similar to: generating RANDOM ROWS from matrix

I am trying to find out if R can recognize specific criteria for removing rows (i.e. a prexisting function) I have a matrix myMatrix that is 12000 by 20 I would like to remove rows from myMatrix that have: -999 across all columns -999 across all columns but one -999 across all columns but two -999 across all columns but three -999 across all columns but four -999 across all columns but five

I have two objects, one matrix and one vector. I want to use my vector to subset certain values out of my matrix. For example: I want to tell R, to select out all rows in myMatrix into a new myMatrix2 IF that corresponding row is less than a 0.5 in myVector. So: myMatrix = a matrix of 8000 by 20 myVector = vector of 8000 myMatrix2 = a matrix of < 8000 by 20 (based on selection criteria in

I have a data frame with the following dimensions 217 x 5 I want to create two data frames from the original. 1) One containing every tenth row of the original data frame 2) Other containing the rest of the rows. How do I do this? I've tried subset() and calling the index. thank you in advance, [[alternative HTML version deleted]]

How do I manipulate the read.table function to read in only the 2nd column??? [[alternative HTML version deleted]]

dear r-users, i have got a problem which i am trying to solve: i have got the following commands: Mymatrix <- matrix(1:9,ncol=3) Z <- list("V1"=c("a","",""),"V2"=c("b","",""),"V3"=c("c","",""),"V4"=c("d","","")) Mymatrix <-

Hi, I have a dataframe (say myData) and want to get a list (say myList) that contains a matrix for each row of the dataframe myData. These matrices are calculated based on the corresponding row of myData. Using a for()-loop to do this is very slow. Thus, I tried to use apply(). However, afaik apply() does only return a list if the matrices have different dimensions, while my matrices have

Dear All, Thanks in advance for any help. I have a square matrix of measures of interactions among individuals and would like to calculate a values from a function (colSums for example) with a single individual (row) excluded in each instance. That individual would be returned to the matrix before the next is removed and the function recalculated. I can do this by hand removing rows based upon

Hi Martin, Fair enough for R functions in general. But the behaviour of apply violates the expectation that apply(m, 1, fun) calls fun n times when m has n rows. That seems pretty basic. Also, I understand from your argument why it makes sense to call apply and return a special result (presumably NULL) for an empty argument; but why should apply call fun? Cheers David On Mon, 30 Jul 2018 at

hi all, i'm trying to get elements in a matrix into a vector. i need a "streamlined" way to do it as the way i'm doing it is not very serviceable. an example is a 3x3 matrix like 0 0 3 2 0 0 0 4 0 to a vector like 3 2 4 thanks...mj [[alternative HTML version deleted]]

Hi, Here i have a Matrix MyMatrix <- Name Age --------- ------- ANTONY 27 IMRAN 30 RAJ 22 NAHAS 32 GEO 42 and here i have an array with Minimum and Maximum values. MinMaxArray <- data.frame(MIN = 25,MAX=35) MIN MAX ------ -------- 25 35

Hello Spse I have a matrix, say 1 2 3 4 5 6 7 8 9 and I would like to expand it by repeating rows within the matrix, to get, if the repeating factor is 2, say: 123 123 456 456 789 789 (or columnwise as well) . There must be a smart way of doing that? Many thanks Juhana Vartiainen juhana.vartiainen at labour.fi -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-

Hi there, I have a question, which I thought is very easy to solve, but somehow I can't find a solution. Probably someone could help me quickly? Here it is: I have two matrices: a [,1] [,2] [,3] [1,] 1 4 9 [2,] 2 6 10 [3,] 3 6 11 [4,] 4 8 12 b [,1] [,2] [1,] 1 4 [2,] 2 5 [3,] 3 6 Now I want to find

Hello, I am running a program in r that calls a function, which calls another function, which calls another etc. These functions are of the form: example<- function(x,y,z) {x, y, and z are defined within curly braces like this} Here's my question. To start the main function, I input as an initial parameter a matrix of regressors of the form: MyMatrix<-cbind(this.one,that.one)

Dear list, I would like to subset a large expression matrix based on rownames. That is, I have a list (as a txt-file) with gene names that matches some of the rows in my matrix. I've loaded my matrix as well as gene list using the read.table() command. myMatrix <- read.table("name_of_file.txt", header=T, row.names=1) list_to_keep <- read.table("name_of_file.txt",

Quick question: how can I reverse the order of the rows in a matrix? i.e. make the last row first and the first row last, etc.?

Hello, I have a double matrix that I want to represent in a line chart. Although I have seen some examples I still don't manage to get it. My data is this (a double matrix called mymatrix) : Blogs Wikis Redes Etiq. SPC LMS Menor de 30 57.14 28.57 14.29 28.57 57.14 28.57 de 31 a 40 63.83 61.70 29.79 17.02 59.57 70.21 de 41 a 50 72.64 70.75 47.17 20.75 55.66 75.47 Mayor de

I have a very large matrix. I would like to display the variance across each row. In other words, I want to output a vector containing the values of variance across row. When I use the function var(), it seems to give me the variability of each column. Any ideas?? [[alternative HTML version deleted]]

Hi R, I have a matrix x with repeating row names. > dim(x) [1] 862 19 zz<-matrix(0,4,4) rownames(zz)=c("a","a","b","b") data.frame(zz) (?) I need to use x in a linear regression lm(as.formula(paste("final_dat[,5]~",paste(colnames(x),collapse="+"))),x ) this gives me a error Error in

Hello, If a matrix with 5 columns has been defined and the first two columns need to be sorted in ascending order, how can this be achieved whilst ensuring the other 3 columns data are in relative position to the sorted columns? Glen Jones [[alternative HTML version deleted]]

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