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Hello, I am a new r-user, and after a great effort I have made this fantastic figure with qplot: qplot(ROI, CBF, fill=factor(Carrier), data=combinedboxplot_dataset_se_1_CBF, geom="boxplot", position="dodge",xlab=NULL,ylab=("CBF,white matter-normalized"),main=("Differences between carriers and non-carriers on baseline"))+theme_bw() Can anyone help me

Hi I am using R version 2.7.2. on a windows XP OS and have a question concerning an analysis of covariance with count data I am trying to do, I will give details of a scaled down version of the analysis (as I have more covariates and need to take account of over-dispersion etc etc) but as I am sure it is only a simple problem but I just can't see how to fix it. I have a data set with count

Dear all, Suppose I have a nested, repeated measure lme model. Which of the following formulae is correct? (assuming data are sampled from several plots in an agricultural experiment) (1) y~explanatory.variables,random=~time|block/plot/subplot/individual (2) y~explanatory.variables,random=~time|unique.ID.of.every.individual I have read that (2) is the only approach that works. But how could I

Good morning reader, I have encountered a, probably, simple issue with respect to the *formulae* of a *regression model* I want to use in my research. I’m researching alliances as part of my study Business Economics (focus Strategy) at the Vrije Universiteit in Amsterdam. In the research model I use a moderating variable, I’m looking for confirmation or help on the formulation of the model.

I would like to be able to grab x and y columns out of a dataframe and then grab all of the columns that are not equal to x or y. I am sure that I am missing something easy. ftbr_UTM_downstream <- (structure(list(site = c("Jennie_Creek_Main_Stem", "Wolf_Pit_Creek_Main_Stem", "Little_Rockfish_Main_Stem_North", "Big_Muddy_Creek_Main_Stem",

I have a model Q=K*A*(R^r)*(S^s) A, R, and S are data I have and K is a curve fitting parameter. I have linearized as log(Q)=log(K)+log(A)+r*log(R)+s*log(S) I have taken the log of the data that I have and this is the model formula without the K part lm(Q~offset(A)+R+S, data=x) What is the formula that I should use? Thanks for all of your help. I can provide a subset of data if necessary.

R 2.12.2 on Scientific Linux 6.4 #works chron(times.="15:00:00", format=c(times="h:m:s")) #doesn't work chron(times.="15:00", format=c(times="h:m")) From chron Manual: The times format can be any permutation of "h", "m", and "s" separated by any one non-special character. The default is "h:m:s". what am I

Hi, Is there a way I can specify linear contrasts in glm? I'm looking for something equivalent to SAS' contrast statement. I'd like to do the following, suppose I have a categorical input with 4 levels (a,b,c,d), I'd like to test something like: (i) a+b=c+d, (ii) a=b, (iii) a=b+d, etc... Thanks in advance for your help! Leo. [[alternative HTML version deleted]]

Hi, I'm trying to fit a smooth line in a plot(y ~ x) graph. x is continuous variable y is a proportion of success in sub-samples, 0 <= y <= 1, from a Monte Carlo simulation. For each x there may be several y-values from different runs. Each run produces several sub-samples, where "0" mean no success in any sub- sample, "0.5" means success in half of the

I just had a manuscript returned with the biggest problem being the analysis. Instead of using principal components in a regression I've been asked to analyze a few variables separately. So that's what I'm doing. I pulled a feather from young birds and we quantified certain aspects of the color of those feathers. Since I often have more than one sample from a nest, I thought I

Dear all, I have a dataset where I reduced the dimensionality, and now I have a response variable with probabilities/proportions between 0 and 1. I wanted to do a logistic regression on those, but the function glm refuses to do that with non-integer values in the response. I also tried lrm, but that one interpretes the probabilities as different levels and gives for every level a different

Hello all, I'm very satisfied to say that my grip on both R and statistics is showing the first hints of firmness, on a very greenhorn level. I'm faced with a problem that I intend to analyze using ANOVA, and to test my understanding of a primitive, one-way ANOVA I've written the self-contained practice script below. It works as expected. But here's my question: How can I not

I have just upgraded to R 2.13 and have library(ggplot2) in my .Rprofile (among other things). when i start R I get an error message. Has something in the start up scripts changed? Is there a better way to specify the library calls in .Rprofile? Thanks for all of the help in advance. Error: Loading required package: grid Loading required package: proto Error in rename(x, .base_to_ggplot) :

s <- 1.00 max(s) returns 1 is there anyway that I can get it to return 1.00. I am using the results of this max statement in a grep statement and it returns the wrong numbers, I will provide more information and code if it would make more sense in context. -- Stephen Sefick ____________________________________ | Auburn University? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?? | | Department of

Dear all, I have built glm models based on presences/absences and a number of predictor maps and would like to compute habitat suitability based on the modelled coefficients. I thought this is pretty straight forward and wanted to use predict() and supply the new data in a data frame, with one column for each predictor. However, I do get an error msg warning me that the number of rows for

Is there a way to use a continuous variable to pch in qplot? I believe this worked in previous version. I need to specify certain values of a shape for particular points so that multiple graphs all show the same shapes for the same streams. I have gone to the original data and added a pch column that I would like to use to specify the shapes to pch in qplot. Any help would be greatly

I have found a regression model, and i would like to predict value in different points. I have tried to use predict function but it doesn't work. I have used predict function like this: newdata<-seq(from=0.1, to=0.32,by=0.02) data<-predict(fm,newdata) where fm is a regression model. The predict function return me that: Error in eval(predvars, data, env) : numeric argument

All, 1. I will try and make this clear and concise. Please let me know any information that would be helpful in figuring out this problem (I don't know the relevant information to post). I am on linux- see below for session information. 2. Problem: working directory: home an old version of a function is sourced into the R session and doesn't work working directory: Desktop the

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

With ggplot2, I can plot the glm stat_smooth for binomial data when the response is binary or a two-level factor as follows: data("Donner", package="vcdExtra") ggplot(Donner, aes(age, survived)) + geom_point(position = position_jitter(height = 0.02, width = 0)) + stat_smooth(method = "glm", family = binomial, formula = y ~ x, alpha = 0.2, size=2) But how can I