search for: inds

Dear R users, I have the "any" function of R  any(ind.c, ind.r, ind.sgn) all are logical factors  it works fine when any of three is true but when they are combined it doesnt work. i tried this any(ind.c, ind.r, ind.sgn,((ind.c==TRUE) & (ind.r==TRUE)), ((ind.c==TRUE) & (ind.sgn==TRUE)),((ind.r==TRUE) & (ind.sgn==TRUE)),  ((ind.c==TRUE) & (ind.r==TRUE) &

Hi All, Just hoping some one can give me a hand with a problem... I have a dataframe (DF) with about 5 million entries that looks something like the following: >DF ID Cl Co Brd Ind A AB AB 1 S-3 IND A BR_F BR_F01 1 0 0 2 S-3 IND A BR_F BR_F01 1 0 0 3 S-3 IND A BR_F BR_F01 1 0 0 4 S-3 IND A BR_F BR_F01 1 0 0 5 S-3 IND A BR_F BR_F01 1 0 0 6 S-3 IND A BR_F

###Dear R users ###I have been using SensoMineR package from CRAN for most of my work in sensory data analysis and from my usage experience, I encountered some areas for improvement and considered ###modifying the function in SensoMineR package for my personal use. I felt that it could be useful to share this to the community for enabling adoption by other users where they might require a

Hi, I'm trying to understand if my journal is fragmented. Here's some output from `debugfs -R "stat <8>" /dev/sda3`: Inode: 8 Type: regular Mode: 0600 Flags: 0x0 Generation: 0 User: 0 Group: 0 Size: 104857600 File ACL: 0 Directory ACL: 0 Links: 1 Blockcount: 205016 Fragment: Address: 0 Number: 0 Size: 0 ctime: 0x3c0442fd -- Tue Nov 27

I' ve seen that several people are looking for a function that creates a barplot with an error indicators (I was one of them myself). Maybe you will find the following code helpful (There are some examples how to use it at the end): # Creates a barplot. #bar.plot() needs a datavector for the height of bars and a error #indicator for the interval #many of the usual R parameters can be set:

Hi, There is any way to figure out where physically is the journal on a ext3 fs and it's size? Thanks! Jordi

I have the following lines of code: ind <- rollapply(GSPC, 200, mean) signal <- ifelse(diff(ind, 5) > 0 , 1 , -1) signal[is.na(signal)] <- 0 I never get a value of -1 for signal even though I know diff(ind , 5) is less than zero frequently. It looks like when diff(ind , 5) is less than zero, signal gets set to 0 instead of - 1. Any ideas why ? Here's some information on ind and

I've not seen an answer to this in any forum. I make a call through Asterisk, with a VOIP phone, doesn't matter which. The call gets made, I leave a voicemail, or complete the call in some manner, and the other side hangs up. I hear a busy signal on the phone on my end. If I have an extension that looks like this, after the hangup() is executed, my phone gives busy signals until I

Dear R users, I am trying to extract the rownames of a data set for which each columns meet a certain criteria. (condition - elements of each column to be equal 1) I have the correct result, however I am seeking for more efficient (desire vectorization) way in implementing such problem as it can get quite messy if there are hundreds of columns. Arbitrary data set and codes are shown below for

...<- dims[2] if(r > k) stop("symarray needs dimension larger than array rank") len <- choose(k, r) out <- data[1:len] attr(out, "dims") <- dims class(out) <- "symarray" out } # Index calculation: indx <- function(inds, k){ r <- length(inds) if(r==1) return(inds) else { if(inds[1]==1) { return( indx(inds[-1]-1, k-1 ) ) } else { return( indx(c(inds[1]-1, seq(from=k-r+2, by=1, to=k)), k) + indx( inds[-1]-inds[1], k-inds[1] )) } } } # end indx...

Hello, I am trying to plot time-series data with certain weeks highlighted using symbols. require(ggplot2) #plotting time series data timescale <- seq(as.Date("01/01/09","%m/%d/%y"), length.out=12, by=7) data.all <- data.frame( id = c(rep('111',12),rep('222',12),rep('333',12)), week=c(timescale,timescale,timescale),

Hi, Here is my data, the first 10 rows > u=regCond_all[1:10,] > dput(u) structure(c(999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 99, 99, 99, 99, 99, 99, 99, 99, 99, 99, 99, 99, 99, 99, 99, 99, 99, 99, 99, 99, 1.9, 2, 1.97, 1.99, 1.83, 1.78, 1.6, 1.52, 1.52, 1.36, 10.53, 9.88, 9.88, 10.53, 10.53, 10.53, 5.26, 9.88, 10.53, 10.53, 5.4, 5.57, 5.46, 5.34, 5.5, 5.59, 5.62, 5.76, 6.23, 6.19,

Dear R community members and R experts I am stuck at a point and I tried with my colleagues and did not get it out. Sorry, I need your help. Here my data (just created to show the example): # generating a dataset just to show how my dataset look like, here I have x variables # x1 .........to X1000 plus ind and y ind <- c(1:100) y <- rnorm(100, 10,2) set.seed(201) P <-

Hi I have the following array: test <- array(c(1:16), dim = c(3,4,3)) test ## I call some enries using an index array test.ind <- array(rbind(c(1,2,1), c(3,3,2)), dim = c(2,3)) test[test.ind] ## suppose I want all values in the 2nd row and 4th col over ## all three 3rd dimensions test[2,4,] how to specify a test.ind array with the last index left with ',' i.e test.ind should be

Consider the following little "benchmark" > require(Matrix) > tmp <- Matrix(c(rep(1,1000),rep(0,9000)),ncol=1) > ind <- sample(1:10000,10000) > system.time(tmp[ind,]) user system elapsed 0.004 0.001 0.005 > ind <- sample(1:1000,10000,replace=TRUE) > system.time(tmp[ind,]) user system elapsed 0.654 0.006 0.703 >

I have a large dataset grouped by a factor and I want to perform a regression on each data subset based on this factor. There are many ways to do this, posted here and elsewhere. I have tried several. However I found one method posted on the R wiki which works exactly as I want, and I like the elegance and simplicity of the solution, but I don't understand how it works. Its all in the formula

Hi all, This is my first post to the developers list. As I understand it, aggregate() currently repeats a function across cells in a dataframe but is only able to handle functions with single value returns. Aggregate() also lacks the ability to retain the names given to the returned value. I've created an agg() function (pasted below) that is apparently backwards compatible (i.e.

Hi everybody, I see a missed optimization opportunity in LLVM that GCC catches and I'd love to hear community's input. Here's the original C code: 1 char arr[2]; 2 char *get(unsigned ind) { 3 if (ind >= 1) { 4 return 0; 5 } 6 return &(arr[ind]); 7 } The variable `ind` is unsigned so, based on the comparison, if it is not greater or equals to one, than it is

Hello,everybody. My name is Chuan Zun Liang. I come from Malaysia. I am just a beginner for R. Kindly to ask favor about median filter. The problem I facing as below: > x<-matrix(sample(1:30,25),5,5) > x [,1] [,2] [,3] [,4] [,5] [1,] 7 8 30 29 13 [2,] 4 6 12 5 9 [3,] 25 3 22 14 24 [4,] 2 15 26 23 19 [5,] 28 18 10 11 20 This

Hi, I am new in R and stumbled on a problem my (more experienced) friends can not help with with. Why isnt this code working? The function is working, also with the loop and the graph appears, only when I build another loop around it (for different values of p) , R stays in a loop? Can't it take more then 2 loops in one program? powerb<-function(x,sp2,a,b,b1,m) {