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I have two matrices (see example below) and I want the differences for the matching row numbers, but the row numbers are not identical in the two matrices. There are probably many ways to do this. Anyone know of any easy way to do this? I could loop over them, but you know what they say about for loops... Thanks, Roger > out.r[1:5, 1:3] 1 2 3 1100 -0.0992 -0.0802 -0.0653 1200 -0.1242

Hello, i have the following model, poi1<-glm(F~S+T+L+C,family=poisson,x=T) where F,S,T,L are metric and C is a factor variable with the levels "0", "1", "2", "3", "4", "5" and "6" if i do summary(poi1), i get the following Call: glm(formula = F ~ S + T + L + C, family = poisson, x = T) Deviance Residuals: Min

I have a fairly large set of data with the following attributes: >str(raw.data) `data.frame': 1429 obs. of 16 variables: $ TStamp :`POSIXlt', format: chr "2001-11-25 02:00:00" "2001-11-25 01:55:00" "2001-11-25 01:50:00" "2001-11-25 01:45:00" ... $ iPDT.AHU14.14: num 0.0122 0.0125 0.0120 0.0120 0.0122 ... $ iPDT.AHU14.15: num 0.0121

Dear all, I was wondering if you could give me any suggestions/help on the following issue. So I carried out the analysis of my data using generalized linear model (glm). After that, to check for multiple comparisons, I applied the glht function from the multcomp package in R. The output, however, gave me a warning (please see below). So my question is whether this warning is smth that I should

Hi! I try to explain the efffect of (1) forest where i took samples's soils (* Lugar*: categorical variable with three levels), (2) nitrogen addition treatments (*Tra*: categorical variable with two levels) on total carbon concentration's soil samples (*C: *continue* *variable) during four months of sampling (*Time:* categorical and ordered variable with four levels). I fitted the

Using the small reproducible example below, I'd like to know if one can somehow use the matrix "sig" (defined below) to add a black outline (with lwd=2) to all pixels with a corresponding value of 1 in the matrix 'sig'? So for example, in the ggplot2 plot below, the pixel located at [1,3] would be outlined by a black square since the value at sig[1,3] == 1. This is my first

I am running a multilevel growth curve model to examine predictors of social anhedonia (SA) trajectory through ages 12, 15 and 18. SA is a continuous numeric variable. The age variable (Index1) has been coded as 0 for age 12, 1 for age 15 and 2 for age 18. I am currently using a time varying predictor, stress (LSI), which was measured at ages 12, 15 and 18, to examine whether trajectory/variation

I am running a multilevel growth curve model to examine predictors of social anhedonia (SA) trajectory through ages 12, 15 and 18. SA is a continuous numeric variable. The age variable (Index1) has been coded as 0 for age 12, 1 for age 15 and 2 for age 18. I am currently using a time varying predictor, stress (LSI), which was measured at ages 12, 15 and 18, to examine whether trajectory/variation

Hi Eric, you can use an annotate-layer, eg ind<-which(sig>0,arr.ind = T) ggplot(m1.melted, aes(x = Month, y = Site, fill = Concentration), autoscale = FALSE, zmin = -1 * zmax1, zmax = zmax1) + geom_tile() + coord_equal() + scale_fill_gradient2(low = "darkred", mid = "white", high = "darkblue",

Hello, I would like to perform a Partial least square discriminate analysis (PLSDA) in R. To do this I use the package mixOmics. I could perform the PLSDA in R. however I would also like to perform a leave-one-out cross validation in order to assess the performance of my model. My supervisor told me that I should focus on the R2/Q2 ratios. However when I read the instruction for running the

Dear all, I'm using R 2.8.1 under Windows Vista on a dual core 2,4 GhZ with 4 GB of RAM. I'm trying to reproduce a result out of "Analysis of Financial Time Series" by Ruey Tsay. In R I'm using the fGarch library. After fitting a ar(3)-garch(1,1)-model > model<-garchFit(~arma(3,0)+garch(1,1), analyse) I'm saving the results via > result<-model

Dear Srinidhi, You are trying to fit 1 random intercept and 2 random slopes per individual, while you have at most 3 observations per individual. You simply don't have enough data to fit the random slopes. Reduce the random part to (1|ID). Best regards, Thierry ir. Thierry Onkelinx Statisticus / Statistician Vlaamse Overheid / Government of Flanders INSTITUUT VOOR NATUUR- EN BOSONDERZOEK

Dear Srinidhi, You are trying to fit 1 random intercept and 2 random slopes per individual, while you have at most 3 observations per individual. You simply don't have enough data to fit the random slopes. Reduce the random part to (1|ID). Best regards, Thierry ir. Thierry Onkelinx Statisticus / Statistician Vlaamse Overheid / Government of Flanders INSTITUUT VOOR NATUUR- EN BOSONDERZOEK