search for: lst

I'm trying to run a function inside a function but get an error message. lst <- list(roots = c("car insurance", "auto insurance"), roots2 = c("insurance"), prefix = c("cheap", "budget"), prefix2 = c("low cost"), suffix = c("quote", "quotes"), suffix2 = c("rate", "rates"),...

...) (note that all parameters in call refine1 have been defined previosly) Then the following steps look like this ##################################### # Refinement process # ##################################### refine1 <- function(M, A, B, p_A, p_B, FAIL, elim, i, j, k, sc, h, lst, x){ #print("refine 1") # elim marks if there was eliminated a 1 (and changed to 0) lst <- vector(mode = "numeric") elim <- 0 i <- 1 refine2(M, A, B, p_A, p_B, FAIL, elim, i, j, k, sc, h, lst, x) } refine2 <- function(M, A...

R-help, I am trying to create a function that i pass a data set to and have the function return some calculations based on data. Allow me to illustrate: myfunc <- function(lst,mn,sd){ lst <- sort(lst) mn <- mean(lst) sd <- sqrt(var(lst)) return(lst,mn,sd) } data1 <-c (1,2,3,4,5) data2 <- c(6,7,8,9,10) myfunc(data1,data1mn,data1sd) myfunc(data2,data2mn,data2sd) This snippet errors that data1mn not find and warns that return is deprecating!!!...

I'd like to pass a list object created by one function as an argument of another function. once inside the second function, I'd like to break the list up to it's individual elements, each then identifiable by the 'names' of the list. The list looks something like lst<-list(a=1, b=2, df=5, g=7) then inside the function I've been writing a sequence of statements that extract the objects within lst like do_something<-function(L){ a<-lst$a b<-lst$b df<-lst$df g<-lst$g a+b+df+g } do_something(lst) My question is, is it possible to avoid the...

Hello, it seems an mismatch of schema file provided by dovecot-2.3.5.2 [root at mail conf]# pwd /var/solr/data/dovecot/conf # Below solrconfig.xml is from Solr-8.0.0. [root at mail conf]# grep text_general solrconfig.xml <str name="queryAnalyzerFieldType">text_general</str> field types. Text content will be indexed as "text_general" as <str

...ng 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, collapse="+") } lst <- split(rep(LETTERS, 100), rep(1:1300, 2)) library(microbenchmark) microbenchmark(fun0(lst), times=10) ## Unit: milliseconds ## expr min lq median uq max neval ## fun0(lst) 4.153287 4.180659 4.513539 5.19261 5.280481 10 setGeneric("paste...

...;useColdSearcher>false</useColdSearcher> > </query> > <requestDispatcher> > <httpCaching never304="true"/> > </requestDispatcher> > <requestHandler name="/select" class="solr.SearchHandler"> > <lst name="defaults"> > <str name="echoParams">explicit</str> > <int name="rows">10</int> > </lst> > </requestHandler> > <requestHandler name="/query" class="solr.SearchHandler"...

Dear R-experts, Here below the reproducible example. I am trying to get the average of the 100 results coming from the "lst" function. I have tried lst$mean and mean(lst). It does not work. Any help would be highly appreciated. #################### ?## R script for getting MedAe and MedAeSQ from HBR model on Testing data install.packages("robustbase") install.packages( "MASS" ) install.packag...

I'm writing a function and keep getting the following error message. myfunc <- function(lst) { lst <- list(roots = c("car insurance", "auto insurance"), roots2 = c("insurance"), prefix = c("cheap", "budget"), prefix2 = c("low cost"), suffix = c("quote", "quotes"), suffix2 = c("rate", "rates&q...

...</listener> <useColdSearcher>false</useColdSearcher> </query> <requestDispatcher> <httpCaching never304="true"/> </requestDispatcher> <requestHandler name="/select" class="solr.SearchHandler"> <lst name="defaults"> <str name="echoParams">explicit</str> <int name="rows">10</int> </lst> </requestHandler> <requestHandler name="/query" class="solr.SearchHandler"> <lst name=&qu...

