similar to: means across list of data frames

Displaying 20 results from an estimated 20000 matches similar to: "means across list of data frames"

2009 Jul 13
1
Add grand mean to every entry in a matrix
Hi, I have a matrix: mymat <- matrix(runif(10*4), ncol=4) I wish to subtract the column means, down the colums, subtract the row means from the rows and add back the grand mean - all the means should be the means of mymat rather than of the new matrix. How can I do this? Any help much appreciated. Thanks Tom _________________________________________________________________
2010 Jul 20
2
Means from selected columns in a data frame
Hi all, I have a dataframe with survey data. Now I want to calculate means from several but not all columns (e.g. a1, a2, a3) and save them in a new separate column (e.g. a). Like: a = mean(a1, a2, a3) Can someone help me or give me the right key word to look for? Thanks [[alternative HTML version deleted]]
2010 Apr 18
3
xtabs() of proportions, and naming a dimension (not a row)
Hi, xtabs() creates a table of counts. I want a table of proportions -- that is, I want to divide every vector (along a particular dimension) by its sum. The tiny example below does that. The call to xtabs() creates a matrix "A" with dimensions ("x1","x2","y"). I transform "A" using aperm() and aaply() to get the matrix "B". The
2013 Mar 22
1
Median across matrices
Hey all, I have a list of matrices. I'd like to calculate the median across all those matrices for each element. What I'd like to end up with is a matrix containing the median of all [1,1] [1,2] etc. elements across all matrices. Is there a concise way of doing that? Thanks!
2018 Jan 30
1
Simulation based on runif to get mean
Hello, Another way would be to use ?replicate and ?colMeans. set.seed(2511) # Make the results reproducible fun <- function(n){ f <- function(){ a <- runif(5, 1, 10) b <- runif(5, 10, 20) colMeans(cbind(a, b)) } replicate(n, f()) } fun(10) Hope this helps, Rui Barradas On 1/30/2018 8:58 AM, Daniel Nordlund wrote: > On 1/29/2018
2018 Jan 30
0
Simulation based on runif to get mean
On 1/29/2018 9:03 PM, smart hendsome via R-help wrote: > Hello everyone, > I have a question regarding simulating based on runif.? Let say I have generated matrix A and B based on runif. Then I find mean for each matrix A and matrix B.? I want this process to be done let say 10 times. Anyone can help me.? Actually I want make the function that I can play around with the number of simulation
2018 Jan 30
2
Simulation based on runif to get mean
Hello everyone, I have a question regarding simulating based on runif.? Let say I have generated matrix A and B based on runif. Then I find mean for each matrix A and matrix B.? I want this process to be done let say 10 times. Anyone can help me.? Actually I want make the function that I can play around with the number of simulation process that I want. Thanks. Eg: a <- matrix(runif(5,1, 10))
2002 Aug 14
3
t-test via matrix operations
I need to calculate a large number of t statistics, and would like to do so via matrix operations. So far I have figured out a way to calculate the mean of each row of the matrix: d <- matrix(runif(100000,1,10), 1000, 10) # some test data s <- rep(1,ncol(d)) # a sum vector to use for matrix multiplication means <- (d%*%s)/ncol(d) This is at least 1 order of magnitude faster than
2010 Jun 01
2
Problem using apply
Hello , I can not get apply function to do what I want when doing stuff on rows. Let's say I want to divide the rows of matrix by their mean, the below show you get the same result weather you use 1 or 2, i.e. same result for columns than for means..:( Thanks a lot for the help, m = matrix( c(1,4,7,4,5,8,7,8,9), nrow = 3 ) divideByMean = function( v ){ return(
2012 Sep 02
3
Help on finding specific columns in matrix
Dear All,   I have a matrix with 33 columns and 5000 rows. I would like to find 2 specific columns in the set: the one that holds the highest values and the one that holds the lowest values. In this case the column's mean would be apropriate to use to try to find those specific columns because each columns mean is different and they all change together based on the same "change of rate
2006 Sep 07
2
Matrix package in R-2.4.0alpha
In a newly downloaded version (today) of R-2-4-0alpha, with all packages from CRAN also installed today, I get: > library(Matrix) Erro en loadNamespace(package, c(which.lib.loc, lib.loc), keep.source = keep.source) : in 'Matrix' methods specified for export, but none defined: BIC, anova, coef, confint, deviance, fitted, fixef, formula, head, lmer, logLik, mcmcsamp, plot,
2006 Feb 28
2
Elegant way to express residual calculation in R?
