similar to: how to calculate the mean in a period of time?

Hi, dat1<- read.table(text=" id??????????????? t???????????????????? scores 2???????????????? 0??????????????????????? 1.2 2???????????????? 2???????????????????????? 2.3 2???????????????? 3??????????????????????? 3.6 2???????????????? 4??????????????????????? 5.6 2???????????????? 6??????????????????????? 7.8 3???????????????? 0??????????????????????? 1.6 3????????????????

Hi, I am not sure I understand your question correctly. dat1<- read.table(text=" id??????????? responsed_at???????????????? number_of_connection????????????????? scores 1????????????????? 12-01-2010?????????????????????????????????? 1????????????????????????????????????????????? 2 1????????????????? 15-02-2010??????????????????????????????????

Hi, I guess you meant this: dat2<- read.table(text=" patient_id????? t???????? scores 1????????????????????? 0??????????????? 1.6 1????????????????????? 1??????????????? 2.6 1????????????????????? 2???????????????? 2.2 1????????????????????? 3???????????????? 1.8 2????????????????????? 0????????????????? 2.3 2?????????????????????? 2???????????????? 2.5 2?????????????????????

HI, You can do this: dat1<- read.csv("dat7.csv",header=TRUE,stringsAsFactors=FALSE,sep="\t") dat.bru<- dat1[!is.na(dat1$evnmt_brutal),] fun2<- function(dat){? ????? lst1<- split(dat,dat$patient_id) ??? lst2<- lapply(lst1,function(x) x[cumsum(x$evnmt_brutal==0)>0,]) ??? lst3<- lapply(lst2,function(x) x[!(all(x$evnmt_brutal==1)|all(x$evnmt_brutal==0)),])

Hi, May be this helps: dat1<- read.csv("dat7.csv",header=TRUE,stringsAsFactors=FALSE,sep="\t") dat.bru<- dat1[!is.na(dat1$evnmt_brutal),] fun2<- function(dat){?? ????? lst1<- split(dat,dat$patient_id) ??? lst2<- lapply(lst1,function(x) x[cumsum(x$evnmt_brutal==0)>0,]) ??? lst3<- lapply(lst2,function(x) x[!(all(x$evnmt_brutal==1)|all(x$evnmt_brutal==0)),])

Hi, Try this: dat1<- read.csv("dat7.csv",header=TRUE,stringsAsFactors=FALSE,sep="\t") dat.bru<- dat1[!is.na(dat1$evnmt_brutal),] fun1<- function(dat){??? ? ??? lst1<- split(dat,dat$patient_id) ??? lst2<- lapply(lst1,function(x) x[cumsum(x$evnmt_brutal==0)>0,]) ??? lst3<- lapply(lst2,function(x) x[!(all(x$evnmt_brutal==1)|all(x$evnmt_brutal==0)),]) ???

Hello R experts, I've used metaMDS to run NMDS on some fish abundance data, and am also working on correlating environmental data to the NMDS coordinates. I'm fairly new to metaMDS and NMDS in general, so I have what are probably some very basic questions. My fish abundance data consists of 66 sites for which up to 20 species of fish were identified and counted. I ran metaMDS on this data

HI Elisa, You could use ?cut() vec1<-c(33,18,13,47,30,10,6,21,39,25,40,29,14,16,44,1,41,4,15,20,46,32,38,5,31,12,48,27,36,24,34,2,35,11,42,9,8,7,26,22,43,17,19,28,23,3,49,37,50,45) label1<-unlist(lapply(mapply(c,lapply(seq(0,45,5),function(x) x),lapply(seq(5,50,5),function(x) x),SIMPLIFY=FALSE),function(i) paste(i[1],"<x<=",i[2],sep="")))

Hi, Try: x<- 1:4 ?y<- c("*","/","-","+") res<-sapply(y,function(i) {x1<-expand.grid(x,x); unlist(lapply(paste0(x1[,1],i,x1[,2]),function(u) eval(parse(text=u))))}) row.names(res)<- as.character(interaction(expand.grid(x,x),sep="_")) head(res) #??? *?? /? - + #1_1 1 1.0? 0 2 #2_1 2 2.0? 1 3 #3_1 3 3.0? 2 4 #4_1 4 4.0? 3 5 #1_2 2 0.5