...ntel.com > diff -r 61ec78692b13 tools/python/xen/util/pci.py --- a/tools/python/xen/util/pci.py Wed Jun 17 07:39:27 2009 +0100 +++ b/tools/python/xen/util/pci.py Wed Jun 17 18:35:31 2009 +0800 @@ -547,12 +547,12 @@ class PciDevice: else: dev = {} lst = parent.split(':') - dev['domain'] = int(lst[0], 16) - dev['bus'] = int(lst[1], 16) + dev['domain'] = '%04x' % int(lst[0], 16) + dev['bus'] = '%02x' % int(lst[1], 16)...

hello, I have programmed this function to calculate the Neuman-Keuls test but I have a problem the function return an empty list and I don't know why. summary(fm1) E <- sqrt((summary(fm1)[[1]]["Residuals","Mean Sq"])/length(LR)) lst <- list() lst1 <- list() lst2 <- list() NK <- function (x) { if (length(x) == 2) { Tstudent <- t.test(subset(exple, groupe == names(x)[1])$vd,subset(exple, groupe == names(x)[2])$vd) t <- as.numeric(Tstudent$statistic) if (t >= Tstudent$conf.int[1:2][1] & t <= Ts...

I'm studying lists and I came to an example where > L $name [1] "Fred" $wife [1] "Mary" $no.children [1] 4 $child.ages [1] 4 7 9 then following the instructions to extend the list with a new component, I executed: > L[5] <-list(NewName="something") and the new list I got was: > L $name [1] "Fred" $wife [1] "Mary"

...<- new( "antsMatrix", "float" ) b <- as.data.frame( a ) this all works fine and i am pleased. fyi, the implementation of as.data.frame is: setMethod( f = "as.data.frame" , signature( x = "antsMatrix" ) , definition = function( x ) { lst = .Call( "antsMatrix_asList" , x ) names(lst)[ 1 : (length(lst)) ] <- lst[[ length(lst) ]] lst[[ length(lst) ]] <- NULL as.data.frame(lst) } ) now the problem comes when i try to access the same functionality in a function that is within my package R so...

Hi,   it seems a simple problem, but I can not find a clear way. I have a list: lst=list(m=c('a','b','c'),n=c('c','a'),l=c('a','bc')) > lst $m [1] "a" "b" "c" $n [1] "c" "a" $l [1] "a"  "bc" how can I get list elements that include a given subset? for e...

Sorry to ask such a simple question, but I can't find the answer after extensive searching the docs and the web. How do you remove a component from a list? For example say you have: lst<-c(5,6,7,8,9) How do you remove, for example, the third component in the list? lst[[3]]]<-NULL generates an error: "Error: more elements supplied than there are to replace" Also, how do you remove a row from a data frame? For example, say you have: lst1<-c(1,2,3,4,5...

Hallo, I build a list by the following way: Lst = list(name="Fred", wife="Mary", no.children=3, cild.ages=c(4,7,9)) I know how I can extract the information one by one. But now I want to add a new entry which looks like name="Barney", wife="Liz", no.children=2, cild.ages=c(3,5) How can I add this inform...

On 5/8/2018 12:26 PM, varin sacha via R-help wrote: > > Dear R-experts, > > Here below the reproducible example. I am trying to get the average of the 100 results coming from the "lst" function. I have tried lst$mean and mean(lst). It does not work. > Any help would be highly appreciated > > #################### > > ?## R script for getting MedAe and MedAeSQ from HBR model on Testing data > install.packages("robustbase") > install.packages(...

...ction to split the ip address as character into class A,B,C and D components. It works but is horribly inefficient in terms of speed. I can't quite see how one of the l/s/m/t/apply functions could be brought to bear on the problem. Does anyone have any thoughts? for(i in 1:4861469) { lst <-unlist(strsplit(data$ComputerName[i], "\\.")) data$IPA[i] <-lst[[1]] data$IPB[i] <-lst[[2]] data$IPC[i] <-lst[[3]] data$IPD[i] <-lst[[4]] rm(lst) } Andrew Andrew Roberts Children's Orthopaedic Surgeon RJAH, Oswestry, UK [[alternative HTML version...

Dear R-gurus, I would like to use a lapply on a kind of "bivariate" problem. I have a vector and a list, components of which are vectors, e.g. vec <- c(1,2,3) lst <- list(1, c(2,3), c(4,5,6)) I want to apply a function to each component of the list, using the corresponding component of the vector as a parameter. E.g. I want a list in the form list(lst[[1]] + vec[1], lst[[2]] + vec[2], .... ) I think this can be achieved with a cycle and probably using...