similar to: binary

Un texte encapsul? et encod? dans un jeu de caract?res inconnu a ?t? nettoy?... Nom : non disponible URL : <https://stat.ethz.ch/pipermail/r-help/attachments/20130315/7300f19c/attachment.pl>

Dear All, I apologize for this trivial question, I can not find the solution I try to use t.test function per line in the data.frame, But i dont'understand the error message my program is as follows group1<-gl(2,20) fun<-function(x){ m<-data.frame(group1,x) #anova(aov(x ~ group1, m)) t.est(x ~ group1,m) } ttest<- apply(data, 1, fun) Error in t.test.default(x = c(1, 1,

Un texte encapsul? et encod? dans un jeu de caract?res inconnu a ?t? nettoy?... Nom : non disponible URL : <https://stat.ethz.ch/pipermail/r-help/attachments/20110525/4926c5ee/attachment.pl>

Un texte encapsul? et encod? dans un jeu de caract?res inconnu a ?t? nettoy?... Nom : non disponible URL : <https://stat.ethz.ch/pipermail/r-help/attachments/20110525/acc78a27/attachment.pl>

Un texte encapsul? et encod? dans un jeu de caract?res inconnu a ?t? nettoy?... Nom : non disponible URL : <https://stat.ethz.ch/pipermail/r-help/attachments/20111215/d9983681/attachment.pl>

Dear all, Lets say I have a data frame as follows: > Date <- as.Date(c('2006-08-23', '2006-08-30', '2006-09-06', '2006-09-13', '2006-09-20')) > Income <- c(73.79, 72.46, 76.32, 72.43, 72.62) > data.frame(Date, Income) Date Income 1 2006-08-23 73.79 2 2006-08-30 72.46 3 2006-09-06 76.32 4 2006-09-13 72.43 5 2006-09-20 72.62 >

Dear All, I have matrix (50000 X 60) of subjects and their responses to a set of questions. All responses are classified into categories (500). I would like to average all subject's responses for each category. I wrote a code using a for loop but is not working. Could please tell me what's wrong with the code? I guess, there is a elegant R way of doing the same thing. Thanks in advance.

HI, Dear R community, My original file has 1932 lines, but when I read into R, it changed to 1068 lines, how comes? cdu@nuuk:~/operon$ wc -l id_name_gh5.txt 1932 id_name_gh5.txt > gene_name<-read.table("/home/cdu/operon/id_name_gh5.txt", sep="\t", skip=0, header=F, fill=T) > dim(gene_name) [1] 1068 3 -- Sincerely, Changbin -- Changbin Du DOE Joint Genome

Hi, I''m having a dataframe ''Subset1'' with a number of factor variables and 160 numerical variables Now I want to make sums for all rows that have the same values for the different factor variables, except for the factor variables: VAR1,VAR2,VAR3 who may have the same values. With the formula given below this works great, but in a situation with 15000 rows and 13

Dear r-help, Could you help me to find a solution for this error: Il y a eu 50 avis ou plus (utilisez warnings() pour voir les 50 premiers) > warnings() Messages d'avis : 1: In if ((data[pa, k] == df[, j]) & (data[ch, k] == i)) { ... : la condition a une longueur > 1 et seul le premier élément est utilisé 2: In if ((data[pa, k] == df[, j]) & (data[ch, k] == i)) { ... : la

Hi all, I wanted to have a seasonality study like whether a particular month has significant effect as compared to others. Here is my data : 0.10499 0 0 1 0 0 0 0 0 0 0 0 0.00259 0 0 0 1 0 0 0 0 0 0 0 -0.06015 0 0 0 0 1 0 0 0 0 0 0 0.10721 0 0 0 0 0 1 0 0 0 0 0 0.03597 0 0 0 0 0 0 1 0 0 0 0 0.10584 0 0 0 0 0 0 0 1 0 0 0 0.02063 0 0 0 0 0 0 0 0 1 0 0 -0.03509 0 0 0 0 0 0 0 0 0 1 0 -0.03485 0 0 0

Hi, I'm trying to find an easy way to do this. I want to select the top three values of a specific column in a subset of rows in a data.frame. I'll demonstrate. A B C x 2 1 x 4 1 x 3 2 y 1 5 y 2 6 y 3 8 I want the top 3 values of B from the data.frame where A=X and C <2 I could extract all the rows where C<2, then sort by B, then take

Hi, I'm trying to get maximum likelihood estimates of \alpha, \beta_0 and \beta_1, this can be achieved by solving the following three equations: n / \alpha + \sum\limits_{i=1}^{n} ln(\psihat(i)) - \sum\limits_{i=1}^{n} ( ln(x_i + \psihat(i)) ) = 0 \alpha \sum\limits_{i=1}^{n} 1/(psihat(i)) - (\alpha+1) \sum\limits_{i=1}^{n} ( 1 / (x_i + \psihat(i)) ) = 0 \alpha \sum\limits_{i=1}^{n} (

I have a data frame called "pose": DESCRIPTION QUANITY CLOSING.PRICE 1 WHEAT May/10 1 467.75 2 WHEAT May/10 2 467.75 3 WHEAT May/10 1 467.75 4 WHEAT May/10 1 467.75 5 COTTON NO.2 May/10 1 78.13 6 COTTON NO.2 May/10 3 78.13 7 COTTON NO.2 May/10 1 78.13

Hola, Tengo una base de datos con estructura jerárquica en la que quiero clasificar observaciones en distintas categorías. En el caso más simple, tengo una variable con dos categorías (variable Y1) y dentro de cada una de ellas hay otras dos categorías (variable Y2). Además tengo una variable explicativa cuantitativa discreta X. El banco de datos sería de este tipo: X Y1 Y2 5 0 1 9 0 0 2

Hi Listers, seems I have forgotten some basics re offset in glm: data: counts (y) from locations off different size (area), explanatory variable: x Model: y ~ x+offset(area) Predictions (pred) using Poisson errors plot(x,y) and points(x,pred) gives neat "line" of estimated values. However, for ease of understanding graphs are better using plot(x,y/area). Question: How to display

Dear List: My question is more statistical than R oriented (although it originates from my work with nlme). I know statistical questions are occasionally posted, so I hope my question is relevant to the list as I cannot turn up a solution anywhere else. I will frame it in the context of an R related issue. To illustrate the problem, consider student achievement test score data with multiple

In the logistic regression model, there is no residual log (pi/(1-pi)) = beta_0 + beta_1*X_1 + ..... But glm model will return residuals What is that? How to understand this? Can we put some residual in the logistic regression model by replacing pi with pi' (the estimated pi)? log (pi'/(1-pi')) = beta_0 + beta_1*X_1 + .....+ ei Thanks! [[alternative HTML version deleted]]

Howdy, R Grues I have enjoyed R, but I cannot solve one problem easily. Please help my problem. When I tried the R script, I got the following Error. This error results from input data file exported through a Excel spreadsheet software. Error in step(lm(pop.rate ~ as.numeric(year) + as.factor(policy) + as.numeric(nation.grant) + ?: ??number of rows in use has changed: remove missing values?

Hello, Let's assume that one has a list: my.list <- list(a=c(1,2,3,4),b=4,c=c(1:7)) I want to save my.list as a matrix in .txt file. Because of different length we can put NA or 0. Effect (row.names a, b and c are not necessary) a 1 2 3 4 0 0 0 b 4 0 0 0 0 0 0 c 1 2 3 4 5 6 7 How to do it? Thank you. Robert