Re: [R] help needed in least square curve fit equation

As I understand, it is an optimization problem. LeastSquare=sum[y-f(x)]^2 Minimize least square by solving equations about a, b, and c. An iterative method could be developed to get the result or some R functions might be found useful. Please refer http://r.789695.n4.nabble.com/Is-there-any-Gaus

Re: [R] non-linear fourth-order differential equations

Hi Ravi, Thank you for your correction. I hope I didn't mess up anything:) Cheers. Wu - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/non-linear-fourth-order-differential-equations-tp3062805p3063761.html Sent from the R help mailing list archive at Nabble.co

Re: [R] periodic time series

Hi Andy, If you could provide a sample data set, it would help others to give a solution. I suggest look at the data and select a model, then anova. Take group as one variable, record time (1 to 24 ) as the second variable and the week day (Monday to Friday) as the third variable. Then test the

Re: [R] non-linear fourth-order differential equations

Hi Yanika, Please try ?uniroot and ?ployroot f <- function(x) x^4-16 uniroot(f, lower= -3, upper=0) polyroot(c(-16,0,0,0,1)) - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/non-linear-fourth-order-differential-equations-tp3062805p3062894.html Sent from the R

Re: [R] Matrix Operations

Hi Romildo, Merge two table by id2 first, then reshape to the wide format, sum by source at last. Hope it helps. ### Data simulation lsp.text <- " id source destiny id2 caminho order 1 1 2 4 7 0 0 2 2 6 10 4 0 0 3 3 6

Re: [R] use sample function to create new data frame

Hi, I would try to sample by row index and column index. Hope it helps. df <- data.frame(matrix(1:18,nrow=3,ncol=6,byrow=TRUE)) colnames(df) <- c(letters[1:3],LETTERS[1:3]) ## Generate a new data frame with 10 rows one <- two <- three <- numeric(10) for (i in 1:10) { ## Sample from type I col

[R] How to plot a normal distribution curve and a shaded tail with alpha?

I want to create a graph to express the idea of the area under a pdf curve, like http://r.789695.n4.nabble.com/file/n3032194/w7295e04.jpg Thank you for any help. - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/How-to-plot-a-normal-distribution-curve-and-a-sh

Re: [R] transfer string to expression

Hi Duncan: I'm curious about the environment setting. ?eval says: "If envir is not specified, then the default is parent.frame() (the environment where the call to eval was made). " So what's the difference between set envir=parent.frame() or not? Thank you. Wu - A R learner. -- View t

Re: [R] Randomly split a sample in two equal subsamples

firsthalf <- myframe[v1,] or firsthalf <- subset(myframe, number %in% v1) - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/Randomly-split-a-sample-in-two-equal-subsamples-tp3021140p3021353.html Sent from the R help mailing list archive at Nabble.com. __

Re: [R] Randomly split a sample in two equal subsamples

Hi Yoan, Please try ?sample. Suppose you have 1:n ids of total observations where n is even, you want to randomly split it into two subsamples, the following code should work. n <- 20 one.sample <- sort(sample(1:n, n/2)) another.sample <- (1:n)[-one.sample] Good luck. Wu - A R learner.

Re: [R] Help: Using vectorization method for vectors comparision

Hi Bruclee, ?rle may help. a <- c(5, 10, 13, 19, 23) b <- c(1, 4, 7, 9, 15) ab <- data.frame(value = c(a,b), type=c(rep(0,length(a)),rep(1,length(b ab <- ab[order(ab$value),] ab$v2 <- cumsum(ab$type) ab$matched <- rep(ab$value[ab$type==1],rle(ab$v2)$lengths) (result <- ab[ab$type==0,c("valu

Re: [R] sampling from normal

Hi Solafah, You are right that two commands are equivalent when p= pnorm(a). You can check the results by following codes. n <- 5 a <- -1 set.seed(123456) qnorm(runif(n,0,pnorm(a))) p <- pnorm(a) set.seed(123456) qnorm(p*runif(n)) Anyway, the elements of the lower tail are not chosen equally by

Re: [R] Data contamination

Hi, You are right, runif(5,1,50) could generate a same value twice. And I think that your code runs runif(5,1,50) twice too. Try ?sample selected <- sample(1:50,3) data.sim[selected,1] <- data.sim[selected,1] + rnorm(5, mean=20, sd=1) Hope it helps. - A R learner. -- View this message in

Re: [R] how to convert string to object?

