similar to: greek letters in plots

Hi, I'd like to find which values of x will give me a y. In other words, in the example of a gaussian curve, I want to find the values of x that will give me a density, let's say, of 0.02. curve(((1/(sqrt(2*pi)*10))*exp(-((x-50)^2)/(2*10^2))),xlim=c(0,100)) Thanks for any help, Antonio Olinto ------------------------------------------------- WebMail Bignet - O seu provedor do

Hi, I have doubts in using subset command. I have a list of, lets say, 15 species and I want to make a subset with only 2 of them. I??m the command data2 <- subset(data1, species=="sp1"|species=="sp2") Nevertheless, when I ask for the summary (summary(data2)) the others species names still apearing. Also I tried data2<-data1(data1$species %in%

Hello, I have a doubt in using the function step (step wise) to select glm models. Usually I apply the gamma distribution to analyze fishery data. To select the terms I use a routine where I first compare single term models to the null model (eg. U~1 vs. U~depth; U~1 vs. U~latitude; etc. ? where U= abundance) and, by means of the result given by a likelihook function applied for each comparison,

Hello all, Here goes one of my first functions. I want to make a nonparametric multiple sample comparison with unequal sample sizes (see Zar?s Biostatistical Analysis, 3rd. Ed., pg. 201 Example 10.11, pg. 288 Example 11.10). In the real world, I want to compare samples of fish length captured with different fishing gears. After using the Kruskal-Wallis test I want to check the differences

Hi, I have a dataframe with 3 columns: month (1 to 12), length and length class. I'm trying to make a cross table with the counts of lengths per length class and month. I don??t have all classes per month. | 1 2 3 ... ------------- 120 | 0 1 0 150 | 5 6 0 170 | 3 0 7 ... I have already tryed many commands, without success - unfortunately. May someone help me? Thanks in advance!

Thanks for your reply Nassar. What I really want is to overlay (like add=TRUE) the plots 1 and 2. With plots "at the same screen" I meant overlaid and not one beside the other, i.e., only one plot with all the points (1 to 10) in blue (or any other color), the points 5 to 9 in red and a regression line from points 5 to 9. Best regards, Antonio Olinto ----- Original Message -----

Hi, I?m trying to make a barplot with the following dataframe, with information on relative frequency per sediment type (ST) for some species: Species ST1 ST2 ST3 SP_A 10 60 30 ... At x-axis are (should be ...) the species names and at y-axis the frequency per sediment, in stacked bars. I tried to use barplot command but with no results. Could anyone help me on this? Thanks in

Hello, This is my first post to this list. Well, I tried to install Google Earth using WineTools but I got the "Unknown Graphics Card" error. I can open the program but Earth doesn't appear. I'm running Debian Sarge on a ASUS CUW(E)-FX motherboard (Intel 810 Chipset) and Gnome 2.10.2 inteface. Thanks for any help. Best regards, Antonio -- Ant?nio Olinto Biologist /

Dear R community, In the excellent nlme package the default diagnostic plot graphs the innermost residuals against innermost fitted values. I recently fit a mixed-effects model in which there was a very clear positive linear trend in this plot. I inferred that this trend occurred because my fixed effect was a two-level factor, and my random effect was a 12-level factor. The negative residuals

Dear R-list members Here goes a question on data frames: I want to create a new data frame excluding some records (rows) from an existing one. Considering the data frame "fish.dat" at the end of this message, I can make a boxplot(Ring~Radius) and identify the points at rows 11 and 25 as outliers. With the command fish.dat[c(11, 25),] I can identify which fishes are related to

Hello R-list members, I’m trying to make a min-avg-max plot (like a boxplot, without box and outliers, i.e., for each class of "x" a vertical line indicating the range (max-min) and an horizontal line or point indicating the mean) with data like: Year Avg Min Max 98 10 7 13 99 12 10 15 01 11 6 14 Which command should be used? Many thanks, Antonio Olinto

Hi, I have a data set with values (L) varying from 1 to 700, with repeated numbers, and I want to sample them according to the probabilities given by a logistic curve P=1/(1+e(-r*(L-Lc))) where r gives "S" the inclination and Lc the "rotation point". P ranges from 0 to 1. As I could see I cannot use this P as prob in sample function. I'd need the probabilities as given by

I don't get the same result, do you have a package loaded that would change the default behavior (such as Hmisc)? > list(grp.1) [[1]] [1] 1 2 Levels: 1 2 > list(grp.1[mask]) [[1]] [1] 1 Levels: 1 2 > library(Hmisc) <<snip>> > list(grp.1[mask]) [[1]] [1] 1 Levels: 1 --Matt > version _ platform i386-pc-mingw32 arch i386 os

Hi all, I'm using R 1.5.1 on Windows 2000. The following snippet of code doesn't seem to do anything - no error is reported, and there is no name change. names(myFrame[,c(1:3)]) <- c("name1", "name2", "name3") This code however works nicely: names(myFrame)[c(1:3)] <- c("name1", "name2", "name3") Can anyone suggest why

Greetings all, I'm trying to figure out how to calculate the inverse CDF (i.e. a quantile) for a non-central F distribution. I could put together a quick numerical solver routine using the CDF, but I wonder if there's a function that I've missed that would be more efficient? Thank-you, Andrew Andrew Robinson Ph: 208 885 7115 Department of Forest Resources Fa: 208 885

Hello, I have been searching for ways to obtain these for combinations of fixed factors and levels other than the 'baseline' group (contrasts coded all 0's) from a mixed-effects model in lme. I've modelled the continuous variable y as a function of a continuous covariate x, and fixed factors A, B, and C. The fixed factors have two levels each and I'd like to know whether

Dear R-list members, Some months ago I wrote a message on the usage of gnls (nlme library) and here I come again. Let me give an example: I have a 10 year length-at-age data set of 10 fishes (see growth.dat at the end of this message) and I want to fit a von Bertalanffy growth model, Li= Linf*(1-exp(-k*(ti-t0))) where Li = length at age i, Linf= asymptotic length, k= curvature parameter, ti=

I would appreciate some thoughts on using the bootstrap functions in the library "bootstrap" to estimate confidence intervals of ICC values calculated in lme. In lme, the ICC is calculated as tau/(tau+sigma-squared). So, for instance the ICC in the following example is 0.116: > tmod<-lme(CINISMO~1,random=~1|IDGRUP,data=TDAT) > VarCorr(tmod) IDGRUP = pdLogChol(1)

I have two lists: xa <- list( X=c(1,2,3), Y=c(4,5,6), Z=c(7,8,9) ) xb <- with( barley, tapply( X=seq(1:nrow(barley)), INDEX=site , FUN=function(z)yield[z])) I can convert xa to a dataframe easily with: as.data.frame(xa) But if i try the same with xb I get: as.data.frame(xb) Error in as.data.frame.default(xb) : can't coerce array into a data.frame What

Why does stop("we are done") print "Error in eval.with.vis(expr, envir, enclos) :" ? It would seem to me that a plain stop() is not an error, and that it would make more sense to have an error() function that is different from a stop(). Is there a rationale here that I am missing? sincerely, /iaw