similar to: novice questions about programming in "R"

Dear All, I have to use loop over an array so I am using following procedure count<-1 repeat{ count<-count + 1 c(g[count],1:i[count]) ->qw if(count>5)break } as a result qw is [1] 0.9643836 1.0000000 2.0000000 3.0000000 4.0000000 5.0000000 [7] 6.0000000 7.0000000 8.0000000 9.0000000 10.0000000 11.0000000 [13] 12.0000000 13.0000000 14.0000000 15.0000000 16.0000000

Hello all! I have a problem with ggplot2 library. I want to do an heat map and the y variables are the year months. If I use the following code, he y values are in alphabetical order, but I want it in month order. The code is: library(reshape) library(ggplot2) library(scales) p <- ggplot(data.m, aes(variable, Month)) + geom_tile(aes(fill = value),

Hi R, y1 <- zoo(matrix(1:10, ncol = 2), 1:5) colnames(y1)=c("a","b") y2 <- zoo(matrix(rnorm(10), ncol = 2), 6:10) colnames(y2)=c("b","a") > y1 a b 1 1 6 2 2 7 3 3 8 4 4 9 5 5 10 > y2 b a 6 0.9070204 0.3527630 7 1.2405943 0.8275001 8 -0.1690653 -0.1724976 9 -0.6905223 -1.1127670 10

Buenas Tengo un Data table de la siguiente manera: datos<-data.table(grupo=rep(c("a","b"),5),x=c(1:10),y=rnorm(10,2,1)) Lo que quiero es añadir una fila por cada grupo y en esa nueva fila, al valor de la x ponerle el valor anterior de la y Lo que hago es añadir una nueva fila por grupo, con: datos[,.SD[1:(.N+1)],by=grupo] Y para añadir el valor anterior uso la función

Hello, I have 2 functions (a and b) a = function(n) { matrix (runif(n*2,0.0,1), n) } > > > b = function (m, matrix) { > n=nrow (matrix) > p=ceiling (n/m) > lapply (1:p, function (l,n,m) { > inf = ((l-1)*m)+1 > if (l<p) sup=((l-1)*m)+m > else sup=n >

Since format, i.e., format.default(.), is a pretty basic function I thought I'd ask before just changing it... aa <- cbind(1:7, rnorm(7)) format(aa) or format(aa, digits=7) looks like [,1] [,2] [1,] " 1.0000000" " 0.2406669" [2,] " 2.0000000" "-0.4973221" [3,] " 3.0000000" " 0.4672260" [4,] "

is there a way to do element-by-element multiplication as in Gauss and MATLAB, as shown below? Thanks. --- a 1.0000000 2.0000000 3.0000000 x 1.0000000 2.0000000 3.0000000 2.0000000 4.0000000 6.0000000 3.0000000 6.0000000 9.0000000 a.*x 1.0000000 2.0000000 3.0000000 4.0000000

Hello, I have a matrix mat1 of dim [1,8] and mat2 of dim[30,8], I want to replace the first row of mat2 with mat1, this is what I do: mat2[1,]<-mat1 but it transforms mat2 in a list I don't understand, I want it to stay a matrix... ----- Anna Lippel -- View this message in context: http://n4.nabble.com/Simple-question-on-replace-a-matrix-row-tp1427857p1427857.html Sent from the R help

Fellow R-helpers, Suppose we create a histogram as follows (although it could be any vector with zeroes in it): R> lenh <- hist(iris$Sepal.Length, br=seq(4, 8, 0.05)) R> lenh$counts [1] 0 0 0 0 0 1 0 3 0 1 0 4 0 2 0 5 0 6 0 10 0 9 0 4 0 [26] 1 0 6 0 7 0 6 0 8 0 7 0 3 0 6 0 6 0 4 0 9 0 7 0 5 [51] 0 2 0 8 0 3 0 4 0 1 0 1 0 3

Hello r-experts, I sure could us a little help. I have an ever updating text file with timestamped data in it. I can reformat in anyway I want if need be but currently I have chosen to make columns of date, time and measuresed value (comma delimeted and with the dates and times in quotes to interpret them as strings). Here is a small section of my text data file:

Hi folks, can anyone point me to an R add-on package that will work with the MS-windows IIS server as is? I know from trying that the "CGIwithR" package was not written with IIS in mind. I'm hoping to find package that will work with IIS without having to do any deep magic. much obliged :-) - revansx

below is my data frame. I would like to compute summary statistics for mgl for each river mile (mean, median, mode). My apologies in advance- I would like to get something like the SAS print out of PROC Univariate. I have performed an ANOVA and a tukey LSD and I would just like the summary statistics. thanks stephen RM mgl 1 215 0.9285714 2 215 0.7352941 3 215 1.6455696 4 215

How can I get the result of, e.g., poly(1:3. degree=2) to give me the unnormalized integer coefficients usually used to explain orthogonal polynomial contrasts, e.g, -1 1 0 -2 1 1 As I understand things, the columns of x^{1:degree} are first centered and then are normalized by 1/sqrt(col sum of squares), but I can't see how to relate this to what is returned by poly(). >

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Hello I am able to construct a url that points to some data online in the JSON format. See an example at [0]. I would like to work with this data as a dataframe in R. I know that there is a package for handling json data [1] but it assumes the data is in a local file but It is not clear to me how to request the data from the web in an R script and get the json data converted into a data frame

hello, please can anyone help me out. Am a new user of R program. Am having problem with this code below, not getting the expected results. Each m, the cumulative sum should be 1.000 but the 2nd and 3rd m returned 2.000 and 3.000 instead of 1.000. thanks Aruike pp=function(x,n,M){z=1.0;a=2.3071430;b=7.266064;H=3 out.h=c() out.y=c() out.m=c() out.prob=c()

I am doing analysis on hourly precipitation on a file that is disorganized. However, I managed to clean it up and store it in a dataframe (called CA1) which takes the form as followed: Station_ID Guage_Type Lat Long Date Time_Zone Time_Frame H0 H1 H2 H3 H4 H5 H6 H7 H8 H9 H10 H11 H12 H13 H14 H15 H16 H17 H18 H19 H20 H21 H22 H23 1

There's something peculiar that I do not understand here. However, did you realize that the thing you are assigning into parts of `a' is NULL? Check you're my.test.boot.ci.1: It's NULL. Be that as it may, I get: > a <- data.frame(matrix(1:4, nrow=2), X3=NA, X4=NA) > a X1 X2 X3 X4 1 1 3 NA NA 2 2 4 NA NA > a[a$X1 == 1,]$X3 <- NULL > a X1 X2 X3 X4 1 1

Dear R helpers, I am having trouble with an array inside a loop. I wish to accumulate the results of a function in an array within the function, over several loops of a program outside the function. The problem is that the array seems to re-set at every entry to the function. Here is an example, so you can see what I mean.

Dear all, here is the result of R.Version(): > R.Version() $platform [1] "powerpc-apple-darwin8.6.0" $arch [1] "powerpc" $os [1] "darwin8.6.0" $system [1] "powerpc, darwin8.6.0" $status [1] "" $major [1] "2" $minor [1] "3.1" $year [1] "2006" $month [1] "06" $day [1] "01" $`svn rev`