimitri Liakhovitski <ld7...@gmail.com> wrote:
> I would like to thank once more everyone who helped me with this question.
> I compared the speed for different approaches. Below are the results
> of my comparisons - in case anyone is interested:
>
> ### Building an EXAMPLE FRAME with N rows - with groups and a lot of NAs:
> N<-100000
> set.seed(1234)
> frame<-data.frame(group=rep(paste("group",1:10),N/10),a=rnorm(1:N),b=rnorm(1:N),c=rnorm(1:N),d=rnorm(1:N),e=rnorm(1:N),f=rnorm(1:N),g=rnorm(1:N))
> frame<-frame[order(frame$group),]
>
> ## Introducing 60% NAs:
> names.used<-names(frame)[2:length(frame)]
> set.seed(1234)
> for(i in names.used){
> i.for.NA<-sample(1:N,round((N*.6),0))
> frame[[i]][i.for.NA]<-NA
> }
> lapply(frame[2:8], function(x) length(x[is.na(x)])) # Checking that it worked
> ORIGframe<-frame ## placeholder for the unchanged original frame
>
> ####### Objective of the code - divide each value by its group mean ####
>
> ### METHOD 1 - the FASTEST - using ave():##############################
> frame<-ORIGframe
> f2 <- function(frame) {
> for(i in 2:ncol(frame)) {
> frame[,i] <- ave(frame[,i], frame[,1],
> FUN=function(x)x/mean(x,na.rm=TRUE))
> }
> frame
> }
> system.time({new.frame<-f2(frame)})
> # Took me 0.23-0.27 sec
> #######################################
>
> ### METHOD 2 - fast, just a bit slower - using data.table:
> ##############################
>
> # If you don't have it - install the package - NOT from CRAN:
> install.packages("data.table",repos="http://R-Forge.R-project.org";)
> library(data.table)
> frame<-ORIGframe
> system.time({
> table<-data.table(frame)
> colMeanFunction<-function(data,key){
> data[[key]]=NULL
> ret=as.matrix(data)/matrix(rep(as.numeric(colMeans(as.data.frame(data),na.rm=T)),nrow(data)),nrow=nrow(data),ncol=ncol(data),byrow=T)
> return(ret)
> }
> groupedMeans = table[,colMeanFunction(.SD, "group"), by="group"]
> names.to.use<-names(groupedMeans)
> for(i in
> 1:length(groupedMeans)){groupedMeans[[i]]<-as.data.frame(groupedMeans[[i]])}
> groupedMeans<-do.call(cbind, groupedMeans)
> names(groupedMeans)<-names.to.use
> })
> # Took me 0.37-.45 sec
> #######################################
>
> ### METHOD 3 - fast, a tad slower (using model.matrix & matrix
> multiplication):##############################
> frame<-ORIGframe
> system.time({
> mat <- as.matrix(frame[,-1])
> mm <- model.matrix(~0+group,frame)
> col.grp.N <- crossprod( !is.na(mat), mm ) # Use this line if don't
> want to use NAs for mean calculations
> # col.grp.N <- crossprod( mat != 0 , mm ) # Use this line if don't
> want to use zeros for mean calculations
> mat[is.na(mat)] <- 0.0
> col.grp.sum <- crossprod( mat, mm )
> mat <- mat / ( t(col.grp.sum/col.grp.N)[ frame$group,] )
> is.na(mat) <- is.na(frame[,-1])
> mat<-as.data.frame(mat)
> })
> # Took me 0.44-0.50 sec
> #######################################
>
> ### METHOD 5- much slower - it's the one I started
> with:##############################
> frame<-ORIGframe
> system.time({
> frame <- do.call(cbind, lapply(names.used, function(x){
> unlist(by(frame, frame$group, function(y) y[,x] / mean(y[,x],na.rm=T)))
> }))
> })
> # Took me 1.25-1.32 min
> #######################################
>
> ### METHOD 6 - the slowest; using "plyr" and
> "ddply":##############################
> frame<-ORIGframe
> library(plyr)
> function3 <- function(x) x / mean(x, na.rm = TRUE)
> system.time({
> grouping.factor<-"group"
> myvariables<-names(frame)[2:8]
> frame3<-ddply(frame, grouping.factor, colwise(function3, myvariables))
> })
> # Took me 1.36-1.47 min
> #######################################
>
>
> Thanks again!
