similar to: Extract information from the names of a list within lapply

Dear list, I am running some linear regressions through lapply, >lapply(c('EMAX','EC50','KOUT','GAMMA'),function(x)confint(lm(get(x)~RR0,dataset2))) I got results like [[1]] 2.5 % 97.5 % (Intercept) 0.6595789212 0.8821691261 RR0 -0.0001801771 0.0001489083 [[2]] 2.5 % 97.5 % (Intercept) -63.83694930

Hi, I have run a groups of t test and obtained a list of the t-test results (about 30). How do I extract p-values from the 30 t-test results and make a table output? Thanks Jun Example t-test output: ============================ $Lambda_z Welch Two Sample t-test data: x[PA1.BS.DN$Onecode == 2] and x[PA1.BS.DN$Onecode == 3] t = -0.1128, df = 33.635, p-value = 0.9109 alternative

Dear All, to produce output of several columns of a data frame, I tried to use lapply and also l_ply. In both cases, I would like to print a header line containing also the name of the respective column in the data frame. For example, I would like the following lapply(data.frame(a=1:3, b=2:4), function(x) print(deparse(substitute(x)))) to produce: [1] "a" [1] "b" and

Dear all, List g has 2 elements > names(g) [1] "2009-10-07" "2012-02-29" and the list plot lapply(g, plot, main=names(g)) results in equal plot titles with both list names, whereas distinct titles names(g[1]) and names(g[2]) are sought. Clearly, lapply is passing 'g' in stead of consecutively passing g[1] and then g[2] to process the additional 'main'

Hi all, I would like to iterate through a list with named elements and access the names within an lapply / sapply call. One way to do this is iterate through the names and index the list with the name. Is there a way to iterate through the list elements themselves and access the element names within in the function? For example, mylist <-

|Hello, Given a function with several arguments, I would like to perform an lapply (or equivalent) while holding one or more arguments fixed to some common value, and I would like to do it in as elegant a fashion as possible, without resorting to wrapping a separate wrapper for the function if possible. Moreover I would also like it to work in cases where one or more arguments to the original

Gregexpr - extract results with lapply Hello, I need to extract sequences of three upper case letters in a string. In other words, in this string: str <-c("ABC", "this WOUld be gOOD") The result I'm looking for is ABC WOU OOD. With gregexpr, I can get the position and length of the sequences gregexpr('[A-Z]{3}',str,perl=TRUE) [[1]] [1] 1

Hi, When using lapply (or sapply) to loop over a list, can I somehow access the index of the list from inside the function? A trivial example: df1 <- split( x=rnorm(n=100, sd=seq(from=1, to=10, each=10)), f=letters[seq(from=1, to=10, each=10)] ) str(df1) #List of 10 # $ a: num [1:10] -0.801 0.418 1.451 -0.554 -0.578 ... # $ b: num [1:10] -2.464 0.279 4.099 -2.483 1.921 ... # $ c: num

Hello I am using R 2.2.0 with Windows XP. I've got a five element list object, each element containing two dataframes of equivalent size. > str(mylist) List of 1 $ data1:List of 2 ..$ data1a :`data.frame': 77 obs. of 63 variables: .. ..$ var1 : num [1:77] 0.41375 0.00056 1.43040 1.43528 0.61730 ... .. ..$ var2 : num [1:77] 1.154 1.686 0.673 0.800 0.760 ... ..

Hi all, I have a data frame called "rst", see below: ------------------------------------------------------------------------------------------ # This is a paste able example # In case you don't have "KernSmooth" package installed, please uncomment below line. # install.packages("KernSmooth") library(KernSmooth) rst <- data.frame(hsp = rnorm(23), dal =

Sometimes I'm iterating over a list where names are keys into another data structure, e.g. a related list. Then I can't use lapply as it does [[]] and loses the name. Then I do something like this: do.one <- function(ldf) { # list-dataframe item key <- names(ldf) meat <- ldf[[1]] mydf <- some.df[[key]] # related data structure r.df <-

I have a question regarding accessing the names of list components when applying a function with lapply. Here is an example that demonstrates what I'd like to do. I have a list like this one: mylist <- list(a=letters[1:10], b=letters[10:1], c=letters[1:3]) Now I would like to append the names of the list components to their corresponding vectors with the c() function. I thought this

Dear list, Let's say we have a data frame as follows, >expand.grid(a=1:5,b=c(1,5,10,20),DV=c(0.1,0.2,0.3))->df1 >df1$DV<-rgamma(60,shape=10) columns a and b are two levels. DV is the column with values we are interested in. Then another data frame df2 with values as follows data.frame(a=c(2,2,3,4,5),b=c(5,10,1,5,20))->df2 Now I want to a subset of df1 that match df2 at the

Would like to try my luck to see if I can catch your eyes. I was trying to do some contrasts within ANOVA. I searched the archive and found a clue posted by Steffen Katzner ( http://tolstoy.newcastle.edu.au/R/help/06/01/19385.html) I have three levels for a factor names "StdLot" and want to make three comparisons, 1 vs 2, 1 vs 3 and 2 vs 3. First,

I have applied the same linear model to several different subsets of a dataset. I recently read that in R, code should never be repeated. I feel my code as it currently stands has a lot of repetition, which could be condensed into fewer lines. I will use the mtcars dataset to replicate what I have done. My question is: how can I use fewer lines of code (for example using a for loop, a function or

Dear all, In a plot command like xyplot(Y~X|ID*PERIOD,data=...) xyplot will generate all the possible ID*PERIOD combinations. But not all of them have data in there. So I have a lot of empty plots. How can I suppress those empty plots and ask xyplot only to generate plots actually with data. Thanks. Jun [[alternative HTML version deleted]]

Hi. I have a list where each object in the list has multiple parts. I'd like to take the mean of just one part of each object. Is it possible to do this with lapply? If not, can you recommend another function? Thanks. eric > x1 <- c(0,1,2,3) > x2 <- c(7,8) > x3 <- c(2,6,6,8) > x4 <- c(4,8) > > Lst1 <- list(label1 = x1,label2 = x2) > Lst2 <-

Dear list, I want to show all the objects starting with "d" and ending with a digit. How do I specify these conditions in the pattern argument I can do one condition but not two ls(pattern='^d') ls(pattern='[[:digit:]]$') are working. But, ls(pattern='^d'&'[[:digit:]]$') is not working. Appreciate any comment. Jun Shen [[alternative HTML

tmp <- matrix(1:2) tmp tmp[,1,drop=FALSE] See the above example. Is there a way to make 'drop=FALSE' as global default, so that when I say 'tmp[,1]', R will treat it as 'tmp[,1,drop=FALSE]'?

Long story short, I have a big iterative procedure that produces a long list of data.frames such as the one called "results" here. Is there an easy way to produce a similar list of data.frames comprised of the mean of each of the columns in results, such that it ends up like the one I've shown in "resultsmean" below? I've tried apply and lapply, still not got the