Hi, Please try ?parse and ?eval. Here is an example: text <- "3*6" parse(text=text) eval(parse(text=text)) - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/how-to-convert-string-to-object-tp2999281p2999336.html Sent from the R help mailing list archive at Nabb

[R] 95% CI of Kaplan-Meier survival in package survival

Hi: My result using Kaplan-Meier estimate in survival package was inconsistent with that from Minitab. The survival probabilities are same, but their 95% CI are different from other calculation. Could you help me find what is wrong with the code? Thanks in advance. library(survival) raw.a <- c(8

Re: [R] paired samples, matching rows, merge()

Hi Cecília, Assuming that you want to pair two samples as more as possible, let's relabel the k1 indices. The x$k1 c(1,1,2,3,3,5) could be relabeled as c(1.01,1.02,2.01,3.01,3.02,5.01), so as y$k1. x$k1 <- x$k1+sequence(rle(x$k1)$lengths)/100 y$k1 <- y$k1+sequence(rle(y$k1)$lengths)/100 merge(x

Re: [R] functions and multiple levels

Hi Chris, Try ?ave will help you. Anyway, I guess you are computing a statistic. strs <- " level.1 level.2 observation 1 1 0.5 1 1 0.2 1 2 0.6 1 2 0

Re: [R] Fwd: R SOFTWARE

Hi Nokuzola, I guess the capital "T". 'c:/temp/M5/limestone.dat' C:\Temp\M5\limestone.dat Regards, Wu - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/Fwd-R-SOFTWARE-tp2329861p2329887.html Sent from the R help mailing list archive at Nabble.com. __

Re: [R] identical()

Hi John, They are different as Erik said. > identical(a,b) [1] FALSE > row.names(a) [1] "2" "4" "6" "8" > row.names(b) [1] "1" "3" "5" "7" > row.names(a) <- NULL > row.names(b) <- NULL > identical(a,b) [1] TRUE Regards, Wu array chip wrote: > >> a > a b > 2 10011048 L > 4 100110

Re: [R] print numbers

Hi John, formatC will do your work. Hope it helps. x <- y <- c(50.00,25.00,10.00,1.00,0.05,0.01) plot(x,y,log = "xy",axes = F) axis(1, x, formatC(x)) axis(2, y, formatC(y)) Regards, Wu - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/print-numbers-tp2327378p

Re: [R] Scatterplot - Overlap Frequency

Hi Marcio, Your friend has given the answer. x <- rnorm(10) y <- rnorm(10) ind <- c(3,0,1,0,3,0,2,2,0,0) plot(x, y, col = grey(0:max(ind)/max(ind))[ind], pch = 16) Mestat wrote: > > I am working o a scatterplot where I would like to plot the variables > according with another frequency var

Re: [R] Append to csv without header

Hi, It seems that write.csv doesn't support append now. Use write.table() write.table(dataF, file = outputFilePath, append=T,col.names =F) Regards, Wu - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/Append-to-csv-without-header-tp2322115p2322172.html Sent

Re: [R] How to calculate the concentration

Hi, The code seems complicate, but it's understandable. ## food=c('fruit','fruit','fruit','drink','drink','drink') type=c('apple','apple','orange','water','soda','soda') value=c(2,3,1,5,7,6) data=data.frame(food,type,value) share=c((2+3)/(2+3+1),5/6,1/6,5/(5+7+6),13/18,13/18) market_con=c(re

Re: [R] How to calculate the concentration

Hi Yi, It would be helpful for others to provide a solution if you give your formulas that calculating the value of share and concentration. ?apply will helps. Regards, Wu - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/How-to-calculate-the-concentration-