Hi All, I am illustrating a simple, two-way ANOVA using the following data and I'm having difficulty in expressing the predicted values succinctly in R. X<- data.frame(read.table(textConnection(" Machine.1 Machine.2 Machine.3 53 61 51 47 55 51 46 52 49 50
2010 Nov 07
2
remove data frame from list of data frames
I have a list of data frames like this: a<- data.frame(x=runif(10), y = runif(10), Acc = 1) b<- data.frame(x=runif(10), y = runif(10), Acc = 0) ls<- list(a,b) and I want to remove the data frames from ls that have Acc values other than 1. How do I do that? Thanks for any help! Matthew
2008 Oct 17
2
function help
Hi everyone, I have dataset which I make a sample of it couple of times and each time I get the mean and standard deviation of each row for each sample. I have a function for that, which takes the name of the file and number of times to sample and then returns the mean and standard deviation for each row in each sample. Sample=function(name, n){
2005 Jul 19
3
extracting row means from a list
Hello: I'm reading in a series of text files (100 files that are each 2000 rows by 6 columns). I wish to combine the columns (6) of each file (100) and get the row mean. I'd like to end up with a data.frame of 2000 rows by 6 columns. foo <- list() for(i in 1:10){ # The real data are read in from a series of numbered text files foo[[i]] <- data.frame(x1 = rnorm(100), x2 =
2018 Feb 20
0
Take the maximum of every 12 columns
Ista, et. al: efficiency? (Note: I needed to correct my previous post: do.call() is required for pmax() over the data frame) > x <- data.frame(matrix(runif(12e6), ncol=12)) > system.time(r1 <- do.call(pmax,x)) user system elapsed 0.049 0.000 0.049 > identical(r1,r2) [1] FALSE > system.time(r2 <- apply(x,1,max)) user system elapsed 2.162 0.045 2.207 ##
2012 Dec 24
2
colmeans not working
[text file is also attached in case you find the format of email difficult to understand] Dear useRs,You must all the planning for the christmas, but i am stucked in my office on the following issue i had a file containg information about station name, year, month, day, and discharge information. i opened it by using following command > dat1<-read.table("EL.csv",header=TRUE,
2018 Feb 20
2
Take the maximum of every 12 columns
On Tue, Feb 20, 2018 at 11:58 AM, Bert Gunter <bgunter.4567 at gmail.com> wrote: > Ista, et. al: efficiency? > (Note: I needed to correct my previous post: do.call() is required for > pmax() over the data frame) > > > x <- data.frame(matrix(runif(12e6), ncol=12)) > > > system.time(r1 <- do.call(pmax,x)) > user system elapsed > 0.049 0.000
2006 Aug 16
3
separate row averages for different parts of an array
I have an array with 44800 columns and 24 rows I would like to compute the row average for the array 100 columns at a time, so I would like to end up with an array of 24 rows x 448 columns. I have tried using apply(dataset, 1, function(x) mean(x[])), but I am not sure how to get it to take the average 100 columns at a time. Any ideas would be welcomed. thanks, Spencer [[alternative HTML
2024 Jun 08
1
Can't compute row means of two columns of a dataframe.
John, Maybe you can clarify what you want the output to look like. It took me a while to realize what you may want as it is NOT properly described as wanting rowsums. There is a standard function called rowMeans() that probably does what you want if you want the mean of all rows as in: > rowMeans(xxxz) [1] 84.33333 87.00000 89.66667 92.33333 95.00000 97.66667 100.33333 103.66667