similar to: npmc package

Hi. Please, how can I put together 2 or more lines at the same scatterplot ? Example: measures of protein intake (quantitative) of 4 children over 30 days, by day. How to plot all children at same graphic: Protein X Time ? Is there any command like "overlay" ? Thank you, Mauricio

Hi everyone ! Could you please help me with this problem ? I??ve trying to write a code that assign to a variable the content from another, but all I??ve got is a message error. For example: if (age <=10) {group == 1} else if (age > 10 & age <= 20) {group == 2} else {group == 3} Syntax error Or if (age <=10) {group == 1} else (age > 10 & age <= 20) {group == 2}

Hello. I asked before and it was great, cause as a beginner I learned a lot. But, if I have this in R (1 and 2 are codes for sex): > sex<-c(1,2,2,1,1,2,2,2) > sex [1] 1 2 2 1 1 2 2 2 I´d like to obtain the proportion according to sex.So I type: > prop.table(sex) [1] 0.07692308 0.15384615 0.15384615 0.07692308 0.07692308 0.15384615 0.15384615 [8] 0.15384615 The result is OK, but I

Please, how to fill the area under the curve? x <- c(1:10) y <- c(rnorm(10)) plot(x,y) lines(x,y) Thanks, Mauricio Cardeal

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Hi all, I'm trying to use lapply on a list with the following command: out<-lapply(mylist,myfun,par1=p,par2=d) (1) where myfun<-function(x,par1,par1) {.....} (2) now this function is in fact a wrapper for some Fortran code I have written so I think this might be the problem. When I call lapply() as in (1) I get the following message: Error in get(x,

Please, I??d like to store this sample matrix as a new object. How can I do this ? pulse <- c(67, 67, 68, 68, 68, 69, 69, 69, 69, 69, 70, 70, 70, 70, 71, 71, 72, 72, 73, 74) m <- NULL x <- 0 for (i in 1:5) { x <- sample(pulse,3) m <- mean(x) cat(x,m,"\n") } Thanks, Mauricio

Is it possible to make disappear the id numbers from scatter.dudi (mc analysis) ? a <- as.factor(c(1, 2, 3, 2, 1)) b <- as.factor(c(3, 2, 3, 1, 1)) x <- as.factor(c(1, 2, 2, 1, 3)) y <- as.factor(c(2, 2, 3, 1, 1)) dat <- data.frame(a=a, b=b,x=x,y=y) summary(dat) dat require(ade4) dat.acm <- dudi.acm(dat, scann = FALSE, nf = 2) scatter.dudi(dat.acm) Thank you very much !

HI, I am a new user of R and want to analyse some data using npmc. My data have several levels of factor (Site, Year and Season) and several variable (Percentages). I have tried to use npmc but I always get an error message. My data are in a table following this example: Site YEar Season Var1 Var2 .... A 2009 Dry 10 56 B ........ here is the error

Hi folks, sorry if this has been answered before, I searched long and hard before deciding to make a thread. I'm trying to include multiple variables in a non-parametric analysis (hah!). So far what I've managed to figure out is that the NPMC package from CRAN MIGHT be able to do what I need, but I can't get it to. First I created a dataset as NPMC calls for. > Ind=Individual

Hi there, I am having a problem with package npmc. If I use the included datasets (brain, kronen) everything is fine, but if I try and use my dataset I receive an error message: Error in probval.GenzBretz(algorithm, n, df, lower, upper, infin, corr, : NAs in foreign function call (arg 2) In addition: Warning message: In probval.GenzBretz(algorithm, n, df, lower, upper, infin, corr, : NAs

Hi, I'm having trouble getting the function `npmc' working. I have a 2 column 30 row table called `mydata': > npmc(mydata, df=2, alpha=0.05) Error in sort(unique.default(x), na.last = TRUE) : `x' must be atomic > is.atomic(mydata) [1] TRUE Is there anything I have overlooked? Am I invoking the function correctly? I wasn't able to find any hints in the archives

Hi all # I have this data set and how can I assign NA?s in just one command ? And why the summary(dat) function preserves the value 9 as real. ? x <- c(1,2,3,9,4) y <- c(3,6,9,2,3) z <- c(9,9,2,2,8) w <- c(6,5,3,0,9) dat <- cbind(x,y,z,w) summary(dat) x[x==9] <- NA y[y==9] <- NA z[z==9] <- NA w[w==9] <- NA summary(dat) summary(x) summary(y) summary(z) summary(w)

I followed the threads that enquired about doing post-hoc tests after doing Kruskal testing. It took me to npmc. But npmc is giving an output I do not understand. I noticed a thread entitled "npmc function: 'x' must be atomic" but there never appeared to be a resolution. > npmc(npmcinput) Error in sort(unique.default(x), na.last = TRUE) : 'x' must be atomic

Hi, I sent an earlier message concerning this, but perhaps I wasn't clear enough. Up until about a month ago, I downloaded packages by clicking on the filename of the package. Now this action leads me to a directory listing, e.g.: drwxr-xr-x hornik/users 0 2002-03-20 18:00:54 npmc/ -rw-r--r-- hornik/users 314 2002-03-20 11:33:42 npmc/DESCRIPTION -rw-r--r-- hornik/users 202

Hello, I am trying to run a nonparametric comparison of multiple datasets (A-E). After importing my csv file I sorted the variable value by the letter class (representing height along a streambank). a=Z[Letter=="A"] b=Z[Letter=="B"] c=Z[Letter=="C"] d=Z[Letter=="D"] e=Z[Letter=="E"] Next I tested normality and variance assumptions and the

Hi, I recently tried to conduct non-parametric multiple comparison with npmc package, and encountered some problem; sometimes it stopped with following error. Error in uniroot(f = function(arg) p - z.dist(arg, corr = corr, df = df, : f() values at end points not of opposite sign Although I'm not quite sure about calculation of z.quaintile in npmc.R, it seems increment of

Dear R Developers, the following CRAN packages do not cleanly pass R CMD check for quite some time now and did not have any updates since the time given. Several attempts by the CRAN admins to contact the package maintainers had no success. norm, 1.0-9, 2002-05-07, WARN sound, 0.6, 2002-08-31, WARN Bhat, 0.9-07, 2003-11-04, WARN HI, 0.1, 2003-11-04, WARN mmlcr, 1.3.2, 2003-11-15, WARN mvnmle,

Hi, I am using R 2.10.1 and I have a doubt. Do you know how many cases can R handle? I want to use the library npmc but if I have more than 4,500 cases I get an error message. If I use less than 4500 cases I don´t have problems with this library. Is there any way to increase the number of cases in order to use this library. Thanks in advance Julio Flores [[alternative HTML version

Carlos ?mca states that mca works on a dataframe. As you've written it is.data.frame(de) returns FALSE Try de <- data.frame(d,e) instead of de <- factor(c(d,e)) HTH Peter Alspach > -----Original Message----- > From: r-help-bounces at stat.math.ethz.ch > [mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of Carlos > Mauricio Cardeal Mendes > Sent: Wednesday,