Dear R help, I have a doubt regarding the write.csv function.? When I save a file (write.csv(res1,"res1.csv")), it is saving with the read-only permission.? I tried to change the permission as a root user with chmod u+x.? Though, it says that "-rwxr--r--? 1 root root???? 164 2012-05-27 10:18 res1.csv", when I open the file it is still read-only.? Is there any way to save the

Hi, You could try: A<- matrix(unlist(read.table(text=" 1 2 3 4 5 6 7 8 9 9 8 7 6 5 4 3 2 1 ",sep="",header=FALSE)),ncol=3,byrow=FALSE,dimnames=NULL) library(matrixStats) ?res1<-t(sapply(split(as.data.frame(A),as.numeric(gl(nrow(A),2,6))),colProds)) ?res1 #? [,1] [,2] [,3] #1??? 4?? 10?? 18 #2?? 63?? 64?? 63 #3?? 18?? 10??? 4

Hello R community, I am trying to combine two CSV files that look like this: File A Row_ID_CR, Data1, Data2, Data3 1, aa, bb, cc 2, dd, ee, ff File B Row_ID_N, Src_Row_ID, DataN1 1a, 1, This is comment 1 2a, 1, This is comment 2 3a,

Hi the list, I write a short function to draw lines in 3D, showing then turning. At some point, I add "delais" to slow down the rotation. So two questions: 1) I try to find a library to draw animate lines in 3D but I did not find. That surprise me since it is very simple to do. Did I forget to look somewhere ? If it does not exists and I have to use my own function : 2) Is it

Hi Gustav, Try this: lapply(1:length(models),function(i) lapply(models[[i]],function(x) summary(x)$coef[2,]))[[1]] #1st list component [[1]] #??? Estimate?? Std. Error????? z value???? Pr(>|z|) # pm10 #5.999185e-04 1.486195e-04 4.036606e+00 5.423004e-05 #[[2]] #??? Estimate?? Std. Error????? z value???? Pr(>|z|) #ozone #0.0010117294 0.0003792739 2.6675428048 0.0076408155 #[[3]] #???

Thanks. Yes. Your approach can identify: Glaxy ace S 5830 and S 5830 Glaxy ace But you can not identify using same program: Iphone 4S 16 G Iphone 4S 16G How should I solve both in same time. Kind regards,Tammy [[alternative HTML version deleted]]

Hi, Suppose, if I create 15 files in my working directory. set.seed(48) lapply(1:15,function(i) {m1 <- matrix(sample(1:20,1686*2,replace=TRUE),nrow=1686,ncol=2); write.table(m1,paste0("file_",i,".txt"),row.names=FALSE,quote=FALSE)}) ?D <-dir() D1 <- D[order(as.numeric(gsub("\\D+","",D)))] D1 ?res <- t(sapply(D1,function(x) {x1<-

I have and XTS time series object that has date and time. I started with 1 minute data and used apply.daily(x, sum) to sum the data to one cumulative value. This function works just fine however it leaves a time for the last summed value which looks like this 2006-07-19 14:58:00. I need to just have the date and to remove the time value of 14:58:00 just leaving the date value of 2006-07-19 .

Hello, Hope that someone could help me plotting longitudinal data below: 7213 3333330001 0.8300 13.05.09 1 1 3333330001 0.8700 09.02.05 NULL 4797 3333330001 0.7700 21.03.07 NULL 2399 3333330001 0.7800 12.04.06 NULL 2400 3333330002 NULL 27.03.06 NULL 7230 3333330002 0.8200 14.05.09 0 2 3333330002 0.8400 09.02.05 NULL 4798 3333330002 0.8700 20.03.07 0 4799 3333330003 0.9000 20.03.07 13 2401

Dear R People: I have two data frames, a.df and b.df as seen here: > a.df[1:10,] DATE GENDER PATIENT_ID AGE SYNDROME 1 4/16/2009 F 23686 45 RASH ON BODY 2 4/16/2009 F 13840 35 CANT URINATE 3 4/16/2009 M 12895 30 BLURRED VISION 4 4/16/2009 M 18375 33 UNABLE TO VOID 5 4/16/2009 M 2237 44

Thank you very much for your reply. Your code work well with this example. I modified a little to fit my real data, I got an error massage. Error in split.default(x = seq_len(nrow(x)), f = f, drop = drop, ...) : Group length is 0 but data length > 0 On Thu, Jan 31, 2013 at 12:21 PM, arun kirshna [via R] < ml-node+s789695n4657196h87@n4.nabble.com> wrote: > Hi, > Try this: >