similar to: how to checking whether elements of a vector changed or not.

Hello, I have a set of data that has a Date column looks like this: 12/9/2007 12/16/2007 1/1/2008 1/3/2008 1/12/2008 etc. I'd like the date to look something like the follow (so that I could sort by date easily). 20071209 20071216 20080101 20080103 20080112 How to do it? Thank you very much Julia -- View this message in context:

Hello, I have a data set that looks like this. ID value 111 5 111 6 111 2 178 7 178 3 138 3 138 8 138 7 138 6 . . . I'd like to calculate the mean and var for each object identified by the ID. I can in theory just loop through the whole thing..., but is there a easier way/command which let me calculate the mean/var by ID? Thanks, Julia -- View this

Dear R users, I am building a hierarchical model on a large data set. It can take quite some time to finish one fit, I was just wondering whether it is possible to store the fit object (the result) to a file for later (offline) analysis. thanks Julia -- View this message in context: http://www.nabble.com/how-to-store-lme-lmer-fit-result-tp19910951p19910951.html Sent from the R help mailing

Hello, I have a data file (.csv) that has a size of about 2.6 GB. I am not able to read in the whole data set because of the memory limit. I actually only need some columns (3 columns) of the data set, is there a way to read in specified columns? I am using windows. Thanks, Julia -- View this message in context:

Dear R users, I need to compare two scatter plots, plot(x1, y1) plot(x2, y2) and would like to plot them in the same figure. How do I do it? Thank you. -- View this message in context: http://www.nabble.com/two-scatter-plots-in-one-tp21963375p21963375.html Sent from the R help mailing list archive at Nabble.com.

Dear R users: I have a data set that looks something like this: ID time y 1 2/01/2008 4 1 2/09/2008 12 1 9/01/2008 8 2 1/06/2008 3 2 3/01/2008 4 2 3/09/2008 9 2 6/03/2008 4 3 1/02/2008 3 3 1/10/2008 8 3 2/02/2008 7 3 2/10/2008 3 I'd like to sum up the weekly data to the monthly level, so that it looks something

Hello, I have a time series plot drawn using 3 different colored lines, each line corresponds to different category group. I'd like to put legends on the plot. I am using "legend" to do this, however, I can either specify lty or col in legend. Will I be able to do the following in the legend box: a solid black line then g1, a solid red line then g2, a solid blue line then g3?

Dear R users: I am running lmer(), and I had the following warning message: "In mer_finalize(ans): iteration limit reached without convergence (9) What is the default limit for lmer()? And how do I increase the limit? Many thanks, Julia -- View this message in context: http://www.nabble.com/How-to-increase-iteration-limit-for-lmer%28%29--tp20252974p20252974.html Sent from the R help

Dear R users, Suppose I have a vector that consists of characters like ABC, A02, RCA, etc., and there are about 700 of possible characters. For example, x <- c("ABC", "ABC", "ABC", "A02", "ABC", "RCA", "ABC", "ABC") I'd like to get a frequency matrix that looks something like this: ABC 6 A02 1 RCA 1 I

Hi, I would like to operate on certain columns in a dataframe, but not others. My data looks like this: x1 x2 x3 1 2 3 4 5 6 7 8 9 I want to create a new column named x4 that is the sum of x1 and x2, but NOT x3. I looked at colSums and apply, but those functions seem to use all the columns in a dataframe. How do I only use select columns? If it helps, in Stata this would be gen x4

Dear all, I need to run a simple linear regression such that: y = b0 + b1*x1 + (1-b1)*x2 + e which I know I can use: lm(y ~ I(x1 - x2) + offset(x2)). However, I also need to restrict the coefficient b1 to be between 0 and 1. Is there any way to include such restriction in the linear regression estimation? I saw suggestion related with the function Solve.QP, but I really did not understand such

Dear colleagues, this seems like an easy problem, and I found some suggestions which I've incorporated in the help list, but I can't quite get it right. I want to add a series of years to a second x-axis category label. I generate them with test and test_2 below, format them with some spacing (which is the suggestion I took from the R-list) and concatenate them and then write them with

n=10 x1<-(1:n)/n y1<-rnorm(n,x1^2,1) m=20 x2<-(1:m)/m The value of y2 will be rnorm (m, x2^2,1) if none of the elements of x2 is same as x1, but for every same elements in x1 and x2, the value of y2 will be same as y1. I know the following is correct, but for large vectors, this won't work efficiently as this is taking longer time. Can you please let me know how to write the

Dear R users, I am running lmer() and having memory problem: "reached total allocation of 1535Mb". The lmer() function was successful (no errors or warnings). However when I do "summary(fit.lmer), I got this "Error: cannot allocate vector of size 19.5 Mb). I used memory.size() and got 1880. I used memory.limit(2500) to increase the memory size to 2500Mb, however I still get

Dear R users, I would like to test the randomness in a series of N values (N>=2). I know that runs.test works for dichotomous factor only: x <- rep(c(1,2), 50) runs.test(factor(x)) However it doesn't work for series that can take any N values (N>2): x <- rep(c(1,2,5,4),50) runs.test(factor(x)) Error in runs.test(factor(x)) : x does not contain dichotomous data Are there any R

Dear R users, I have this vector that consists numeric numbers. Is there a command that detects the repeated numbers in a vector and returns the index of the repeated numbers (or the actual numbers)? For example, v <- c(3,4,5,7,4). The command would return me index 2 and 5 (or the repeated number, 4). Thank you very much, Julia -- View this message in context:

Thanks for all the replies which are really helpful. one more question regarding transformation: 4. Given an instruction %x1 = I1, can I replace the rhs "I1" with "I2" and get a new instruction %x1 = I2? Alternatively, I can add a new instruction %x2 = I2, and replace all uses of %x1 with %x2, and then delete %x1 = I1, but it seems the former is simpler or faster if

Hello people. I wish to reorder a simple vector of numbers by another vector of the order (and then do the same, but with a data frame rows) I try this (which doesn't work) : > aa <- c(3, 1 ,2 ) > aa[aa] [1] 2 3 1 The same won't work if I try to order a data frame: > data.frame(matrix(c(3,1,2), 3,2))[c(3,1,2),] X1 X2 3 2 2 1 3 3 2 1 1 How should I do this? p.s:

On Wed, Apr 14, 2010 at 5:45 PM, Neal N. Wang <neal.wang at gmail.com> wrote: > Thanks for all the replies which are really helpful. > > one more question regarding transformation: > > 4.  Given an instruction  %x1 = I1,  can I replace the rhs "I1"  with  "I2" > and get a new instruction %x1 = I2?   Alternatively, I can add a new > instruction %x2 =

Dear all, I am trying to compare the estimated coefficients of a quantile regression model between two different samples. It is a Wald test, but I cannot find one way to do that in R.The samples are collected conditional on a specific characteristic and I would like to test whether such characteristic indeed affect the estimators. The problem in the test anova.rq is that the response variable