similar to: taking daily means from hourly data

I have a data set for different time intervals. The data has three comment lines before data for each time interval. For each time interval there are 500 data points. I want to change the dataset such that I have the following format: t1 t2 t3 ................ 0.00208 0.00417 0.00625 ................. a1 a2 a3 ...................

Consider this matrix: > sg X1 X2 X3 X4 X5 1 3.240 2.592 2.592 2.592 2.592 2 2.592 3.240 2.592 2.592 2.592 3 2.592 2.592 3.240 2.592 2.592 4 2.592 2.592 2.592 3.240 2.592 5 2.592 2.592 2.592 2.592 3.240 If I compute the eigenvalues of the 'sg' matrix using R Version 1.5.0 (2002-04-29) under Linux (or using Version 1.4.0 (2001-12-19) under Solaris), I obtain: >

Dear R-helpers, I'm stuck with a little problem that surely has an easy solution but I can't think of a way to solve it. I'd really appreciate any help you can offer me! I'll provide a small example. Given a dataframe data.txt that looks like this: ID freq Var Var_mean Ratio_mean Var_median Ratio_median Var_sum Ratio_min Var_max Ratio_max Var_min

I tried to compare if cch() and coxph() can generate same result for same case cohort data Use the standard data in cch(): nwtco Since in cch contains the cohort size=4028, while ccoh.data size =1154 after selection, but coxph does not contain info of cohort size=4028. The rough estimate between coxph() and cch() is same, but the lower and upper CI and P-value are a little different. Can we

#how do I break these up into first two letters (RM), number, and then the last part #is there an easily accessible regex tutorial on the internet? v = (structure(1:122, .Label = c("RM215Temp", "RM215SpCond", "RM215DO.Conc", "RM215Depth", "RM215pH", "RM215ORP", "RM215Turbidity.", "RM215Battery", "RM215DO.",

I have a file, each column of which is a separate year, and each row of each column is mean precipitation for that month. Looks like this (except it goes back to 1964). month X2000 X2001 X2002 X2003 X2004 X2005 X2006 X2007 X2008 X2009 1 1.600 1.010 4.320 2.110 0.925 3.275 3.460 0.675 1.315 2.920 2 2.960 3.905 3.230 2.380 2.720 1.880 2.430 1.380

I keep getting the same error in my survival analysis. I have access to a very large database but am just using small subsets to get some results. In this particular subset there is 50 explanatory variables(both factors of many levels and covariates) and 117 data pieces with some of the data being censored. I am using the stepAIC command to find my model. My initial model is built from all

Dear all, I'm pleased to announce the release on CRAN of the "mmand" package (for Mathematical Morphology in Any Number of Dimensions). It provides functions for performing mathematical morphology (erode, dilate, etc.), smoothing, and other kernel-based operations on array-like objects of any dimensionality. The package is centred around a flexible function called morph(), which can

Dear all, I'm pleased to announce the release on CRAN of the "mmand" package (for Mathematical Morphology in Any Number of Dimensions). It provides functions for performing mathematical morphology (erode, dilate, etc.), smoothing, and other kernel-based operations on array-like objects of any dimensionality. The package is centred around a flexible function called morph(), which can

After applying the "step" command to a long list of predictors I came up with the following showing there still are some non significant coefficients. I'd like to try the command "leap". However, I don't quite understand how it returns the spared predicting variables. Can please, someone help ? Tank you in advance. Mara EM > summary(stepmod) Call: lm(formula =

I am trying to do some verification across a large dataset, cuData, that has 23 columns. Column 23 (similarity) is the outcome 0 or 1 and the other columns are the features. I do this: verificationglm.model <- glm(formula = similarity ~ ., family=binomial, data=cuData[1:1000,]) and produce the model: > summary(verificationglm.model) Call: glm(formula = similarity ~ ., family =

I am switching from Matlab to R, but I found that R is 200 times slower than matlab. Since I am newbie to R, I must be missing some important programming tips. Please help me out on this. Here is the function: ## make the full pair-wise permutation of a vector ## input_fc=c(1,2,3); ## output_fc=( 1 1 1 2 2 2 3 3 3 1 2 3 1 2 3 1 2 3 ); grw_permute = function(input_fc){ fc_vector = input_fc

Hi R users, I am using the optim funciton to maximize a log likelihood function. My code is as follows: p<-optim(c(-0.2392925,0.4653128,-0.8332286, 0.0657, -0.0031, -0.00245, 3.366, 0.5885, -0.00008, 0.0786,-0.00292,-0.00081, 3.266, -0.3632, -0.000049, 0.1856, 0.00394, -0.00193, -0.889, 0.5379, -0.000063, 0.213, 0.00338, -0.00026, -0.8912, -0.3023, -0.000056), f,

Hello all, I have just started using fitting the PCM (Partial Credit Model) using eRm and have 2 problems which I cannot solve, I have checked everywhere on the net, but no joy: Firstly, I have fit a PCM model to 10 variables with differing response categories (3 for the first 6 items, 4 for the following 2 and 2 for the last two items). mat1 <- matrix(c(rasch_bart$bart_bowel,

R Help - I'm trying to run a correlation matrix with a covariate of "age" and will at some point will also want to covary other variables concurrently. I'm using the "psych" package and have tried other methods such as writing a loop to extract semi-partial correlations, but it does not seem to be working. How can I accomplish this? library(psych) > set.cor(y =

https://bugzilla.mindrot.org/show_bug.cgi?id=1494 Summary: doesn't look on path for $SHELL Classification: Unclassified Product: Portable OpenSSH Version: 5.1p1 Platform: Other URL: http://bugs.debian.org/cgi-bin/bugreport.cgi?bug=49272 8 OS/Version: Linux Status: NEW Severity:

Hi, I am trying to find m breakpoints of a linear regression model. I used the segmented package. It works fine for small number of predicators and breakpoints.(3 r.v. 3 points). However, my model has 14 variables it even would not work even for just one breakpoints!. The error message is always estimated breakpoints are out of range. Since my problem is time related problem. So I