similar to: Looping an lapply linear regression function

Ira, obj_name<- load("arun.RData") Pred1<- get(obj_name[1]) Actual1<- get(obj_name[2]) dat2<- data.frame(S1=rep(Pred1[,1],ncol(Pred1)-1),variable=rep(colnames(Pred1)[-1],each=nrow(Pred1)),Predict=unlist(Pred1[,-1],use.names=FALSE),Actual=unlist(Actual1[,-1],use.names=FALSE),stringsAsFactors=FALSE) dat2New<- dat2[!(is.na(dat2$Predict)|is.na(dat2$Actual)),] ?dat3<-

Hi, May be you can try this: dat1New<-? dat1[!(duplicated(dat1$gene)|duplicated(dat1$gene,fromLast=TRUE)),] dat2<-dat1[duplicated(dat1$gene)|duplicated(dat1$gene,fromLast=TRUE),] ?lst1<-split(dat2,dat2$gene) dat3<-unsplit(lapply(lst1,function(x) {x1<- sum(apply(x[,6:32],2,function(y) y[1]>=y[2]));x2<- sum(apply(x[,6:32],2, function(y) y[1]<=y[2])); if(x1>x2) x[1,] else

In one particular situation predict.coxph gives an error message. Namely: stratified data, predict='expected', new data, se=TRUE. I think I found the error but I'll leave that to you to decide. Thanks, Chris ######## CODE library(survival) set.seed(20121221) nn <- 10 # sample size in each group lambda0 <- 0.1 # event rate in group 0 lambda1 <- 0.2 # event rate in group 1

It's generally a very good idea to examine the structure of data after you have read it in. str(data2) would have shown you that read.csv() turned your strings into factors, and that's why the == operator no longer does what you think it does. use ... data_2 <- read.csv("excel_data.csv", stringsAsFactors = FALSE) ... to turn this off. Also, the %in% operator will achieve

Combining and completing the advice from Greg and Boris the complete solution is two lines: data_2 <- read.csv("excel_data.csv", stringsAsFactors = FALSE) match_list <- match( data_2$data1, data_2$data2 ) The vector match_list will have the matching position when it exists and NA's otherwise. Its length will be the same as the length of data_2$data1. You should get

Hi, I have two columns that contain numbers along with letters (as shown below) and have different lengths. Each entry in the first column is likely to be found in the second column at most once. For each entry of the first column, if that entry is found in the second column, I would like to get the corresponding index. For instance, if the first entry of the first column is 5th entry in the

HI, You can do this in many ways: dat1<-read.table(text=" med1,med2,med3???? ?1,0,1?????? 0,1,1??? 2,0,0 ",sep=",",header=TRUE)?? #1st method library(reshape) dat2<-melt(dat1) dat3<-aggregate(dat2$value,by=list(dat2$variable),sum) ?colnames(dat3)<-c("name","sum(n11)") ?dat3 #? name sum(n11) #1 med1??????? 3 #2 med2??????? 1 #3 med3??????? 2

hi... i am a phd student using r. i am having difficulty plotting average effects. admittedly, i am not really understanding what each of the commands mean so when i get the error i am not sure where the issue is. here is my code... i will include the points at which there are errors.... > dat2 <- dat3 <- dat > dat2$popc100 <- dat2$popc100 + 1000 >

Can somebody tell me that, if I do some arithmetic calculation over 2 matrices then how the column names and row names are preserved? It seems that, for multiplication, column names and row names of the 2nd matrix are preserved and for additional, there seems not having any explicit rule: > set.seed(1) > dat1 <- matrix(rnorm(25), 5); colnames(dat1) = rownames(dat1) =

HI Utpal, Alight, I will look into it.? I was under the impression that this is what you wanted: dat1<- structure(list(V1 = c(1L, 1L, 1L, 1L, 1L, 1L, 1L), V2 = c(1L, 1L, 1L, 1L, 1L, 1L, 0L), V3 = c(1L, 1L, 1L, 1L, 1L, 1L, 1L), ??? V4 = c(1L, 1L, 0L, 1L, 1L, 1L, 1L), V5 = c(1L, 1L, 1L, 1L, ??? 1L, 0L, 1L), V6 = c(1L, 1L, 1L, 0L, 1L, 1L, 1L), V7 = c(1L, ??? 1L, 1L, 1L, 1L, 1L, 1L), V8 = c(1L,

I have a series of animals that are being radio-tracked. I have written some code that goes through each animal and calculates the total distance that each animal is traveling. I will need to run these data on a regular basis and want to write the result to a .php webpage file so that the page updates automatically when I re-run the code. I thought I'd be able to use the

I've been trying with no success to model mixtures of Gamma distributions using the package flexmix (see examples below). Can anyone help me get it to model better? Thanks very much. -Ben ## ## Please help me get flexmix to correctly model mixtures of ## Gamma distributions. See examples below. ## library('flexmix') ## ## Plot a histogram of dat and the Gamma mixture model given

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Well, here is one way but it seems a bit clumsy. In words, I created a new data.frame with "250" in the Chloride vector and "SMCL" in the Detections vector and supplessed one legend. Warning: For my convenience I am using different data.frame names . library(ggplot2) MyData <-read.csv("http://doylesdartden.com/Stats/TimeSeriesExample.csv", sep=",")

Hi, From your example data, dat1<- read.table(text=" id1?? id2?? value a????? b?????? 10 c????? d??????? 11 b???? a???????? 10 c????? e???????? 12 ",sep="",header=TRUE,stringsAsFactors=FALSE) #it is easier to get the output you wanted dat1[!duplicated(dat1$value),] #? id1 id2 value #1?? a?? b??? 10 #2?? c?? d??? 11 #4?? c?? e??? 12 But, if you have cases like the one

Hi Elisa, You need to check your data.? For some 'st', the data is repeated/duplicated (I am assuming, didn't check it) especially for a particular year. dat1<-read.csv("elisa.csv",sep="\t") dat1$st<- as.character(dat1$st) ?str(dat1) #'data.frame':??? 506953 obs. of? 5 variables: # $ st?????? : chr? "AGOMO" "AGOMO"

I care a lot about R's speed. So I decided to give REvolution's R (http://revolution-computing.com/) a try, which bills itself as an optimized R. Note that I used the free version. My machine is a Intel core 2 duo under Windows XP professional. The code I run is in the end of this post. First, the regular R 1.9. It takes 2 minutes and 6 seconds, CPU usage 50% Next, REvolution's R.

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, May be you can try this: obj_name<- load("arun.RData") Pred1<- get(obj_name[1]) Actual1<- get(obj_name[2]) library(reshape2) dat<-cbind(melt(Pred1,id.vars="S1"),value2=melt(Actual1,id.vars="S1")[,3])? # to reshape to long form colnames(dat)[3:4]<- c("Predict","Actual") dat$variable<- as.character(dat$variable) #not that

Hi all Say I have a function: myname=function(dat,x=5,y=6){ res<<-x+y-dat } for various input such as myname(dat1) myname(dat2) myname(dat3) myname(dat4) myname(dat5) how should I modify the 'res' line, to have new informative variable name correspondingly, such as dat1.res dat2.res dat3.res dat4.res dat5.res stored in the workspace. This is only an example of a complex