similar to: Write columns from within a list to a matrix?

Silly question, but, can I test to see if any value of list a is contained in list b without doing a loop? A loop is easy enough, but wanted to see if there was a cleaner way. By way of example: List 1: a, b, c, d, e, f, g List 2: z, y, x, w, v, u, b Return true, since both lists contain b List 1: a, b, c, d, e, f, g List 2: z, y, x, w, v, u, t Return false, since the lists have no mutual

Dear R users, I have a matrix composed of lists: m <- matrix( list(), nrow=1, ncol=3 ) m[[ 1, 1 ]] <- list("A", "B") m[[ 1, 2 ]] <- list("A", "C") m[[ 1, 3 ]] <- list("A", "B") and want to get the sub-matrix where cells contain "B". But m[ , "B" %in% m[ 1, ], drop=F ] as well as m[ , "B"

Dear all, I would like to know how to merge columns like: Input file: V1 V2 V3 V4 V5 V6 1 G A G G G G 2 A A G A A G Desired output file: V1 V2 V3 1 G/A G/G G/G 2 A/A G/A A/G So for every 2 consecutive columns merge their content into one. Thanks in advance. [[alternative HTML version deleted]]

Hello. Please forgive me if this problem has already been posted (and solved) by someone else ... I can't find it anywhere though it seems so very basic. Here it is: I have a list comprised of several matrices, each of which has two columns. > list [[1]] [,1] [,2] [1,] 1 3 [2,] 2 4 [[2]] [,1] [,2] [1,] 5 7 [2,] 6 8 [[3]] [,1] [,2]

Hello- Sorry to ask a basic question, but I've spent many hours on this now and seem to be missing something. I have a loop that looks like this: mainmat=data.frame(matrix(data=0, ncol=92, nrow=length(predata$Words_MH))) for(i in 1:length(predata$Words_MH)){ for(j in 1:92){ mainmat[i,j]=ifelse(j %in% as.numeric(unlist(strsplit(predata$Words_MH[i], split=","))),

Hi, I want to apply a function to a matrix, taking the columns 3 by 3. I could use a for loop: for(i in 1:3){ # here I assume my data matrix has 9 columns j = i*3 set = my.data[,c(j-2,j-1,j)] my.function(set) } which looks cumbersome and possibly slow. I was hoping there is some function in the apply()/lapply() families that could take 3 columns at a time. I though of turning mydata in a

Consider a matrix like > ma = matrix(10:15, nr = 3) > ma [,1] [,2] [1,] 10 13 [2,] 11 14 [3,] 12 15 I want to rearrange each column according to row indexes (1 to 3) given in another matrix, as in > idx = matrix(c(1,3,2, 2,3,1), nr = 3) > idx [,1] [,2] [1,] 1 2 [2,] 3 3 [3,] 2 1 The new matrix mb will have for each column the

Dear all, #Last week, I asked about merge x and y as list. #Now I have a dataset with list of list like: x <- list(list(matrix(1:20, 5, 4),matrix(1:20, 5, 4)), list(matrix(1:20, 5, 4),matrix(1:20, 5, 4))) y <- list(list(c(1, -1, -1, 1, 1),c(1, 1, -1, -1, -1)), list(c(1, 1, 1, 1, 1),c(1, 1, -1, 1, -1))) x y #I need merge x and y, I have tried with list.uni <-

How would one go about putting titles in each of several plots that are generated from within a call to tapply? For example I'd like the following two barplots to have titles 'Group 1' and 'Group 2', where '1' and '2' come from the levels of 'group'. group <- gl(2, 10) result <- sample(c('A', 'B'), size=length(group),

Dear R-users I am trying to "vectorize" a function so that it can handle two vectors of inputs. I want the function to use phi (a function), k[1] and t[1] for the first price, and so on for the second, third and fourth price. I tried to do the mapply, but I dont know how to specify to R what input I want to be vectors (k and t)(see in the bottom what I tried). I have read the help file,

Hi : I think I need to use sapply but I can't figure this out. Suppose I have two vectors : tempa ( 4, 6,10 ) and tempb ( 11,23 ,39 ) I want a function that returns 4:11,6:23 and 10:39 as vectors. I tried : sapply(1:length(tempa) function (z) seq(tempa[z],tempb[z]) but i got 3 really strange vectors back in the sense that the numbers in them did not make no sense to me. obviously, i

I have a data frame whose first colum contains the names of the variables and whose second colum contains the values to assign to them: : kkk <- data.frame(vars=c("var1", "var2", "var3"), vals=c(10, 20, 30), stringsAsFactors=F) If I do : assign(kkk$vars[1], kkk$vals[1]) it works : var1 [1] 10 However, if I try with mapply

Gracias, Carlos. Habia pensado en algo similar usando sapply(): sapply(seq(1, ncol(vtmp), by = 2), function(i) c(rbind(as.character(vtmp[, i]), as.character(vtmp[, i+1])))) Dependiendo de la dimension de los datos, quizas mapply() sea mas eficiente que sapply(). Saludos cordiales, Jorge.- 2015-02-25 1:01 GMT+11:00 Carlos Ortega <cof en qualityexcellence.es>: > Hola, > > Este

Hi group! Suppose I have 2 matrices A and B of equal dimensions. I want to apply a function f to all corresponding pairs of rows from A and B in an efficient manner. Basically, I want mapply(f, data.frame(A), data.frame(B)) but for rows. How do I do it? Thanks, Andrey

Hi R-helpers, I have a dataframe with 60columns and I would like to convert several columns to factor, others to numeric, and yet others to dates. Rather than having 60 lines like this: data$Var1<-as.factor(data$Var1) I wonder if it's possible to write one line of code (per data type, e.g. factor) that would apply a function (e.g., as.factor) to several (non-contiguous) columns. So, I

Dear All,   as a follow up to my previous e-mail (I think I am getting closer...):   I am trying to apply the trapezoidal functions to a matric column by column. I have the following code:   a <-matrix(c(1:100),ncol=10) b <-matrix(c(2,4,6,8,10,12,14,16,18,20))   apply(a,2,function(b,a) sum(diff(b)*(a[-1]+a[-length(a)]))/2)   for some reason i get an error message: Error in FUN(newX[[, i],

Hello all I would like to ask your help use mapply. I have a function called findCell that takes two arguments(x,sr) where x is a vector of size two (e.g x<-c(2,3) and sr is a matrix. I would like to call many times the findCell function (thus I need mapply) for different x inputs but always for the same sr. as x is a vector of size two (two cells) I want to pass inside inside the following

To use the gauss.quad function: gauss.quad(n,type) which returns two lists $nodes and $weights whose length will each equal n. I'd like to do this for n=1 to 40 (type will not change) and have a dataset with 40 rows and 81 columns with all the nodes and weights. The first record would have N1 and W1 only and N2--N40 and W2--W40 would be missing. The last record would be full. I've

R pattern-matching and replacement functions are vectorized: they can operate on vectors of targets. However, they can only use one pattern and replacement. Here is code to apply a different pattern and replacement for every target. My question: can it be done better? sub2 <- function(pattern, replacement, x) { len <- length(x) if (length(pattern) == 1) pattern <-

Hi, I have 2 questions on list object:   1. Suppose I have a matrix like: dat <- matrix(1:9,3)   Now I want to replicate this entire matrix 3 times and put entire result in a list object. Means, if "res" is the resulting list then I should have:   res[[1]]=dat, res[[2]]=dat, res[[3]]=dat   How can I do that in the easilest manner?   2. Suppose I have 2 list objects: list1 <- list2