similar to: Concecutive zeros and ones

Dear Group, I have a following dataset: > a A B C D 1 22 3 31 40 2 26 31 36 32 3 3 7 49 16 4 24 40 27 26 5 20 45 47 0 6 34 43 11 18 7 48 48 24 2 8 3 16 39 48 9 20 49 7 21 10 17 36 47 10 > dput(a) structure(list(A = c(22L, 26L, 3L, 24L, 20L, 34L, 48L, 3L, 20L, 17L), B = c(3L, 31L, 7L, 40L, 45L, 43L, 48L, 16L, 49L, 36L), C = c(31L, 36L, 49L, 27L, 47L, 11L, 24L,

Hello. I want to create some sequence of numbers . So far I used sequence which does not work always seq(CRagent[[1]]$xy[1],CRagent[[2]]$xy[1],by=0.01) Error in seq.default(CRagent[[1]]$xy[1], CRagent[[2]]$xy[1], by = 0.01) : wrong sign in 'by' argument Calls: seq -> seq.default if the parameters are in descending form. The ideal would be to be able to use seq like this

Dear all, I would like to place into a same plot two plots. Their y values though are different a lot (first series has around -100 y values and the second one slightly over 1). The x value are the same. I would like to have in the same plot (under the same x,y axis the two plots). Is it possible then to have two y labels, one for example in the left side and an extra y label at the right. Each y

Dear R helpers Suppose x  <- c(1:3) y  <- matrix(1:12, ncol = 3, nrow = 4) > y      [,1] [,2] [,3] [1,]    1    5    9 [2,]    2    6   10 [3,]    3    7   11 [4,]    4    8   12 I wish to multiply 1st column of y by first element of x i.e. 1, 2nd column of y by 2nd element of x i.e. 2 an so on. Thus the resultant matrix should be like > z      [,1]   [,2]    [,3] [1,]    1   

Dear R list, I have one very elementary question regrading correlation between two variables. x = c(44,46,46,47,45,43,45,44) y = c(44,43,41,41,46,48,44,43) > cov(x, y) [1] -2.428571 However, if I try to calculate the covariance using the formula as covariance = sum((x-mean(x))*(y-mean(y)))/8       # no of of paired obs. = 8 or     covariance = sum(x*y)/8-(mean(x)*mean(y)) gives

Hi, I have this lame question. I want to convert a list (each with varies in length) to matrix with same row length by eliminating vectors outside the needed range. For example: l<-list(NULL) l[[1]]=1,2,3.7 l[[2]]=3,4,5,6,3 l[[3]]=4,2,5,7 l[[4]]=2,4,6,3,2 l[[5]]=3,5,7,2 #so say I want to only have 4 rows and 5 column in my matrix (or data.frame) and eliminating the 5th index value in l[[2]]

Dear All, I would appreciate any help with the following: given the vector 'x' x <- c("Ass1", "Ass.s1", "Ass2", "Ass.s2") I would like to pick up the positions where the character string contains "Ass" but does not contain "Ass.s", so for 'x' that would be positions 1 and 3. I guess this could be programmed around

Dear all, I would like to task you if you know a rle version that can work also in a non consecutive way too. B.R Alex [[alternative HTML version deleted]]

Hello List, I have a data frame like: V130 V131 V132 V133 V134 V135 V136 1 0 0 0.9 0 0.9 0 0 2 0 0 0 0 0 0.8 3 0 0 0 0 0.9 0 0 4 0.9 0 0 0 0 0 0.9 5 0 0 0 6 0 0 0 0.9 0 0 0.9 7 0 0 0.8 0 0 0 0 8 0.9 0 0 0.9 0.8 0 9 0 0 0 0.9 0.9 0 0 10 0 0 0 0 0 0 0.9

Dear all I have data like this: x y [1,] 59.74889 3.1317081 [2,] 38.77629 1.7102589 [3,] NA 2.2312962 [4,] 32.35268 1.3889621 [5,] 74.01394 1.5361227 [6,] 34.82584 1.1665412 [7,] 42.72262 2.7870875 [8,] 70.54999 3.3917257 [9,] 59.37573 2.6763249 [10,] 68.87422 1.9697770 [11,] 19.00898 2.0584415 [12,] 60.27915 2.5365194 [13,] 50.76850

Hi, Not sure how to go about checking for missing report pages per client. See example below; I like to extract client 730040, because page 46103 is missing. client page 730001 46101 730001 46102 730019 46101 730035 46101 730040 46101 730040 46102 730040 46104 730040 46105 730052 46101 730052 46102 730074 46101 730074 46102 730074 46103 I appreciate any help, Pauline

A list contains several matrices. Over all matrices (list elements) I'd like to access one matrix cell: m <- matrix(1:9, nrow=3, dimnames=list(LETTERS[1:3], letters[1:3])) l <- list(m1=m, m2=m*2, m3=m*3) l[[3]] # works l[[3]][1:2, ] # works l[[1:3]][1, 1] # does not work How can I slice all C-c combinations in the list? S?ren -- S?ren Vogel, Dipl.-Psych. (Univ.), PhD-Student, Eawag,

Hello R-help, Please cc me on all responses, as I only receive summary emails from this list. I'm wondering if anybody has any tips on how to accomplish this efficiently. I have a list of matrices, and I'm trying to get the mean of the [i,j]'th element of each matrix in a list. So if I have a list of matrices, say x <-

Hi everyone. I would like to modify the axis on my plot. First, I would like to make a plot without the box. so I use : plot(x,y, axes = FALSE, type = 'l') Then, I call : axis(1, tck = -0.02) axis(2, tck = -0.02) to have X and Y axis appear. However, I would like them to join at the origin instead of having a space between the 2 axis. I can't find the parameter to modify to get

Dear list, I would like to expand a DF with all the missing levels of a variable. a <- c(2,2,3,4,5,6,7,8,9) a.cut <- cut(a, breaks=c(0,2,6,9,12), right=FALSE ) (x <- data.frame(a, a.cut)) # In 'x' the level "[0,2)" is "missing". AddMissingLevel <- function(xdf) { xfac <- factor( c("[0,2)", "[2,6)", "[6,9)",

Hi All, What is the fastest way of finding if any members of vector x fall in the range of the rows of matrix y? I do not want to use two for loops as this will take forever. Any help will be appreciated, best, salih [[alternative HTML version deleted]]

Hello everyone, I need to fit a bivariate normal regression model to a dataset where the same covariate (say, X) influences two separate but correlated responses (say, Y1 and Y2). So, the bivariate model would look like : Y1 = a1 + b1*X + e1 Y2 = a2 + b2*X + e2 where e1 and e2 are error terms which can be correlated. Is there any package in R which can help me fit this model ? Any help will be

dear R community, i appologize as this is kind of a newbie question, but: I want to assign group letters (like in anova post-hoc tests) to data with combined means and SDs. It's easy to apply t-test formulas to get a materix like (A - D are treatments | A B C D ----------------------- A| T F F B| T F C| F D| where true stands for sign.

Hi Guys, Does anyone know how I could constrain my regression coefficients so that they are positive and add up to one? Any help will be greatly appreciated. Kind Regards, Andre [[alternative HTML version deleted]]

hello, i am familiar with the paste command with which i can paste for exaple: object <- "Hello" paste(object,"World") now i would like to be able to paste all the elements of the same vector together e.g: object <- c("Hello","World") getting as a result also: "Hello World". Does anyone know the solution to this problem? Thank you