similar to: number of zeros in a matrix -row by row

Hi, Can someone tell me how to remove rows of zeros from a matrix? For example if I have the following matrix, 0 0 0 1 2 8 0 0 4 56 I should end up with 0 1 2 8 4 56 -- Thanks, Jim. [[alternative HTML version deleted]]

Hi. I have a 925 by 925 correlation matrix corM. I want to identify all variables that have correlation greater than 0.9. Can anyone suggest an "R way" of doing this? Thank you. -- View this message in context: http://www.nabble.com/search-through-a-matrix-tp23153538p23153538.html Sent from the R help mailing list archive at Nabble.com.

I have a matrix 700x2000 which is sampled in each cycle from another matrix 788x2000 with the numbers 0,1 and 9 There is one specific collumn of this matrix, dart[,1977], that usually, after the samplimg procedure has only 1 and 9 (because the zero frequency in this collumn is low). However, when this happens, I want to include an IF conditional in my code. so basically what i wanted to do was to

Hi Everybody, I have a matrix of about 45 columns. Some of the rows contain zeros. Using >data1<-data[complete.cases(data),], I can remove the "NA" rows. But I am unable to tackle that of zeros. Can anybody give me an idea of how to remove rows containing zeros in a matrix. Thanks so much Best Ogbos [[alternative HTML version deleted]]

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Hi All, If you could help me with this problem I would greatly appreciate it. Suppose I have a matrix A: 1 1 1 1 0 1 1 1 1 1 1 1 0 1 0 1 0 0 1 0 1 0 0 1 0 0 0 1 1 0 0 0 0 1 0 0 I would like, for each row, to sum the number of times a 0 appears in front of a 1. So what I would like is to have Sum 1 1 1 1 0 1 1 1 1 1 1 1 1 0 1 0 1 0 0 2 1 0 1 0 0 1 0 0 0 2 1 1 0 0 0 0 1

Dear all, Can anybody tell me how to export a 3d figure made with the plot3d function? I'm careless about whether it's still interactive or not in another format, as long I can get it out of R. Thanks! Alejandro Gonz?lez Departamento de Biodiversidad y Conservaci?n Real Jard?n Bot?nico Consejo Superior de Investigaciones Cient?ficas Claudio Moyano, 1 28014 Madrid, Spain Tel +0034

Hi All. Imagine you have a large block diagonal matrix. I'd like to replace the zeros in this matrix with small random (runif) numbers. Any ideas for a simple and efficient way to do this? Best regards, Rick DeShon

Hi R-users, How do I calculate a number of NA's in a row of every second column in my data frame? As a starting point: dfr <- data.frame(sapply(x, function(x) sample(0:x, 6, replace = TRUE))) dfr[dfr==0] <- NA So, I would like to count the number of NA in row one, two, three etc. of columns X1, X3, X5 etc. Thanks in advance Lauri [[alternative HTML version deleted]]

I have a big contingency table, approximately of size 60*2*500*500, and I need to count the number of cells containing a count of 1 for each of the factors values defining the first dimension. Here is my attempt: tab1<-with(pisa1,table(CNT,GENDER,ISCOF,ISCOM)) tab2<-apply(tab1,1:4,function(x)ifelse(sum(x)==1,1,0)) tab3<-apply(tab2,1,sum) Computing tab2 is very slow. Is there a faster

Hi. I have an object. I think it is a list. > str(corTFandPCA) num [1:922, 1:5] -0.0226 -0.0504 -0.0208 -0.0582 -0.0257 ... - attr(*, "dimnames")=List of 2 ..$ : chr [1:922] "abdomen.2" "abdomimal.3" "abdominal.4" "aberration.5" ... ..$ : chr [1:5] "PC1" "PC2" "PC3" "PC4" ... I want to order it

Dear expeRts, I have two matrices A and B. They have the same number of columns but possibly different number of rows. I would like to compare each row of A with each row of B and check whether all entries in a row of A are less than or equal to all entries in a row of B. Here is a minimal working example: A <- rbind(matrix(1:4, ncol=2, byrow=TRUE), c(6, 2)) # (3, 2) matrix B <-

I'm trying to: Resample with replacement pairs of distinct rows from a 120 x 65,000 matrix H of 0's and 1's. For each resampled pair sum the resulting 2 x 65,000 matrix by column: 0 1 0 1 ... + 0 0 1 1 ... _______ = 0 1 1 2 ... For each column accumulate the number of 0's, 1's and 2's over the resamples to obtain a 3 x 65,000 matrix G. For those

Hi all, I have a huge matrix of TRUE/FALSE table like following, and I want to count the number of TRUEs in each row. Instead of looping through each row and do length(Z[Z==TRUE]), I am wondering if there is an easier way of doing this. [,1] [,2] [,3] [1,]TRUE FALSE FALSE [2,]FALSE TRUE TRUE Thank you in advance. Wendy -- View this message in context:

Hi all, I am trying to multiply each column of a matrix such to have a unique resulting vector with length equal to the number of rows of the original matrix. In short I would like to do what prod(.) function in Matlab does, i.e. A <-matrix(c(1:10),5,2) V = A[,1]*A[,2] Thank you Graziano [[alternative HTML version deleted]]

Hi everybody, Maybe this question is quite simple but i just don't know how to make it. I have a matrix a somewhat like this one but bigger: > a f g h i j k a NA NA 11 16 21 26 b NA NA 12 17 22 27 c NA 8 13 18 23 28 d NA 9 14 19 24 29 e NA 10 15 20 25 30 And i want to get the rows which at most have 2 Na. Thanks in advance. Shi Jiantao [[alternative HTML version deleted]]

Error in matrix(unlist(value, recursive = FALSE, use.names = FALSE), nrow = nr, : invalid 'ncol' value (too large or NA) the statement was: for (i in v) {for (j in m[0,]){if(v[i]==m[0,j]){M[,-j]}}} where 'v' is a vector, and 'm' and 'M' are both matrices. -- View this message in context:

Hi. Does anyone know of a function which will take as input a number n (or a set of n letters) and will give out, one at a time, the permutations of n (or of those n letters) as a vector? So that I can use the permutations one at a time. And such that it will exhaust all the permutations with no repeats. For example if n is 3, I would want a function which I could use in a loop and the first time

I am just starting with R and have the following problem: given a matrix of ones and zeroes, say mdim=4 m<-matrix(round(runif(mdim^mdim)),mdim,mdim) how to construct the matrix of those rows of m, whose elements sum to 2. Contrary to the random matrix above, the actual matrix always has at least one such row. Serguei Kaniovski

Hey, Thanks for your great inputs. While "index = apply(mat == 0, MARGIN = 1, any)" gives you an idea of the rows containing zero(s), "index<-data[!apply(data==0,MARGIN=1,any),]" does the actual job of removing the rows with zeros. Kind regards Ogbos On 9 March 2010 15:12, Paul Hiemstra <p.hiemstra@geo.uu.nl> wrote: > Dimitris Rizopoulos wrote: > >>