> Dimitri
>
>
> On Wed, Mar 31, 2010 at 8:29 PM, William Dunlap <wdun...@tibco.com> wrote:
>> Dimitri,
>>
>> You might try applying ave() to each column. E.g., use
>>
>> f2 <- function(frame) {
>> for(i in 2:ncol(frame)) {
>> frame[,i] <- ave(frame[,i], frame[,1],
>> FUN=function(x)x/mean(x,na.rm=TRUE))
>> }
>> frame
>> }
>>
>> Note that this returns a data.frame and retains the
>> grouping column (the first) while your original
>> code returns a matrix without the grouping column.
>>
>> Bill Dunlap
>> Spotfire, TIBCO Software
>> wdunlap tibco.com
>>
>>> -----Original Message-----
>>> From: r-help-boun...@r-project.org
>>> [mailto:r-help-boun...@r-project.org] On Behalf Of Bert Gunter
>>> Sent: Tuesday, March 30, 2010 10:52 AM
>>> To: 'Dimitri Liakhovitski'; 'r-help'
>>> Subject: Re: [R] Code is too slow: mean-centering variables
>>> in a data framebysubgroup
>>>
>>> ?scale
>>>
>>> Bert Gunter
>>> Genentech Nonclinical Biostatistics
>>>
>>>
>>>
>>> -----Original Message-----
>>> From: r-help-boun...@r-project.org
>>> [mailto:r-help-boun...@r-project.org] On
>>> Behalf Of Dimitri Liakhovitski
>>> Sent: Tuesday, March 30, 2010 8:05 AM
>>> To: r-help
>>> Subject: [R] Code is too slow: mean-centering variables in a
>>> data frame
>>> bysubgroup
>>>
>>> Dear R-ers,
>>>
>>> I have a large data frame (several thousands of rows and about 2.5
>>> thousand columns). One variable ("group") is a grouping variable with
>>> over 30 levels. And I have a lot of NAs.
>>> For each variable, I need to divide each value by variable mean - by
>>> subgroup. I have the code but it's way too slow - takes me about 1.5
>>> hours.
>>> Below is a data example and my code that is too slow. Is there a
>>> different, faster way of doing the same thing?
>>> Thanks a lot for your advice!
>>>
>>> Dimitri
>>>
>>>
>>> # Building an example frame - with groups and a lot of NAs:
>>> set.seed(1234)
>>> frame<-data.frame(group=rep(paste("group",1:10),10),a=rnorm(1:
>> 100),b=rnorm(1
>>> :100),c=rnorm(1:100),d=rnorm(1:100),e=rnorm(1:100),f=rnorm(1:1
>>> 00),g=rnorm(1:
>>> 100))
>>> frame<-frame[order(frame$group),]
>>> names.used<-names(frame)[2:length(frame)]
>>> set.seed(1234)
>>> for(i in names.used){
>>> i.for.NA<-sample(1:100,60)
>>> frame[[i]][i.for.NA]<-NA
>>> }
>>> frame
>>>
>>> ### Code that does what's needed but is too slow:
>>> Start<-Sys.time()
>>> frame <- do.call(cbind, lapply(names.used, function(x){
>>> unlist(by(frame, frame$group, function(y) y[,x] /
>>> mean(y[,x],na.rm=T)))
>>> }))
>>> Finish<-Sys.time()
>>> print(Finish-Start) # Takes too long
>>>
>>> --
>>> Dimitri Liakhovitski
>>> Ninah.com
>>> dimitri.liakhovit...@ninah.com
>>>
>>> ______________________________________________
>>> R-help@r-project.org mailing list
>>> 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@r-project.org mailing list
>>> 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.
>>>
>>
>
>
>
> --
> Dimitri Liakhovitski
> Ninah.com
> dimitri.liakhovit...@ninah.com
>
> ______________________________________________
> R-help@r-project.org mailing list
> 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.
>
______________________________________________
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and provide commented, minimal, self-contained, reproducible code.