Re: [R] Creating vectors

Hi, Try ?unique please. x <- c(2,2,9,4,6,2,4,4,6,8,6) # Original vector unique(x) #New vector only has unique elements sort(unique(x)) # Ordered Regards, Wu - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/Creating-vectors-tp2321440p2321884.html Sent fro

Re: [R] Colour Point Plot

Hi, ?plot will help you. plot(0:18,col="white") points(0:18,0:18,pch=0:18,col=rainbow(19),cex=2) Turn & Fall wrote: > > How do you change the standard plotting function so that points are solid > and can have a pretty colour? > - A R learner. -- View this message in context: http://r

Re: [R] Running something without a loop when the result from the previous iteration is require for the current iteration

Hi Adrienne, I guess apply should be better than for loop. Code like this: event.gen2 = function(genmat,use1,use2,num,ortho_obs_used){ onerow.gen <- function(one.row, use1){ one.row[num] <- ifelse(...} genmat[,num] <- NA ##Add one row with NA values apply(

Re: [R] Intersecting list vs rows in matrix

Hi GL, Erik has given a data.frame solution. If you data is a list with unequal lengths, try lapply. - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/Intersecting-list-vs-rows-in-matrix-tp2320427p2320463.html Sent from the R help mailing list archive at Nabble.

Re: [R] How to invert a list ?

Hi Carlos, I give a handmade code, hope it helps. y <- list() y$a <- a y$b <- c(b,c) names(y$a) <- "i" names(y$b) <- c("j","k") Carlos Petti wrote: > > a <- 5 > names(a) <- "a" > b <- 9 > names(b) <- "b" > c <- 15 > names(c) <- "c" > x <- list("i" = a, "j" = b, "j" = c) > - A R learner

Re: [R] matrix problem

Hi, I guess you just want to reshape your data to wide format. strs <- "Index Time Value 1 2 0.1 2 3 0.2 3 1 0.3" DF <- read.table(textConnection(strs),header=T) rDF <- reshape(DF, idvar="Index", timevar="Time", direction="wide") rDF[is.na(rDF)] <- 0 - A R learner. -- View this message in

Re: [R] TRUE/FALSE

iginal message which itself might quote the original original message .., seem to be too much redundant to me. I have been trying to be a well-behaved member of the R community:) Best Regards, Wu Gong Uwe Ligges-3 wrote: > > Can you please refer and quote the original messages you are

Re: [R] TRUE/FALSE

I can't understand why you doubled "(" "if((combos[e,f]==1) " - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/TRUE-FALSE-tp2318668p2318684.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-pr

Re: [R] Regular Expression

gsub(pattern = "^[0-9]+ (SPE )*(\\w+) - .*$", "\\2", dat) - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/Regular-Expression-tp2318086p2318101.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r

Re: [R] Bootstrap

Hi, I assume you used boot package. Try this: library(boot) x <- rnorm(136) fun <- function(x,ind){ x <- x[ind] m <- mean(x[63:136])-mean(x[1:62]) m } boot.x<- boot(x, fun, R = 5000, sim = "ordinary") - A R learner. -- View this message in context: http:/

Re: [R] apply family functions

Hi Steven, You code has two problems. One is loop structure which should be for(i in 1:length). The second is that you loop the process twice (for and sapply). Hope followed code will work. for (i in 1:length(ind)) { x <- dat$Close_date[i] dat[["Flag"]][i] <- ifelse((x >= oc[ind[[i

Re: [R] stats::reshape question

It seems that we can't change the order of varying argument or v.names. "Notice that the order of variables in varying is like x.1,y.1,x.2,y.2. " Code can only be: reshape(d,varying=c("x1","y1","x2","y2"),v.names=c("x","y"),dir="long") - A R learner. -- View this message in context: http:

Re: [R] Multiply each depth level of an array with another vector element

It's interesting that sweep is the slowest one comparing to replicate and rep :) - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/Multiply-each-depth-level-of-an-array-with-another-vector-element-tp2315537p2316586.html Sent from the R help mailing list archive a

