similar to: R and SPSS

Is there a colinearty function implemented in R? I have tried help.search("colinearity") and help.search("collinearity") and have searched for "colinearity" and "collinearity" on http://www.rpad.org/Rpad/Rpad-refcard.pdf but with no success. Many thanks in advance, Peter Lauren.

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 questions about MANOVA which I am still not sure if appropriately I should use it. For example I have a data set like this: BloodPressure (BP) Weight Height 120 115 165 125 145 198 156 99 176 I know that BloodPressure is correlated with both Weight and Height, however colinearity exists between Weight and Height. When I use BP = Weight + Height

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 =

I'm sorry to all those who are tired of seeing my email appear in need of help. But, I've never coded in any program before, so this has been a difficult process for me. Is there a simple function to test for colinearity in R? I'm running a logistic regression and a linear regression. Thanks for the help! [[alternative HTML version deleted]]

Hi! I'm having some issues on both conceptual and technical levels for selecting the right combination of variables for this model I'm working on. The basic, all inclusive form looks like lm(mic ~ B * D * S * U * V * ICU) Where mic, U, V, and ICU are numeric values and B D and S are factors with about 16, 16 and 2 levels respectively. In short, there's a ton of actual explanatory

Dear all, I am a newcomer to R. I intend to using R to do stepwise regression and PLS with a data set (a 55x20 matrix, with one dependent and 19 independent variable). Based on the same data set, I have done the same work using SPSS and SAS. However, there is much difference between the results obtained by R and SPSS or SAS. In the case of stepwise, SPSS gave out a model with 4 independent

Hi, Is there any facility in R to perform a stepwise process on a model, which will remove any highly-correlated explanatory variables? I am told there is in SPSS. I have a large number of variables (some correlated), which I would like to just chuck in to a model and perform stepwise and see what comes out the other end, to give me an idea perhaps as to which variables I should focus on. Thanks

Dear R users, I want to estimate a Cox PH model with time-dependent covariates so I am using a counting process format with the following formula: Surv(data$start, data$stop, data$event.time) ~ cluster(data$id) + G1 + G2 + G3 + G4 + G5 +G6 Gs represent a B-spline basis functions so they sum to 1 and I can't estimate the model as is without getting the last coefficient to be NA, which

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 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 two questions regarding install.packages(). Q1: may I run it in non-interactive mode, which means just install the packages I want rather than letting me to choose which mirror etc? Q2: I ran a R code in batch mode, for example, R CMD BATCH a.r In a.r, I have some statements like packages = library()$result[, "Package"]; f(! "plyr" %in% packages) {

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

Hallo, I hope I am posting to the right place. I was advised to try this list by Ben Bolker (https://twitter.com/bolkerb/status/859909918446497795). I also posted this question to StackOverflow (http://stackoverflow.com/questions/43771269/lm-gives-different-results-from-lm-ridgelambda-0). I am a relative newcomer to R, but I wrote my first program in 1975 and have been paid to program in about

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 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 Simon, Yes, if I uses coefficients() I get the same results for lm() and lm.ridge(). So that's consistent, at least. Interestingly, the "wrong" number I get from lm.ridge()$coef agrees with the value from SPSS to 5dp, which is an interesting coincidence if these numbers have no particular external meaning in lm.ridge(). Kind regards, Nick ----- Original Message -----

Thanks, I was getting to try this, but got side tracked by actual work... Your analysis reproduces the SPSS unscaled estimates. It still remains to figure out how Nick got > coefficients(lm(ZDEPRESSION ~ ZMEAN_PA * ZDIVERSITY_PA, data=s1)) (Intercept) ZMEAN_PA ZDIVERSITY_PA ZMEAN_PA:ZDIVERSITY_PA 0.07342198 -0.39650356

Hi John, Thanks for the comment... but that appears to mean that SPSS has a big problem. I have always been told that to include an interaction term in a regression, the only way is to do the multiplication by hand. But then it seems to be impossible to stop SPSS from re-standardizing the variable that corresponds to the interaction term. Am I missing something? Is there a way to perform the