similar to: lapply drops colnames

I am faced with a situation wherein I have to use multiple lapply's. The pseudo-code could be approximated to something as below: For each X from i=1 to n For each Y based on j=1 to m For each F from 1 to f Do some calculation based on Fij Store Xi,Yj = Fij End For F End for Y End for X Is there anyway to optimize the processing logic further? I *guess* using the multiple lapply

I have a list of file names, and a list of data frames contained in those files. mynames <- list.files() mydata <- lapply(mynames, read.delim) Every file contains two columns. > colnames(mydata[[1]]) [1] "Name" "NumReads" > colnames(mydata[[2]]) [1] "Name" "NumReads" I can set the colnames easily enough with a for loop. for (i in

R-help, I have a list whose elements are data frames. I want to change the colnames attribute in each element of this list but an error message comes up: > lapply(LD_strataNew,function(x) dimnames(x)[[2]][-1]) <- as.roman(1:9)[-6] Error in lapply(LD_strataNew, function(x) dimnames(x)[[2]][-1]) <- as.roman(1:9)[-6] : could not find function "lapply<-" >

Hi together I'm pretty new to R, so excuse me if it is a basic question. I have a big dataset (extract of it found in the attachment) of returns from firms. I'd like to compute the Pearson correlation of each firm with the "Market" and the corresponding p-Value. So I thought of making a list of 'cor.test's and then extract the needed values with a for loop. What I did so

> a <- sqlQuery(irrdb, "select count(field) from mytable where field = 1") > print(a) count(field) 1 8 > paste(a) [1] "as.integer(8)" Why the as.integer() representation? I later pass the result into this write.html.table(), and what I get is rows of as.integer()... when all I want is the integer itself. as.integer(31) as.integer(21)

Hi all, I'm trying to use lapply on a list with the following command: out<-lapply(mylist,myfun,par1=p,par2=d) (1) where myfun<-function(x,par1,par1) {.....} (2) now this function is in fact a wrapper for some Fortran code I have written so I think this might be the problem. When I call lapply() as in (1) I get the following message: Error in get(x,

Hi, Would be very glad for help on this problem. Using this code: temp<-function(x, bins, tot) { return(as.numeric(lapply(split(x, bins), wtest, tot))); } wtest <- function(x, y) { return(wilcox.test(x,y)$p.value); } rs <- function(x, bins) { binCount <- length(split(x[,1], bins)); tot <- as.numeric(x); result<-matrix(apply(x, 2, temp, bins, tot),

> From: Marc Schwartz > > On Fri, 2005-04-29 at 13:00 +0100, Gavin Simpson wrote: > > Dear List, > > > > Consider the following example: > > > > dat <- data.frame(var1 = rnorm(100), var2 = rnorm(100), > > var3 = rnorm(100), var4 = rnorm(100)) > > oldpar <- par(mfrow = c(2,2), no.readonly = TRUE) > >

Full_Name: Bert Gunter Version: 2.0.0 patched OS: Win2000 Submission from: (NULL) (192.12.78.250) The TRUE/FALSE options of the colnames argument in sqlFetch (RODBC) seem to be reversed, at least for .xls files. ## z is a connection to an xls workbook opened by odbcConnectExcel() > dat<-sqlFetch(z,'SuccessRates',colnames=FALSE) > dat[1:5,] RecID Name Run Grams Lost Run

Hi: I'm sure this is a very easy problem. I've consulted Data Manipulation With R and the R Book and can't find an answer. Sample list of data frames looks as follows: .xx<-list(df<-data.frame(Var1=rep('Alabama', 400), Var2=rep(c(2004, 2005, 2006, 2007), 400)), df2<-data.frame(Var1=rep('Tennessee', 400), Var2=rep(c(2004,2005,2006,2007), 400)),

Patrick, Thanks. I tried colnames, but it doesn't work. Seems more transformation is needed. But now I got names, that's good enough. Best. Xumin Patrick Burns <pburns@pburns.seanet.com> 03/15/2010 04:04 PM To Xumin Zeng <xumin.zeng@abbott.com> cc Subject Re: [R] assign colnames to data Those are names, not colnames. See the 'More R key actions' page

Dear R-listers, I am trying to assign colnames to a data frame within a loop, but I keep getting a "target of assignment expands to non-language object"-error. I need to split up a large dataset into about 20 smaller ones, and I would like to assign colnames within the loop, so I won't have to type the same thing 20 times over. I have concocted this really goofy example which

Hi, Do you know how to assign colnames from one list to another, for example. a=c(1,2,3) b=c("A","B","C") how can I get the dataset A B C 1 2 3 where A, B and C are colnames. Thanks. Xumin [[alternative HTML version deleted]]

Using the input below, can I do something more elegant (and more efficient) than the loop also listed below to pad strings to a width of 5? The true matrix is about 300K rows and 31 columns. ####################### #INPUT ####################### > temp DX1 DX2 DX3 1 13761 8125 49178 2 63371 v75 22237 3 51745 77703 93500 4 64081 32826 v72 5 78477 43828 87645 >

Hello everyone I'm a new R user and have some questions about lists indexing, names and the lapply function. first have a look to my data (names changed) it's a rainfall record for 67 stations, str(data) List of 67 $ e9999.somename.xx.txt :'data.frame': 456 obs. of 5 variables: ..$ Y : int [1:456] 1960 1960 1960 1960 1960 1960 1960 1960 1960 1960 ... ..$

> From: Weiwei Shi > > it works. > thanks, > > but: (just curious) > why i tried previously and i got > > > is.vector(sample.size) > [1] TRUE Because a list is also a vector: > a <- c(list(1), list(2)) > a [[1]] [1] 1 [[2]] [1] 2 > is.vector(a) [1] TRUE > is.numeric(a) [1] FALSE Actually, the way I initialize a list of known length is by

for a simple example: x <- list() x[["a"]] <- list(a=c(1,2,3),b=c(3,4,5)) x[["b"]] <- list(a=c(6,7,8),b=c(9,10,11)) lapply(x,sum) this fails w/ Error in FUN(X[[1L]], ...) : invalid 'type' (list) of argument Just wondering if I have overlooked something obvious. one can also do: lapply(x,lapply,sum) but that assumes that you already know how many levels

Dear R-help, A colleague of mine was running some code on two of our boxes, and noticed a rather large difference in running time. We've so far isolated the problem to the difference between R 1.7.1 and 1.8.0, but not more than that. The exact same code took 933.5 seconds in 1.7.1, and 3594.4 seconds in 1.8.1, on the same box. Basically, the code calls boot() to bootstrap fitting mixture

My data looks like this: > data name G_hat_0_0 G_hat_1_0 G_hat_2_0 G_0 G_hat_0_1 G_hat_1_1 G_hat_2_1 G_1 1 rs0 0.488000 0.448625 0.063375 1 0.480875 0.454500 0.064625 1 2 rs1 0.002375 0.955375 0.042250 1 0.000000 0.062875 0.937125 2 3 rs2 0.050375 0.835875 0.113750 1 0.877250 0.115875 0.006875 0 4 rs3 0.000000 0.074750 0.925250 2 0.897750 0.102000

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 <-