Re: [R] gsub

i <- "piante_venere.csv" gsub("^.*_(.*)\\.csv$", "\\1", i) - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/gsub-tp2316443p2316550.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org

Re: [R] How to read a file inside a function?

ff <- function(filename) { dat1 <- read.csv(file=filename,header=TRUE) } filename <- "D:\dat.csv" ff(filename) - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/How-to-read-a-file-inside-a-function-tp2316516p2316528.html Sent from the R help mailing list archi

Re: [R] Multiply each depth level of an array with another vector element

I have only achieved a half improvement. x <- array(1:2400*1, dim = c(200,300,400)) y <- 1:400*1 ptm <- proc.time() z <- x*as.vector(t(replicate(dim(x)[1]*dim(x)[2], y[1:dim(x)[3]]))) "replicate:" proc.time() - ptm x <- array(1:2400*1, dim = c(200,300,400)) y <- 1:400*1 ptm <- proc.time

Re: [R] try-error within for loop

Assigning the value directly to result[i] would give a null to result. Like that: result[[i]]<-taxondiveO(abd,taxa) result[[i]]<-as.data.frame(result[[i]])[1,] In fact your code's result should have null elements in it. You lost your nulls through do.call process. I can't find a way to assign t

Re: [R] a question about 'read.table' with or without 'read.table'.(urgent)

read.table("temp.txt",header=T,colClasses = "character") - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/a-question-about-read-table-with-or-without-read-table-urgent-tp2314423p2314445.html Sent from the R help mailing list archive at Nabble.com. _

Re: [R] ticks label of plot

You have almost achieved your goal. plot(1:100, axes=F) axis(1,at=seq(0,100,10)[c(1,3,5,7,11)]) - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/ticks-label-of-plot-tp2314226p2314294.html Sent from the R help mailing list archive at Nabble.com.

Re: [R] aggregate with non-scalar functions

You can't find a way to get the 3 column result by setting the aggregate function. data.frame(res[1], res[[2]],check.names = F) - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/aggregate-with-non-scalar-functions-tp2313987p2314063.html Sent from the R help mai

Re: [R] how to change values of the object in R-lang.

Hi, Try help("$"). dwt <- function(ld, filter='d8', n.levels=lev, boundary="reflaction") { list(W=ld, V=n.levels,filter=filter) } dec <- dwt(1:10, filter='d8', n.levels=10, boundary="reflaction") dec$W dec$V dec$filter dec$W <- sqrt(1:10) dec - A R learner. -- View this messa

Re: [R] discrete ECDF

Just a little difference. ecdf.tbl <- function (.dat) { .dat <- as.matrix(.dat) na.m <- is.na(.dat) .dat[na.m]<-0 # Assign NA a value 0 res <- apply(t(apply(.dat,1,cumsum)),2,cumsum) res[na.m] <- NA # Assign NA back return(res) } -

Re: [R] split / lapply over multiple columns

I don't know what you really want, but reshape will help you. reshape(mydata, timevar="taskid",idvar="userid", direction="wide") - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/split-lapply-over-multiple-columns-tp2312871p2312909.html Sent from the R help mai

Re: [R] Plotting multiple layers(maps) on same page

Do you want to put two figures into one pdf page? pdf(file="testpdf.pdf") par(mfrow = c(2, 1)) plot((1:20)^2,1:20,type="b",lwd=4,col="blue") lines((1:20)^3,1:20,type="b",lwd=4,col="red") plot((1:20)^4,1:20,type="b",lwd=4,col="green") lines((1:20)^5,1:20,type="b",lwd=4,col="yellow") dev.off()

Re: [R] incorrect number of dimensions

Hi, Please check your dimensions of theta. > ff <- function(theta) theta[,1] > theta <- 1:5 > ff(theta) Error in theta[, 1] : incorrect number of dimensions > theta <- data.frame(c(1:5),c(6:10)) > ff(theta) [1] 1 2 3 4 5 - A R learner. -- View this message in context: http://r.789695.n4

