search for: newlist

Morning all, I received this mail this morning from an admin regarding a xen guest (newlists). It would seem that the filesystem is larger than its container? Before I spend too long looking at it, does anyone have any ideas as to how this may have occurred and how it could be fixed? Kind regards, Sam ========================================= ---------- Forwarded message ---------- Fr...

HI all, I'm new to R. Say I have a multi-layered list called newlist. ############ > str(newlist) List of 2 $ :List of 5 ..$ : num [1:8088] NA 464 482 535 557 ... ..$ : num [1:8088, 1:2] NA 464 482 535 557 ... ..$ : num [1:8088, 1:3] NA 464 482 535 557 ... ..$ : num [1:8088, 1:4] NA 464 482 535 557 ... ..$ : num [1:8088, 1:5] NA 464 482 535 557 ... $...

...ist<-list(data.frame(a=1:2,b=2:3), data.frame(a=3:4,b=5:6),data.frame(a=7:8,b=9:10)) (mylist) # I want to grab only one specific column from each list element neededcolumns<-c(1,2,0) # number of the column I need from each element of the list # Below, I am doing it using a loop: newlist<-NULL for(i in 1:length(mylist) ) { newlist[[i]]<-mylist[[i]] [neededcolumns[i]] } newlist<-do.call(cbind,newlist) (newlist) I was wondering if there is any way to avoid the loop above and make it faster. In reality, I have a much longer list, each of my data frames is much larger and I...

This is a minor release of Shorewall. WARNING: This release introduces incompatibilities with prior releases. See http://www.shorewall.net/upgrade_issues.htm. Changes are: a) There is now a new NONE policy specifiable in /etc/shorewall/policy. This policy will cause Shorewall to assume that there will never be any traffic between the source and destination zones. b) Shorewall no longer

...,10)), cc=list( a=matrix(rnorm(100),10,10), b=matrix(rnorm(100),10,10), c=matrix(rnorm(100),10,10)), dd=matrix(rnorm(225),15,15)) str(masterlist) ## Yay! A new list that's just like the master list! newlist = masterlist ## BOO! Can't just set all the values to NA! newlist = masterlist * NA ## This works... but requires knowledge of the structure newlist = masterlist newlist[['aa']] = lapply(masterlist[['aa']], function(x) x * NA) newlist[['bb']] =...

like this, the list is below, I want to remove the last one . not using newlist[-2], but using the function detect its component is numeric(0) and then remove it from the list. newlist [[1]] [1] 2 3 [[2]] [1] numeric(0) [[3]] [1] 7 [[alternative HTML version deleted]]

hello , i''m currently trying to set-up Traffic Shapping with Shorewall and I have strong feelings that I found a bug. I may be mistaken, but I tried everything and can''t get it to work. I''ve turned ON TC_ENABLED=Yes and CLEAR_TC=Yes when i start shorewall ( shorewall start ), i get this message : Setting up Traffic Control Rules... TC Rule "2 eth1 0.0.0.0/0 tcp

...r i in 1:4. It seems that I can create a boxplot of length 48 from the entire list, but I don't seem able to subscript to return 4 boxplots from the list - I have also tried to create 4 new lists (one for each column of each object) by using variations on the following, but none seems to work: newlist<-oldlist[,1] newlist<-oldlist[[]][,1] newlist<-oldlist[[]][,$colone] can anyone please offer some insight?? Thanks in advance, Laura Quinn Institute of Atmospheric Science School of Earth and Environment University of Leeds Leeds LS2 9JT tel: +44 113 343 1596 fax: +44 113 343 6716 mail...

my list al is as below: mylist=list(c(2,3),5,7) > mylist [[1]] [1] 2 3 [[2]] [1] 5 [[3]] [1] 7 How could I get the following FOUR lists: First one [[1]] [1] 3 [[2]] [1] 5 [[3]] [1] 7 Second one [[1]] [1] 2 [[2]] [1] 5 [[3]] [1] 7 Third One [[1]] [1] 2 3 [[2]] [1] 7 Last one [[1]] [1] 2 3 [[2]] [1] 5 Do I have to use 'for' loops? Please give me sone suggestions! Thank you

...llowing and variations thereon to no avail: </code> #for each line of the recovery dataframe #insert the one line of the "book" dataset that corresponds to the MRN AND the course-of-treatment (COURSE) #get the mrn and course from the first line of the recovery dataframe (rec) i=1 newlist=list() colnames(newlist)=colnames(book) for ( i in 1:dim(rec)[1]) { mrn=as.numeric(as.vector(rec$MRN[i])); course=as.character(rec$COURSE[i]); ## find the corresponding row in the book dataframe ## by generating a logical vector and using ## it to access "book" get.vector<-as.ve...

