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I have a simple barplot of 4 mean values, each mean value has an associated 95% confidence interval drawn on the plot as an error bar. I want to make a legend on the plot that uses the error bar symbol, and explains "95% C.I." How do I show the error bar symbol in the legend? I could not find any "pch" values that are appropriate Thanks [[alternative HTML version

Hi there, I've tried to update Dovecot to the latest version and have come upon an error during make. Previous upgrades have all been very easy (thank you!) and it appears that the updated make script includes a wget command and unfortunately OS X does not come with wget by default (curl is the default). Not being completely up to speed with the whole autoconf and make process, I'm a

I have read the R manual and help archives, sorry but I'm still stuck. How would I do a t-test with an adjusted p-value? Suppose that I use t.test ( ) , with the function argument alternative = "two.sided", and data such that degrees of freedom = 20. The function calculates a t-statistic of 2.086, and p-value =0.05 How do I then adjust the p-value? My thought is to do

Hello All; I need to run a multiple regression analysis and use Akaike's Information Criterion for model selection. I understand that this command will give the AIC value for specified models: AIC(object, ..., k = 2) with "..." meaning any other optional models for which I would like AIC values. But, how can I specify (in the place of "...") that I want R to

Hi; I am running generalized linear mixed models (GLMMs) with the lmer function from the lme4 package in R 2.6.2. My response variable is overdispersed, and I would like (if possible) to run a negative binomial GLMM with lmer if possible. I saw a posting from November 15, 2007 which indicated that there was a way to get lmer to work with negative binomial by assigning: family =

I have a matrix with 12 rows (one for each month), 2 columns (baseflow, runoff). I would like to make a barplot similar to Excel’s “clustered column chart”. Here is my matrix ‘x’ 8.258754 13.300710 10.180953 10.760465 11.012184 13.954887 10.910870 13.839839 9.023519 11.511129 7.189241 12.519830 5.925576 17.101491 5.211613 13.585175

I am running R on Windows XP I first tried to install rJava and got this message utils:::menuInstallLocal() package 'rJava' successfully unpacked and MD5 sums checked updating HTML package descriptions then i tried this require(rJava) Loading required package: rJava Error in if (!nchar(javahome)) stop("JAVA_HOME is not set and could not be determined from the registry") :

About a week ago, XL Solutions Corporation posted to this list with a solicitation for course offerings in R. I looked at their site, and many courses do look interesting. Can anyone provide feedback on the quality of their courses? [[alternative HTML version deleted]]

I have a time series of about 1500 measurements. There are sporadic data gaps. I would like to plot the time series with type "line". I only want a line to connect the periods with data; I don't want a line to connect the points across a data gap. Is there a function or recommended method to deal with this? For example there are 2000 days days 1:1000 each have one

This seems like a simple thing, but I have been stuck for some time. My data has 2 columns. Column 1 is the value, and column 2 is the Site where data was collected. Column 2 contains 7 different Sites (i.e. groups). I am only interested in showing 3 groups on a single boxplot. I have tried various methods of subsetting the data, in order to only have the 3 groups in my subset. However

I am trying to plot 2 time series on the same graph. For example, X1 is the vector of dates and times, its class is POSIXt. Y1 is an environmental parameter, e.g. salinity. X2 is a second vector of dates and times, also of class POSIXt. X2 has a different length than X1, but they have the exact same range. Y2 is another environmental parameter, e.g. stream flow rate. My approach is

I have a time series with interval of 2.5 minutes, or 24 observations per hour. I am trying to find a 1 hr moving average, looking backward, so that moving average at n = mean(n-23 : n) The time series has about 1.5 million rows, with occasional gaps due to poor data quality. I only want to take a 1 hour moving average for those periods that are complete, i.e. have 24 observations in the

I found the problem. the date range was not exactly overlapping as i had thought. If the range of the 2 time series exactly overlap, then the two graphs overlay correctly. Unless there is another method or package that fixes this, it seems that if the 2 time series have non-matching ranges (even a small difference will cause a problem), then the 2 graphs do not overlay correctly.