Re: [R] multiple R sessions from one working directory using GNU screen

I don't think one session can get information from another session, except use load. I simulate you situation within one session. Hope it helps. #Simulate the first session object1 <- 1:10 save(object1,file="ob1") rm(object1) object1 #Simulate the second session object2 <- (1:10)*2 save(object2,

Re: [R] Using '[' as a function

I have just learned from the R Language Definiation ##Operators are special functions # Examples "<-"(x,1:6) "["(x,3) "+"(3,5) '-'(3,6) '/'(3,6) '*'(3,6) - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/Using-as-a-function-tp2307292p2312424.html Sent from the R

Re: [R] Plotting multiple layers(maps) on same page

Is this you want? plot((1:20)^2,1:20,type="b",lwd=4,col="blue") lines((1:20)^3,1:20,type="b",lwd=4,col="red") - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/Plotting-multiple-layers-maps-on-same-page-tp2312223p2312414.html Sent from the R help mailing list arc

Re: [R] mixing strings and numeric doubles in an array

Another solution factor(port1[,2], labels=levels(stocks[,2])) - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/mixing-strings-and-numeric-doubles-in-an-array-tp2312091p2312313.html Sent from the R help mailing list archive at Nabble.com. __

Re: [R] Kruskal Walllis test

The apply function coerces the factor results to a character array apply(g,2,class) # gives character The kruskal.test function doesn't take character vector as the group argument. kruskal.test(as.character(plant.height) ~ as.character(g[,8])) #doesn't work kruskal.test(plant.height ~ as.characte

Re: [R] Problems with normality req. for ANOVA

I have been struggling to make the sense of permutation test for weeks. It seems will work for you. - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/Problems-with-normality-req-for-ANOVA-tp2310275p2310967.html Sent from the R help mailing list archive at Nabble.

Re: [R] Dealing with a lot of parameters in a function

Hi, Putting all parameters into a data frame would help. Code like: parameters <- data.frame(mu=1:24,b=(1:24)*2,tau=(1:24)/2,sigma=(1:24)^2,ro=sqrt(1:24)) ll<- function(parameters){ results <- numeric(24) for (i in 1:24){ mu <- parameters$mu[i] b <- parameters$b

Re: [R] Permutation of a sequence to without changing local distribution

It would be helpful if you give a sample data to illustrate the problem. I really can't understand what you want:) - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/Permutation-of-a-sequence-to-without-changing-local-distribution-tp2309283p2309616.html Sent from

Re: [R] Constructing arguments for plotmath

I failed many times until Gabor gave his solution. The magic tick! PART1 <- 'TEXT' PART2 <- '^' PART3 <- '`\u00ae`' ARG <- paste(PART1, PART2, PART3) text(2,8, parse(text=str)) - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/Constructing-arguments-for-plotmat

Re: [R] a problem

The function write.foreign is used to export data to other statistical software. To read data from Excel, R has : library(foreign) read.xport("name.xpt") or library(gdata) read.xls() - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/a-problem-tp2308667p2308676

Re: [R] conditonal "if"

I replicated your error. tau2ca <- NA if (tau2ca==0) {x <- 0.01} else {x <- tau2ca} Hope it helps. And I couldn't replicate your second program, but it seems ok. - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/conditonal-if-tp2308568p2308578.html Sent from

Re: [R] Summing by index

Hi, R has a buildin function ?rowsum rowsum(DF$Data,DF$Id,na.rm=T) - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/Summing-by-index-tp2308332p2308411.html Sent from the R help mailing list archive at Nabble.com. __

Re: [R] how to get higher derivatives with "deriv"

Hi Peter, Here is my homework :) DD <- function(f,order=1){ f.str <- "body(f)" while(order>=1) { f.str <- paste('D(',f.str,',"x")',sep="") order <- order-1} ddf <- f body(ddf) <- eval(parse(text=f.str)) ddf} f <- function(x,alpha) x^alpha

Re: [R] Ouput several lines of cat

Hi, set sep="\n" x <- c("X XXX","X XX.XX","- ( XX.XX XXX )" ) cat(x,file="x.txt",sep="\n") - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/Ouput-several-lines-of-cat-tp2307900p2307945.html Sent from the R help mailing list archive at Nabble.com.

