similar to: install.packages

Dear All, This question sounds very simple but I don't know where I am wrong. I just want to pad leading zeros in some string, for example, "123" becomes "00123". What is wrong if I do following? > sprintf("%05s", "123") [1] " 123" It didn't return "00123", instead it padded with 'blank'. Thank you for your help

Hi All, I have two Rs, one has been installed in Windows system and another one has been installed under UNIX system. Is there any environmental variable or function to tell me which R I am using? The reason that I need to know it is under different system, the data path could be different. I want to do something like if it is R under Windows path =

Hi All, I have a data frame like a = data.frame(date = c(20081201, 20081202, 20081201), product = c("a b c d e", "a b c g h t", "d e h a c e h g"), sales = c(1, 2, 3)) Now I want to aggregate the sales by part of the a$product. 'Product' is the product name, a string separated by a space. The key in my aggregate function is

Hi All, I have a data frame, whose colnames like "A.x, B.x, C.x, A.y, B.y, C.y". There could be many columns like this pattern. I want to compare data in columns A.x with A.y, B.x with B.y and C.x with C.y etc. Suppose my data frame is d, names(d) = c("A.x", "B.x", "C.x", "A.y", "B.y", "C.y"); If I want to D1 =

Hi All, I have a self-cooked package and save it to a zip file after running make, say named xxx.zip. After installing it to R by running "Install packages from local zip files" under Packages menu in R (Windows), I realized I needed to change some source codes and re-make it. My question is that if I need re-install that package or R would automatically update the R

Hi all, I have a dataset. Each time I want to sample N(i) elements from it and I want to repeated sampling M times. N(i) is varied from time to time. For example, dataset = 1:50; a = list(); M = 1000; I want to do something like for(i in 1:M) { a[[i]] = sample(dataset,

Hi All, I know in R there is function named 'step', which does the stepwise regression and choose the model by AIC. However, if I want to choose a model per this logic: 1. Run a full model (linear regression, f = lm(y ~., data = ZZZ), for example) 2. Pick up the variable with biggest p value, delete it from the module and get a new regression model. 3. Repeat step 2

Hi All, I have a matrix like m <- matrix( letters[1:10], ncol=5) How to conver it to 10 * 3 matrix, which the first col is row index of m, second col is colum index of m and third column is the value of matrix, say 1 1 1 "a" 2 1 2 "c" 1 3 "e" etc... Thanks. [[alternative HTML version deleted]]

Hi All, I have a question about package installation in R. We have developed a package, say 'ABC'. We have installed it in two machines, A and B by running 'Install Package(s) from local zip file'. Everything was fine. Right now, suppose that package got damaged in machine A and our zipped file is gone, My question is that may I directly copy ../library/ABC from machine B to

Hi group: I have a data set, which has severe colinearity problem. While running linear regression in R and SPSS, I got different models. I am wondering if somebody knows how to make the two software output the same results. (I guess the way R and SPSS handling singularity is different, which leads to different models.) Thanks. [[alternative HTML version deleted]]

I am trying to replicate the SAS proc univariate in R. I got most of the stats I needed for a by grouping in a data frame using: all1 <- ddply(all,"ACT_NAME", summarise, mean=mean(COUNTS), sd=sd(COUNTS), q25=quantile(COUNTS,.25),median=quantile(COUNTS,.50), q75=quantile(COUNTS,.75), q90=quantile(COUNTS,.90), q95=quantile(COUNTS,.95), q99=quantile(COUNTS,.99) )

I have a data.frame named "df". The dput of df is at the bottom of this e-mail. What I'd like to do is replace the "n/a " values with NA. On Mac OSX, it works to do this: df[df == "n/a"] <- NA However, it does not work on Ubuntu. See below. Thanks in advance, Garrett > x <- df[27, 4] # complete data.frame dput is below > dput(x) "n/a?"

Hi All, Suppose I have the following codes: x = data.frame(A = rnorm(20), B = rnorm(20), C = rnorm(20)) a = list() a[["1.1"]] = x a[["1.2"]] = x b = list() b[["1.1"]] = c("A", "B") b[["1.2"]] = c("B", "C") Now I want to apply b to a like this, for each element of 'a', only select the corresponding columns

This is my first post to R-help and I look forward receiving some advice for a novice like me... I?ve got a simple repeated (4 periods so far) 10-question survey data that is very easy to work on Excel. However, I?d like to move the compilation to R but I?m having some trouble operating on count list data in a neat way. The data C > str(C) 'data.frame': 551 obs. of 13

Hi All, Does somebody know how to know the detail of the line types? For example, lty = 1, means what kind of line?, lty = 2, means what kind of line? Thanks. HXD [[alternative HTML version deleted]]

Hi all, I am planning to parse some information on a website which includes lots of Chinese characters. Does someone know how to read/display Chinese in R? Thanks. url = "http://www.teec.org.cn/html/renwujieshao/" x = readLines(url) I tried encoding = 'UTF-8' already but it didn't help. My R version is $platform [1] "i386-pc-mingw32" $arch [1] "i386"

Hi All, I need to calculate VIF (variance inflation factor) for my linear regression model. I found there was a function named vif in 'HH' package. I have two questions: 1) I was able to install that package in my R under windows. But while trying to install that package in UNIX, I got error: Package HH is not available. Does somebody know why? 2) Does

Hey everybody, I have a matrix with three columns. I want to plot two columns (independent variable) against one column (the defendant). This is my code and the error associated with it: plot(p, q, data=columns) > plot(pprime,q, add=TRUE) Warning messages: 1: In plot.window(...) : "add" is not a graphical parameter 2: In plot.xy(xy, type, ...) : "add" is not a graphical

Hi All, The following is a snippet of my code. It works fine but it is very slow. Is it possible to speed it up by using different data structure or better solution? For 40000 runs, it takes 8 minutes now. Thanks a lot fun_activation = function(s_k, s_hat, x_k, s_hat_len) { common = intersect(s_k, s_hat); if(length(common) != 0) { index = match(common, s_k);

Hello This is Elaine. I am using package lattice to generate boxplots. Using Richard's code, the display was almost perfect except the outlier shape. Based on the following code, the outliers are vertical lines. However, I want the outliers to be empty circles. Please kindly help how to modify the code to change the outlier shapes. Thank you. code package (lattice) dataN <-