similar to: Remove levels

Dear All, I wonder how to compute the age from the date of birth and the date of examination. Suppose that have the following data: df <- as.data.frame(rbind(c(1,"10/08/1950","15/03/1998"), c(1,"10/08/1950","20/07/1998"), c(1,"10/08/1950","23/10/1998"))) names(df) <- c("ID", "date_birth",

Hi R users Here is what I got with help from Petr Pikal (Thanks Petr Pikal). I modified Petr Pikal's code to a little to meet my purpose. I created a function to generate a matrix generate.matrix<-function(n.variable) { mat<-matrix(0,n.variable,(n.variable/2)/5+1) #matrix of zeroes dd<-dim(mat) # actual dimensions mat[1:(dd[1]/2),1]<-1 #put 1 in first half of first column

Hi, I'm looking for some lines of code that does the following: I have a dataframe with 160 Columns and a number of rows (max 30): Col1 Col2 Col3 ... Col 159 Col 160 Row 1 0 0 LD ... 0 VD Row 2 HD 0 0 0 MD Row 3 0 HD HD 0 LD Row 4 LD HD HD 0 LD ... ... LastRow HD HD LD 0 MD Now I want a dataframe that looks like this. As you see

What is the reason, that the levels of the factor returned by cut() are not marked as ordered levels? > is.ordered( cut( breaks=3, sample(10 ) ) ) FALSE > help(factor) ... If 'ordered' is 'TRUE', the factor levels are assumed to be ordered. ... Wolfram

Hi Petr and Richard; Thanks for your responses and supports. I just faced a different problem. I have the following R codes and work well. p <- ggplot(a, aes(x=Phenotypes, y=Metabolites, size=abs(Beta), colour=factor(sign(Beta)))) + theme(axis.text=element_text(size = 5)) p1<-p+geom_point() p2<-p1+theme(panel.grid.major = element_blank(), panel.grid.minor = element_blank(),

Hi After melt you can change levels of your factor variable. Again with the toy example. > levels(temp$variable) [1] "y1" "y2" "y3" "y4" > levels(temp$variable) <- levels(temp$variable)[c(2,4,1,3)] > levels(temp$variable) [1] "y2" "y4" "y1" "y3" > And you will get graphs with this new levels ordering.

Petr, there was a thinko in your response. tmp <- data.frame(m=factor(letters[1:4]), n=1:4) tmp tmp$m <- factor(tmp$m, levels=c("c","b","a","d")) ## right tmp[order(tmp$m),] tmp <- data.frame(m=factor(letters[1:4]), n=1:4) levels(tmp$m) <- c("c","b","a","d") ## wrong tmp[order(tmp$m),] changing levels

Dear all, How can I plot the number of observations per box in a boxplot. Many thanks, Bernard --------------------------------- [[alternative HTML version deleted]]

Hi scale_colour_gradient(?red?, ?blue?) should do the trick. Actually I found it by Google ggplot colour http://www.cookbook-r.com/Graphs/Colors_(ggplot2)/ http://www.sthda.com/english/wiki/ggplot2-colors-how-to-change-colors-automatically-and-manually#gradient-colors-for-scatter-plots question. So you could find it too and probably far more quickly then myself as I have also other duties. Cheers

Dear all I want to display 4 dimensional space by some suitable way. I searched CRAN and found miscellaneous 3 dim graphics packages which I maybe can modify but anyway I am open to any hint how to efficiently display data like: longitude, latitude, height, value Thank you Petr Pikal petr.pikal at precheza.cz

Dear All, I would be grateful if you can help me. My problem is the following: I have a data set like: ID time X1 X2 1 1 x111 x211 1 2 x112 x212 2 1 x121 x221 2 2 x122 x222 2 3 x123 x223 where X1 and X2 are 2 covariates and "time" is the time of observation and ID indicates the

Thank you, but this split the area into two and distorts the shape of the plot. (compared to ``` p + geom_abline(slope = slope_1, intercept = intercept_1 - 1/w[2], linetype = "dashed", col = "royalblue") + geom_abline(slope = slope_1, intercept = intercept_1 + 1/w[2], linetype = "dashed", col = "royalblue") ``` Why there

Hallo all just a small question I did not find an answer in help pages. Is it possible to use identify() after plotting with plot.POSIX to label points and/or to find out some points? Thanks a lot. Best regards Petr Pikal petr.pikal at precheza.cz p.pik at volny.cz

Dear R users I searched some sources but i did not find an answer.Please give me some hint to following problem. I would like to compute a summary statistic for some vector for different factor levels. I know I can use tapply or aggregate but I do not know if there is a way how to use function with several (two) variable input (like weighted.mean). I wrote a simple a function for factor

Hi, I am a newbie to R, just trying to switch from Splus. I am wondering to know if there is any such command in R as "attach" in Splus which can be used to attach the full directory. Also, any idea/referrence about, how to load c functions in R would be highly appreciated. Thanks, Utsav

Dear all I try to refine my nlme models and with partial success. The model is refined and fitted (using Pinheiro/Bates book as a tutorial) but when I try to plot plot(augPred(fit4)) I obtain Error in predict.nlme(object, value[1:(nrow(value)/nL), , drop = FALSE], : Levels (0,3.5],(3.5,5],(5,7],(7,Inf] not allowed for vykon.fac > Is it due to the fact that I have unbalanced

I have to perform ANOVA's on many different data organized in a dataframe. I can run an ANOVA for each sample, but I've got hundreds of data and I would like to avoid manually carrying out each test. in addition, I would like to have the results organized in a simple way, for example in a table, wich could be easy to export. thank you for assistance simone

Hallo I encountered a behaviour which puzzles me (but finally I did get what I wanted). I used structable and strucplot but I wanted to change names of variables in structable object. I tried to subset it, use names but to no avail. So I tried str and expected to get a structure of an object but: > sss<-structable(Titanic) > str(sss) Error in "[.structable"(x, args[[1]],

Petr, It works perfectly! But I still have a question; I have fit the following data; x,y,z 1,10,11 2,11,15 3,12,21 4,13,29 5,14,39 6,15,51 7,16,65 8,17,81 9,18,99 10,19,119 >dat.lm <- lm(z~I(x^2)+y, data=dat) >dat.lm Call: lm(formula = z ~ I(x^2) + y, data = dat) Coefficients: (Intercept) I(x^2) y 1.841e-14 1.000e+00 1.000e+00 How do I create the

> On Oct 13, 2017, at 2:51 AM, PIKAL Petr <petr.pikal at precheza.cz> wrote: > > Hi > > You expect us to solve your problem but you ignore advice already recieved. > > Your data are unreadable, use dput(yourdata) instead. see ?dput > >> test<-read.table("clipboard", heade=T) > Error in scan(file = file, what = what, sep = sep, quote = quote,