Re: [R] how to get higher derivatives with "deriv"

## specify the function string f.str <- "x^alpha" ## higher derivatives DD <- function(f.str, x = 2, alpha=3,order = 1){ expr.s <- parse(text=f.str) while(order>=1) { expr.s <- D(expr.s,"x") order <- order-1} eval(expr.s) } compute DD(f.str,x=1,alpha=0.5,order=1) - A R learner. -- View th

Re: [R] Fwd: duplicates

Hi, Please try ?rle t.x <- x[order(x[,1],x[,2]),] t.x[cumsum(rle(t.x[,1])$lengths),] - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/Fwd-duplicates-tp2306555p2306617.html Sent from the R help mailing list archive at Nabble.com. __

Re: [R] Replace last element in a vector - elegant solution?

Hi, The function head also works. x <- 1:10 head(x,-1) - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/Replace-last-element-in-a-vector-elegant-solution-tp2306315p2306471.html Sent from the R help mailing list archive at Nabble.com.

Re: [R] Help Creating a Stacked Bar Chart with Color Coding

It is a little complicate for me to transform the table. Hope it works for you. x <- read.table(textConnection("Johnson 4 Smith4 Smith2 Smith3 Garcia 1 Garcia 4 Rodriguez 2 Adams 2 Adams 3 Adams 4 Turner 4 Turner 3 "),hea

Re: [R] columns mapping

Hi, please try ?merge DF2$mappedColumn <- DF2$name merge(DF1,DF2,all.x=T,sort=F) - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/columns-mapping-tp2305213p2305250.html Sent from the R help mailing list archive at Nabble.com. __

Re: [R] Permutation test

hi, Library DAAG has onet.permutation function for one-sample permutation test and twot.permutation function for two-sample permutation test. Anyway, p-value is a result of a test, what's your test? - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/Permutation

Re: [R] how to force a table to be square?

Hi William, I'm curious about that you used d[] <- lapply(d, factor... Could you please tell me if there are any differences between d[] and d? Thank you. - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/how-to-force-a-table-to-be-square-tp2298707p2298965.html

Re: [R] how to force a table to be square?

Hi, try ?levels myData <- matrix(sample(c(LETTERS[1:10],NA),100,replace=T),nrow=25) table(factor(as.vector(myData),levels=LETTERS[1:26]),useNA="ifany") - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/how-to-force-a-table-to-be-square-tp2298707p2298833.html Se

Re: [R] function of an integral

Hi, I'm trying to replicate your program. It may be not the same as yours, hope it helps. ## Create a vector of numbers cip <- seq(1.0,2.5,by=0.1) ## Create ecdf function Fn <- ecdf(cip) ## Create f function f <- function(x){(1-Fn(x))^4} ## Create integrate function ## Because the integrate

Re: [R] Y axis break

Sorry, it was a typo:) http://r.789695.n4.nabble.com/break-axis-using-plotrix-td803987.html#a803987 - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/Y-axis-break-tp2295499p2296254.html Sent from the R help mailing list archive at Nabble.com. __

Re: [R] Y axis break

plotrix can do it. There is a post about axis break. http://r.789695.n4.nabble.com/template/NodeServlet.jtp?tpl=reply&node=2295499 - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/Y-axis-break-tp2295499p2295998.html Sent from the R help mailing list archive at

Re: [R] specify blank in a string with special characters

Hi, you can define delimiters strsplit("Split%at blanks", " |%") - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/specify-blank-in-a-string-with-special-characters-tp2295453p2295639.html Sent from the R help mailing list archive at Nabble.com.

Re: [R] multivariate graphs, averaging on some vars

Hi, you can try plot3d library(rgl) plot3d(x=x[,2],y=x[,3],z=rowMeans(x[,2:4])) - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/multivariate-graphs-averaging-on-some-vars-tp2292039p2292076.html Sent from the R help mailing list archive at Nabble.com.

