similar to: randomly generated binary matrix (with conditions)

Hi, I'm using colSums and rowSums to sum the first dimensions of arrays. It works ok but the resulting object is different. See > a3d <- array(rnorm(120, mean=2), dim=c(20,6,1)) > dim(colSums(a3d)) [1] 6 1 > dim(rowSums(a3d)) NULL > class(colSums(a3d)) [1] "matrix" > class(rowSums(a3d)) [1] "numeric" I was expecting rowSums to preserve the array

Working with an NxMxO sized matrix, currently I can do this in my code: if (max(colSums(array)) >= number) But to get an equivalent result using rowSums, I have to do: for (i in 1:10) { if (max(rowSums(array[,,i])) >= number) } I'm running both in a much larger loop that loops millions of times, so speed and such is quite a big factor for me. Currently, the colSums line uses about

Hi, I have a matrix, say m=matrix(c( 983,679,134, 383,416,84, 2892,2625,570 ),nrow=3 ) i can find its row/col sum by rowSums(m) colSums(m) How do I divide each row/column by its rowSum/colSums and still return in the matrix form? (i.e. the new rowSums/colSums =1) Thanks. Casper -- View this message in context:

Hi, I'd like to simply add column-wise using Matrix objects (Csparse). It looks like one can apply mosty any base function to these objects (i.e., apply, colSums), but there is a nasty conversion to traditional matrix objects if one does that. Is there any workaround? I can see colSum listed in the help for Class 'CsparseMatrix' , but I wonder whether I'm using the default

Buenas tardes a tod@s, Me gustaria redondear las entradas de una matriz m manteniendo la suma de filas y columnas constantes (son valores fijos conocidos). En la aplicacion que estoy trabajando (en la que por supuesto m tiene una dimension mayor que en el ejemplo), no son permitidos numeros decimales y por ello debe efectuarse el redondeo. La forma de m es: m <- matrix(c(3.546, 4.5345,

Dear list, I have some trouble generating a frequency table over a number of vectors. Creating these tables over simple numbers is no problem with table() > table(c(1,1,1,3,4,5)) 1 3 4 5 3 1 1 1 , but how can i for example turn: 0 1 0 0 0 1 0 1 0 1 0 0 0 1 0 1 0 0 into 0 0 1 1 1 0 0 2 0 1 0 3 My second problem is, sorting rows and columns of a matrix by the rowSums/colSums. I did it

Dear R-ers! Asking for your help with building the stacked area chart for the following simple data (several variables - with date on the X axis): ### Creating a data set my.data<-data.frame(date=c(20080301,20080402,20080503,20090301,20090402,20090503,20100301,20100402,20100503), x=c(1.1,1.0,1.6,1,2,1.5,2.1,1.3,1.9),y=c(-4,-3,-6,-5,-7,-5.2,-6,-4,-4.9),

Hi [R-2.6.0, macOSX 10.4.10]. The helppage says that rowSums() and colSums() are equivalent to 'apply' with 'FUN = sum'. But I came across this: > a <- matrix(1:30,5,6) > is.integer(apply(a,1,sum)) [1] TRUE > is.integer(rowSums(a)) [1] FALSE > so rowSums() returns a float. Why is this? -- Robin Hankin Uncertainty Analyst National Oceanography Centre,

Gracias, Jorge. Y cual fue la solucion a la que llegaron? --JIV Sent from my phone. Please excuse my brevity and misspelling. On Aug 25, 2013, at 8:36 AM, Jorge Ayuso Rejas <jayusor@gmail.com> wrote: Esto lo hice yo en una práctica en la universidad, Definíamos un problema de optimización entera minimizando el error de redondeo y restringiendo a la suma de filas y columnas. El 23 de

Hi, I am looking for some generalization of colSums and rowSums for general vector valued functions, and for arrays of more than 2 dimensions. So as a concrete example, suppose I have a 3 dimensional array, given by x = array(1:100,c(3,4,5)). and I want to sum the 3rd index of x to obain a 3 by 4 matrix. Using rowSums would return a vector of length 3 because it treats the last two indices as

