similar to: save a list as a matrix

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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 >

Hi all Assume I have a data set xx; Group: 1=group1 ?, 2=group2 IQ: ?1= High, 0 =low fit <- glm(IQ ~group, data = xx, family = binomial()) summary(fit) Results ?????? ????????????Estimate Std. Error z value Pr(>|z|) (Intercept) -2.55456??? 0.210 -12.273? < 5e-16 *** group????????? 0.36180 ?????0.076?? 3.952 ????5.24e-05 *** the odd ratio = exp(0.36180 )= 1.435912 My question

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,

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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

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, 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

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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

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

Hi, Try: x<- 1:4 ?y<- c("*","/","-","+") res<-sapply(y,function(i) {x1<-expand.grid(x,x); unlist(lapply(paste0(x1[,1],i,x1[,2]),function(u) eval(parse(text=u))))}) row.names(res)<- as.character(interaction(expand.grid(x,x),sep="_")) head(res) #??? *?? /? - + #1_1 1 1.0? 0 2 #2_1 2 2.0? 1 3 #3_1 3 3.0? 2 4 #4_1 4 4.0? 3 5 #1_2 2 0.5

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?

Hi, You could try: A<- matrix(unlist(read.table(text=" 1 2 3 4 5 6 7 8 9 9 8 7 6 5 4 3 2 1 ",sep="",header=FALSE)),ncol=3,byrow=FALSE,dimnames=NULL) library(matrixStats) ?res1<-t(sapply(split(as.data.frame(A),as.numeric(gl(nrow(A),2,6))),colProds)) ?res1 #? [,1] [,2] [,3] #1??? 4?? 10?? 18 #2?? 63?? 64?? 63 #3?? 18?? 10??? 4

> On Feb 24, 2015, at 10:50 AM, <luke-tierney at uiowa.edu> wrote: > > The documentation is not specific enough on the indented semantics in > this situation to consider this a bug. The original R-level > implementation of lapply was > > lapply <- function(X, FUN, ...) { > FUN <- match.fun(FUN) > if (!is.list(X)) > X <-

Hi, S4 method dispatch can be very slow. Would it be reasonable to cache the most recent dispatch, anticipating the next invocation will be on the same type? This would be very helpful in loops. fun0 <- function(x) sapply(x, paste, collapse="+") fun1 <- function(x) { paste <- selectMethod(paste, class(x[[1]])) sapply(x, paste,

> On Feb 25, 2015, at 5:35 PM, Benjamin Tyner <btyner at gmail.com> wrote: > > Actually, it depends on the number of cores: Under current semantics, yes. Each 'stream' of function calls is lazily capturing the last value of `i` on that core. Under Luke's proposed semantics (IIUC), the result would be the same (2,4,6,8) for both parallel and serial execution. This is