similar to: Two plots with two different Y labels

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

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 all, I have  a data series (might be vector or matrix) which is composed only from zeros and ones like the following example 0 0 0 1 1 0 1 0 0 1 1 1 1 0 0 0 I want to be able to return back the length of concecutive zeros and the length of concecutive ones. For that I want to have something like that returned: zeros= [3 1 2 3]; ones=[2 1 4]; How I can do that simply in R? I would like

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,

Hi, This might seem like a simple question but at the moment I am stuck for ideas. The columns of my matrix in which some data is stored are of this form: X1 Y1 X2 Y2 X3 Y3 ... Xn Yn with n~100. I would like to look at just the X values (i.e. odd column numbers). Is there an easy way to loop round extracting only these columns? Any help would be appreciated.

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 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

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]]

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 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 is there an alternative to par(new), for ading data to a plot for a different y-axis? My problem with par(new=TRUE) is, that it re-defines all axis and labels (as in example 1) and one has to use xlim=... to fix the x-axis. I am looking for something, which simply resets the y-axis, so that a new plot() (or points()/lines()) keeps the x-axis, but re-defines the y-axis. Is there something

Hi everyone After the following setup sector=2 # Define Number of Sectors sectors=LETTERS[seq( from = 1, to = sector )] # Name sectors No_ent=round(3/runif(sector)) # Number of entities per sector #Tot_No_ent=sum(No_ent) Goal is to get a List like (A1, A2, A3, B1, B2, B3, B4) where A is denoting an industrial sector and then a numbered sequence of companies within the sector. The step

Dear all I have a code like x<-1:10 y1<-x+runif(10)*2 y2<-seq(0,50,length.out=10)+rnorm(10)*10 par(mfrow=c(1,2)) plot(y1~x) plot(y2~x) Now I would like to plot y1 and y2 on the same graph, with its two scales (y1 on left and y2 on rigth side). Any help are welcome. Kind regards Miltinho Brazil [[alternative HTML version deleted]]

Hi dear all, the following code is correct. but I want to use non-numeric x-axis, for example if I replace time <- seq(0,72,6) by month <- c("Jan","Feb","Mar","Apr","May","Jun","Jul","Aug","Sep","Oct","Nov","Dec","Pag") Ofcourse I use factor(month) instead of

Hello, I am stuck with the following problem. Consider the vector: vec <- c(2,4,6,9,10) I now want to use R to manipulate the vector as follows: [1] 2, 4, 2, 6, 2, 9, 2, 10 In words, the first element of the vector should be placed in front of each following number. Which R commands do I need to achieve that? Cheers -- View this message in context:

Hello Dimitris, I was goggling for some help on Sensitivity vs 1-specificity and saw your link. I hope you can be of help to me in one of the issue that I am facing in generating combo chart(bar chart and plot). I am a novice and have some difficulty in getting this logic correct. I am give a dataset (I am attaching a sample dataset). I am using a barplot() and passing values for

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 <-