Re: [R] Deleting a variable number of characters from a string

Hi Davis, Please try ??regex gsub("(\\-|\\+)([0-9]+)(\\w*)(\\w)", replacement="\\4", x) - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/Deleting-a-variable-number-of-characters-from-a-string-tp2291754p2291797.html Sent from the R help mailing list archive at

Re: [R] how to collapse categories or re-categorize variables?

Do you want to replace specific values of a data set? df <- sample(c(0,1,2),600,replace=T) table(df) df[df==2]<-1 table(df) - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/how-to-collapse-categories-or-re-categorize-variables-tp2291704p2291727.html Sent from t

Re: [R] how to skip a specific value when using apply() function to a matrix?

Hi szhan, I think Joshua gives all you wants -- scale is a really good function. You can also make your own function work by setting an argument na.rm. tmp1[tmp1==0]<-NA student<- function(x){ x<-(x-mean(x,na.rm=T))/sd(x,na.rm=T) return (x) } tmp4<-apply(tmp1, 2, student) -

Re: [R] discrepancy matrix

I guess your data frame is a little different from the reference, so your as.logical doesn't work. attach(Q) FUN <- function(X, Y) {abs(X - Y)} round(outer(rank(date)[colour=="b"],rank(date)[colour=="g"],FUN) + outer(rank(number)[colour=="b"],rank(number)[colour=="g"],FUN)) detach(Q) - A

Re: [R] how to skip a specific value when using apply() function to a matrix?

There may exists a good solution, but I can only give a shortcut:) tmp2 <- (tmp1!=0)*tmp3 - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/how-to-skip-a-specific-value-when-using-apply-function-to-a-matrix-tp2290898p2290925.html Sent from the R help mailing li

Re: [R] How to plot a histogram of weekday frequencies in a list of dates?

Correction: ## Minus 7*60*60 seconds to change AM interval from 0-12AM to 7AM-7PM barplot(table(format((dates-7*60*60), "%w %p")),names.arg=labels) - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/How-to-plot-a-histogram-of-weekday-frequencies-in-a-list-of-date

Re: [R] How to plot a histogram of weekday frequencies in a list of dates?

Hope it works. ## Create a sample of dates dates <- sample(seq(ISOdate(2009,1,1), ISOdate(2009,12,31), "hour"),1000) ## Create 14 labels for barplot("Mon AM","Mon PM",etc.) labels <- format(seq(ISOdate(2010,7,11,5), ISOdate(2010,7,18,4), "12 hours"),"%a %p") ## Use table function to calculate

Re: [R] Separating parts of a column of data in R

I don't really understand your question. The function unique can remove duplicate elements. - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/Separating-parts-of-a-column-of-data-in-R-tp2290793p2290840.html Sent from the R help mailing list archive at Nabble.com.

Re: [R] Cut a within elements by length, not value, of vectors

gsub("(.*)(.{2})","\\1/\\2",dates) - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/Cut-a-within-elements-by-length-not-value-of-vectors-tp2290440p2290471.html Sent from the R help mailing list archive at Nabble.com.

Re: [R] Creating Enumerated Variables

If you only want a count, please try table text <- "ID Age School Grade 1 10 1 98 2 10 2 97 3 10 1 92 4 11 1 90 5 11 1 80 6 11 2 70 7 10 1 80 8 10 1 79 9 11 2 70" df <- read.table(textConnection(text),header=T) table(df[,2:3]) If you want sort the data, try order. - A R learner. -- View t

[R] How to see the process of function sample()?