Hi all, I have the following code: set.seed(1) x1 <- matrix(sample(1:12), ncol=3) x2 <- matrix(sample(1:12), ncol=3) x3 <- matrix(sample(1:12), ncol=3) X <- list(x1,x2,x3) tt <- matrix(round(runif(5*4),2), ncol=4) Is there a way I can construct a new list where newlist[[i]] = tt[i,] %*% X[[i]] without using a for loop? Each element of newlist will be 3 x 1 vector. Thanks -- Tina Alexander

I am getting the following message when Shorewall stops can anybody shed any light on this message and where I should be looking? Thanks root@bobshost:~# shorewall stop Loading /usr/share/shorewall/functions... Processing /etc/shorewall/params ... Processing /etc/shorewall/shorewall.conf... Loading Modules... Stopping Shorewall...Processing /etc/shorewall/stop ... IP Forwarding Enabled

I would like to use grep and gsub to manipulate a vector to make the names used consistent, i.e. reduce a level or two. However, here is what I found when I attempted to use grep and gsub: > tmp_test<-c("House 1 Plot Plus +100","House 2 Plot Plus +100","House 3 Plot Plus -100","House 4 Plot Plus -100","House 1 Plus +100","House 2

I am trying to continuously evaluate online created data files using R-algorithms. Is there any simple way to let R iteratively check for new files in a given directory, load them and process them? Any help would be highly appreciated. Best, A. [[alternative HTML version deleted]]

Hi, This question should be simple to answer. I am a new R user. I have a data.frame called appended. I would like to break it into 7 smaller datasets based on the value of a categorical variable dp (which has values 1:7). I would like to name the smaller datasets set1, set2, set3,....,set7. I don't know how to refer to the variable in the for loop, when naming the output datasets. In STATA

...lPointsDataFrameListZoo.R" setMethod("as.yearmon", signature(x = "SpatialPointsDataFrameListZoo"), as.yearmon.SpatialPointsDataFrameListZoo ) # Filename "as.yearmon.SpatialPointsDataFrameListZoo.R" as.yearmon.SpatialPointsDataFrameListZoo=function(x,...) { newlist=mapply(zoo:::as.yearmon,x at list,MoreArgs=list(...),simplify=FALSE) x at list=newlist return(x) } Thoughts? --j -- Jonathan A. Greenberg, PhD Assistant Project Scientist Center for Spatial Technologies and Remote Sensing (CSTARS) Department of Land, Air and Water Resources University of Cali...

Hi all, There are matrices with same column names but arranged in different orders and I desire columns of these matrices to have same order. For example, below are 2 arbitrary data sets with columns arranged in different order. I require columns of these to have same order as specified in "columns" object and the results stored in the original object names. I know this can be done

Dear R users, I wish to utilise processed and saved objects as arguments of a function. Specifically, I have created objects using *"assign"* & *"paste"* functions with an incremental index i, the names of the objects are: fund1, fund2, fund3,....., fund80,..... (where the numerical value increments according to the index i & class of these objects are

I have the following code: list1 <- list() for (i in list.files(pattern="filename1")){ x <- read.table(i) list1[[i]] <- x } list2 <- list() for (i in list.files(pattern="filename2*")){ x <- read.table(i) list2[[i]] <- x } anslist <- vector('list', length(list1)) for(i in 1:length(list1)) if (list1[[i]] & list2[[i]] >1)

...apply this function to all dataframes using lapply() function , I do not get what I want. I wrote this function first :- MyFunction <- function(x) x$NewCol [ x[ ,3] < x[ 1,2] ] <- "C" Then I wrote this code to apply my function to all dataframes in "Mylist" : NewList <- lapply(names(Mylist), function(x) MyFunction(Mylist[[x]])) This returned a list of 7000 elements and each of which contain "C'' letter. Each dataframe has become a vector of "C'' letter which is totally away from what I need. I expected to see a list of my 7000...