similar to: Removing Rows

First I wanted to thank both Marc Schwartz Greg Snow and for their reply. Then I needed to add a level of complexity to the problem. I would be able to create the biggest possible matrix. In other way does it exist a method to ask smthing like the following : max number of rows for a matrix if column=x? Thank you ------------------------------------------------------ Passa a Infostrada.

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 am trying to output an dataframe from R to Excel file. Can anyone tell me how to do it? Thanks a lot. Eg. R dataframe: A B C 1 2 1 3 4 2 . . . [[alternative HTML version deleted]]

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

Hi Thanks for working example. you could use split/ lapply approach, however it is probably not much better than dplyr method. sapply(split(mydf, mydf$type), function(speed, n_vehicles) sum(mydf$speed*mydf$n_vehicles)/sum(mydf$n_vehicles)) gives you averages aggregate(mydf$n_vehicles, list(mydf$type), sum)$x gives you sums Cheers Petr > -----Original Message----- > From: R-help

I have a dataset with 4 years of students, and normally I want to estimate things using each individual year, so I have a for loop as follows for (i in 1:4){} However, the only way I know how to calculate estimates using all four years of data is to put the estimations outside of the loop. Is there anyway to make a for loop that uses all four years at once, then uses each individual year?

> On 9 Nov 2017, at 14:58, PIKAL Petr <petr.pikal at precheza.cz> wrote: > > Hi > > Thanks for working example. > > you could use split/ lapply approach, however it is probably not much better than dplyr method. > > sapply(split(mydf, mydf$type), function(speed, n_vehicles) sum(mydf$speed*mydf$n_vehicles)/sum(mydf$n_vehicles)) > gives you averages > The

I want to make a glm and then use predict. I have a fairly small sample (4000 cases) and I want to train on 90% and test on 10% but I want to do it in slices so I test on every 10th case and train on the others. Is there some simple way to get these elements? Stephen -- 21/10/2005 [[alternative HTML version deleted]]

Hi, I have a set of data like the following: [,1] [,2] [1,] 10 2 [2,] 7 0 [3,] 1 0 [4,] 1 0 [5,] 15 0 [6,] 17 4 [7,] 4 0 [8,] 19 8 [9,] 10 2 [10,] 19 5 I'd like to aggregate it in order to obtain the frequency (the number of occurences) for each couple of values (e.g.: (10,2) appears twice, (7,0) appears once). Something cool

Hello all R users, My simulation function works correctly, but I have problems with plot function. You will find the following code using it. Thank you for your help ##################################################" simulation <- function(k, n){ conc <- seq(0,10,by=0.5) #choixg <- seq(1, length(conc)) choixg <- rep(0,length(conc)) for (i in 1:length(conc)){ choixg[i]

I'd like to remove duplicated rows within a matrix, with missing data being treated as wildcards. For example > x <- matrix((1:3), 5, 3) > x[4,2] = NA > x[3,3] = NA > x [,1] [,2] [,3] [1,] 1 3 2 [2,] 2 1 3 [3,] 3 2 NA [4,] 1 NA 2 [5,] 2 1 3 I would like to obtain [,1] [,2] [,3] [1,] 1 3 2 [2,] 2 1 3

Dear R community, I am working with huge arrays, so I spend a lot of time computing. This is my code: for (x in 1:dim(variable)[1]){ for (y in 1:dim(variable)[2]){ for (z in 1:dim(variable)[3]){ result <- max(variable[x,y,z,]) } } } Is there a more efficient procedure to do this task? Thanks in advance! [[alternative HTML version deleted]]

I have a matrix y: > dimnames(y) $x93 [1] "1" "2" $x94 [1] "0" "1" "2" .................. so on (there are other dimensions as well) I need to access a particular dimension, but a random mechanism tells me which dimension it would. So, sometimes I might need to access dimnames(y)$x93, some other time it would be dimnames(y)$x94.. and so

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(),

Dear all I have a question about quantiles standard error, partly practical partly theoretical. I know that x<-rlnorm(100000, log(200), log(2)) quantile(x, c(.10,.5,.99)) computes quantiles but I would like to know if there is any function to find standard error (or any dispersion measure) of these estimated values. And here is a theoretical one. I feel that when I compute median from given

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.

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

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

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 got this error message > library(dprep) > mardia(Savg) Error in cov(data) : 'x' is empty But with the same data, I got > library(mvnormtest) > mshapiro.test(Savg) Shapiro-Wilk normality test data: Z W = 0.9411, p-value = 0.6739 What does the error message "Error in cov(data) : 'x' is empty" mean? Thanks a lot! Jiao