I have the same question about how to see the process behind a function. If I type sample in R, it really tells nothing about how R selects from a data set and creates samples. Thank in advance for any help. - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/How

Re: [R] Merging columns along time line

Correction: b$label <- cut(b$timestamp, breaks=bks, labels=lbs, include.lowest = T, right=F) - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/Merging-columns-along-time-line-tp2289147p2289401.html Sent from the R help mailing list archive at Nabble.com. ___

Re: [R] Merging columns along time line

I take this case as cut a data set by breaks and assign each segment a label name. a <- data.frame(timestamp=c(3,5,8), mylabel=c("abc","def","ghi")) b <- data.frame(timestamp=c(1:10)) bks <- c(a$timestamp,max(b$timestamp)) lbs <- a$mylabel b$label <- cut(b$timestamp, breaks=bks, labels=lbs, inclu

Re: [R] question about string handling....

Try this: text <- 'var1 var2 1 ab_c_(ok) 2 okf789(db)_c 3 jojfiod(90).gt 4 "ij"_(78)__op 5 (iojfodjfo)_ab' df <- read.table(textConnection(text), head=T, sep=" ",quote="") df$var3 <- gsub("(.*\\()(.*)(\\).*)","\\2",df$var2) - A R learner. -- View this message in context: http://r.789695.

Re: [R] Write value to PHP webpage

I suggest trying to write the data into .php file directly. outfile <-paste(filepath,"distance",".php",sep="") data <- "Distance=num" num <- 1000 data <- sub("num", num,data) write(data,file=outfile) - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/Write-val

Re: [R] Using which function with xts

Correction: data[abs(data$price-avg)<=3*std,] - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/Using-which-function-with-xts-tp226p2289166.html Sent from the R help mailing list archive at Nabble.com. __ R-help

Re: [R] Using which function with xts

Let's say your "data" has 2 columns: one is "date" and another is "price", then avg = mean(data$price, na.rm=T) std = sd(data$price, na.rm=T) The data after those unwanted removed should be: data[data$price-avg<=3*std,] - A R learner. -- View this message in context: http://r.789695.n4.n

Re: [R] problem with comparisons for vectors

I don't know the real reason, but help("==") gives some clues. For numerical and complex values, remember == and != do not allow for the finite representation of fractions, nor for rounding error. Using all.equal with identical is almost always preferable. See the examples. x1 <- 0.5 - 0.3 x2

Re: [R] String truncate

Do you mean substring? sub(".txt","", "mytest.txt") - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/String-truncate-tp2284045p2284062.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-projec

Re: [R] deduplication

Please try this ## Import data id1<-c(4,17,9,1,1,1,3,3,6,15,1,1,1,1,3,3,3,3,4,4,4,5,5,12,9,9,10,10) id2<-c(8,18,10,3,6,7,6,7,7,16,4,5,12,18,4,5,12,18,5,12,18,12,18,18,15,16,15,16) id<-data.frame(id1 = id1, id2 = id2) ## Create same structure table id <- id0 <- unique(id) leng <- nrow(id) n <- 0

Re: [R] two columns into one

Try unique and paste. paste(unique(tes)[,1], unique(tes)[,2], sep = "") - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/two-columns-into-one-tp2242516p2242658.html Sent from the R help mailing list archive at Nabble.com. __

Re: [R] string handling

Hope it helps. text <- "var1var2 9G/G09abd89C/T90 10A/T932C/C 90G/G A/A" x <- read.table(textConnection(text), header = T) x$var1.1 <- sub(".*(.)/.*", "\\1", x$var1) x$var1.2 <- sub(".*/(.).*", "\\1", x$var1) x$var2.1 <- sub(".*(.)/.*", "\\1", x$var2) x$var2.2 <- sub(".*/(.

Re: [R] Re : help to replace variable value

Do you mean replace values of a column? > df <- data.frame("Jan" = 1:3,"Feb" = 11:13) > df Jan Feb 1 1 11 2 2 12 3 3 13 > df$Jan <- 21:23 > df Jan Feb 1 21 11 2 22 12 3 23 13 - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/Re-help-to-replace-

Re: [R] sequential treatment of a vector for formula

Thank you very much David. I'm sorry for this fault , hope it has not confused Frostygoat. I was clueless of recursive reference and I didn't meet any error when I test the code. So I wonder if there are some useful tips to prevent making this kind of faults:) The revised code is followed. ###

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