similar to: boxplot with grouped variables

I tried to draw a point on all boxplots for their means, I did: boxplot( Daten$weight~interaction(Daten$Dosis,Daten$sex, drop=TRUE)) means<-tapply( Daten$weight, Daten$Dosis, mean) points(means, pch=5, col="red", lwd=5) but only the boxplots for male got that point on them, its really weird because I don't think that I separated the sex in the codes above [[alternative HTML

I want to make a stacked bar plot with one bar for two variables from my data "chir", the two variables have about 100 values like no, yes and na. I want to show how many no, yes and na they both have together with the stacked bar. I tried to melt these to variables first like this: melt1=melt(data_chir, measure.vars=c("N1_re", "N2_re"), var="zpd") but it

my data looks like this: PM10 Ref UZ JZ WT RH FT WR 1 10.973195 4.338874 nein Winter Dienstag ja nein West 2 6.381684 2.250446 nein Sommer Sonntag nein ja Süd 3 62.586512 66.304869 ja Sommer Sonntag nein nein Ost 4 5.590101 8.526152 ja Sommer Donnerstag nein nein Nord 5 30.925054 16.073091 nein Winter Sonntag nein nein Ost 6

dear R users, my data looks like this PM10 Ref UZ JZ WT RH FT WR 1 10.973195 4.338874 nein Winter Dienstag ja nein West 2 6.381684 2.250446 nein Sommer Sonntag nein ja Süd 3 62.586512 66.304869 ja Sommer Sonntag nein nein Ost 4 5.590101 8.526152 ja Sommer Donnerstag nein nein Nord 5 30.925054 16.073091 nein Winter Sonntag nein

Hello, I guess a have a very simple problem though up to now couldn't solve it: I want to plot two datasets wihtin one plot like plot(x) provides it for one dataset(type="b" that is: points connected by lines). Example data 'x': Befragung1 Befragung2 Befragung3 Geschlecht 2.25 2.34 1.78 weiblich 1.34 3.45 2.23 maennlich The two rows of the example above

Hello I've stumbled upon a problem for inversion of a matrix with large values, and I haven't found a solution yet... I wondered if someone could give a hand. (It is about automatic optimisation of a calibration process, which involves the inverse of the information matrix) code: ********************* > macht=0.8698965 > coeff=1.106836*10^(-8) >

Hola a todos . ¿cómo puedo ordenar los niveles de un factor? Ejemplo > x<-c("0","0-9","10-49","50-99","100-199",">200") > dosis<-factor(rep(x,10)) > dosis [1] 0 0-9 10-49 50-99 100-199 >200 0 0-9 10-49 [10] 50-99 100-199 >200 0 0-9 10-49 50-99 100-199 >200

Hi everyone: This may be trivial but I just have not been able to figure it out. Imagine the following dataframe: a b c d TRUE TRUE TRUE TRUE FALSE FALSE FALSE TRUE FALSE TRUE FALSE FALSE I would like to create a new dataframe, in which TRUE gets 0 but if false then add 1 to the cell to the left. So the results for the example above should be something like: a b c

Hola compañeros de la lista. Tengo el análisis que muestro abajo, para la estimación de la Dosis Letal Media (DL50) por medio de regresión logística, de dos compuestos similares (denominados 1 y 2) aplicados a grupos de individuos similares, evaluando el número de individuos vivos y muertos al final del ensayo. Mi duda es qué forma me recomiendan para determinar si la diferencia es

Hi, i get error after using a data.frame with subset for "sqlSave". What can i do ? It seems that lines like this in a data.frame structure are after subset deleted and cause the error ? - attr(*, "variable.labels")= Named chr "CUSID" "Welle" "Arbeitgeberfavorit1" "Aktuelle Bewerbungssituation" ... ..- attr(*, "names")=

Hi dear all, the following code is correct. but I want to use non-numeric x-axis, for example if I replace time <- seq(0,72,6) by month <- c("Jan","Feb","Mar","Apr","May","Jun","Jul","Aug","Sep","Oct","Nov","Dec","Pag") Ofcourse I use factor(month) instead of

Dear all, I have data from 1970 to 1990 for people above age 50. Now I want to calculate survival curves by age starting at age 50 using the Kaplan Meier Estimator. The problem I have is that there are already people in 1970 who are older than 50 years. I guess this is called delayed entry or left truncation (?). I thought the code would be: roland <- survfit(Surv(time=age.enter,

Hello, Using the following lines of code, I created the following graph:

Hello, Using the following lines of code, I created the following graph:

I've got this table: A B 1 7.8 215 2 6.8 280 3 9.4 356 4 5.2 38 5 8.2 167 6 4.8 127 7 3.7 8 8 6.2 201 9 9.7 346 10 8.1 241 how can I get the variance and covariance of A and B? -- View this message in context: http://r.789695.n4.nabble.com/how-to-get-variance-tp3232523p3232523.html Sent from the R help mailing list archive at Nabble.com.

Hi! I am looking to choose a condom based on its pleasure score. I received some summarised data from 10 individuals: structure(list(Ramses = c(4, 4, 5, 5, 6, 3, 4, 4, 3, 4), Sheiks = c(5, 5, 6, 4, 7, 6, 4, 5, 6, 3), Trojans = c(7, 8, 7, 9, 6, 3, 2, 2, 2, 3), Unnamed = c(2, 1, 1, 3, 3, 4, 5, 4, 4, 3)), .Names = c("Ramses", "Sheiks", "Trojans", "Unnamed"),

Hi, May be you can try this: obj_name<- load("arun.RData") Pred1<- get(obj_name[1]) Actual1<- get(obj_name[2]) library(reshape2) dat<-cbind(melt(Pred1,id.vars="S1"),value2=melt(Actual1,id.vars="S1")[,3])? # to reshape to long form colnames(dat)[3:4]<- c("Predict","Actual") dat$variable<- as.character(dat$variable) #not that

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

Quiero comparar varias dosis letales 50% (LD50) usando análisis probit. He seguido un ejemplo que viene en paquete DRC, pero no obtengo el resultado esperado. Lo que quiero es saber si las LD50s, son diferentes y si la diferencias son estadísticamente significativas. Gracias de antemano. José Arturo e-mail. jafarfan@uady.mx <grejon@uady.mx> e-mail alterno. jafarfan@gmail.com

I'm pretty sure that I did everything right, but my R is just not drawing the mosaicplot that I want, and there is also no error messege, looks like this right now: > mosaicplot(arthritis) > mosaicplot(~ sex + treatment + improved, data = arthritis, color = TRUE) > -- View this message in context: http://r.789695.n4.nabble.com/problems-with-mosaic-plot-tp3090128p3090128.html Sent