Dear All, I have a problem understanding the difference between the outcome of a fisher exact test and a chi-square test (with simulated p.value). For some sample data (see below), fisher reports p=.02337. The normal chi-square test complains about "approximation may be incorrect", because there is a column with cells with very small values. I therefore tried the chi-square with

I have generated the following: x= E1 E2 E3 D1 0 0 1 D2 1 0 3 D3 0 2 0 y= E1 E2 E3 D1 0 0 1.75 D2 1.75 0 1.3125 D3 0 3.5 0 Where x and y are linked by: y =sum(x) * x / (rowSums(x)%o%colSums(x)) N=x[x[1:3,]>0] R=y[y[1:3,]>0] Now suppose I ONLY

Hello! My code below creates a data frame and a plot for it. However, I can't figure out how to add a legend that is not ON the plot itself, but outside of it (e.g., to the right of my graph or below it). I tried something: I put a line par(xpd=T, mar=par()$mar+c(0,0,0,4)) right before my plot command), but that screwed up all my gridlines - they covered all graph and do not coincide with

Hello, everybody! I have a code below that creates a data set and then a stacked bar chart based on that data set. No need to look at it - just notice please that my horizontal axis is a date varible (x=my.data$date). I have a question about the last 2 lines of this code: grid(nx=NULL,ny=NULL,col = "lightgray", lty = "dotted",lwd = par("lwd")) axis(1, las = 2) Could

I'm using the subset() function in R. dat <- data.frame(one=c(6,7,8,9,10), Number=c(5,15,13,1,13)) subset(dat, Number >= 10) However, I want to find the number of all rows who meet the Number>=10 condition. I've done this in the past with something like colSums or rowSums or another similar function. But I don't remember how to get the number of elements which meet that

Greetings, My goal is to create a Markov transition matrix (probability of moving from one state to another) with the 'highest traffic' portion of the matrix occupying the top-left section. Consider the following sample: inputData <- c( c(5, 3, 1, 6, 7), c(9, 7, 3, 10, 11), c(1, 2, 3, 4, 5), c(2, 4, 6, 8, 10), c(9, 5, 2, 1, 1) ) MAT <- matrix(inputData,

Buenos días quiero definir situación_2 <-tapply(ICE_3$FACTOR_CORR, list(X.1.1Eneltrimestrequeestáacabando=ICE_3$X.1.1Eneltrimestrequeestáacabando, CNAE_1=ICE_3$CNAE_1), f) donde f sea el porcentaje por filas (y/o el porcentaje por columnas) y las funciones rowPercent y colPercent me dan el siguiente error Error en tapply(ICE_3$FACTOR_CORR, list(X.1.1Eneltrimestrequeestáacabando =

Hello! If you run the whole code below, it'll produce a stacked diagram. And it looks good - because the tick-marks are aligned with the grid. However, if I stretch the graph window, grid becomes misaligned with the tickmarks. Or, rather, it seems aligned for the first and the last tick mark, but not for tickmarks in between. Can it be addressed? Thank you! Dimitri ### Creating a data set

Hi all, I'm currently working through the "Beginner's Guide to R" (Zurr et al.) and I'm wondering about the first exercise in chapter 3: I imported the data from BirdFluCases.txt and executed the 'names' and 'str' functions as follows: Bird = read.table(file='C:\\rbook\\BirdFluCases2.txt', header=TRUE) names(Bird) [1] "Year"

Hello all, I recently used the Vegan library quite extensively (in the context of text similarity assessment) on an Ubuntu 6.06 LTS system with R version 2.2.1 (2005-12-20 r36812). The Vegan lib is version 1.6-10. I hit on a problem yesterday, though, when trying to install R and Vegan on two further computers - one Windows XP and one further Ubuntu 6.06 machine, taking either R version 2.4.0