Displaying 20 results from an estimated 2000 matches similar to: "how to calculate means of matrix elements"
2010 Jan 29
7
Simple question on replace a matrix row
Hello, I have a matrix mat1 of dim [1,8] and mat2 of dim[30,8], I want to
replace the first row of mat2 with mat1, this is what I do:
mat2[1,]<-mat1 but it transforms mat2 in a list I don't understand, I want
it to stay a matrix...
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Anna Lippel
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2010 Mar 29
2
Need help on matrix manipulation
Dear all,
Ket say I have 3 matrices :
mat1 <- matrix(rnorm(16), 4)
mat2 <- matrix(rnorm(16), 4)
mat3 <- matrix(rnorm(16), 4)
Now I want to merge those three matrices to a single one with dimension
4*3=12 and 4 wherein
on resulting matrix, row 1,4,7,10 will be row-1,2,3,4 of "mat1", row
2,5,8,11 will be row-1,2,3,4 of "mat2" and row 3,6,8,12 will be row-1,2,3,4
of
2008 Aug 29
2
Newbie: Examples on functions callling a library etc.
Hello
R is pretty new to me. I need to write a function that returns three
matrices of different dimensions. In addition, I need to call a function
from a contributed package with the function. I have browsed several
manuals and docs but the examples on them are either very simple or
extremely hard to follow.
Many thanks
Ed
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2010 Oct 14
1
rbind ing matrices and resetting column numbers
Sorry for the verbose example. I want to row bind two matrices, and all works except I want the column labelled "row" to be sequential in the new matrix, shown as "mat3" here, i.e. needs to be 1:6 and not 1:3 repeated twice. Any suggestions?
Thanks
J
> colnm1 <- c("row","ti","counti")
> colnm2 <-
2010 Jan 27
1
How to sort data.frame
Dear R heleprs
Suppose I have following data
Scenarios
combination_names
series1
series2
Sc1
MAT2 GAU1
7.26554
8.409778
Sc2
MAT2 GAU2
7.438128
8.130275
Sc3
MAT3 GAU1
8.058422
8.06457
Sc4
MAT1 GAU2
8.179855
8.022071
Sc5
MAT3 GAU2
8.184033
8.191831
Sc6
MAT3 GAU2
7.50312
8.232425
Sc7
MAT1 GAU2
7.603291
8.200993
Sc8
MAT1 GAU1
8.221755
8.380097
Sc9
MAT3 GAU2
7.904908
2013 Jan 04
2
Working with Matrix
Hello again,
Let say I have 2 matrices which equal number of columns but different
number of rows like:
Mat1 <- matrix(1:20, 4, 5)
Mat2 <- matrix(1:25, 5, 5)
Now for each column 1-to-5 I need to fetch the corresponding columns
of these 2 matrices and add the corresponding elements (ignoring NA
values if any). Therefore for the 1st column I need to do:
(1+1), (2+2),...,(4+4), (NA+5)
and
2013 Feb 01
2
Nested loop and output help
Hello Everyone,
My name is Thomas and I have been using R for one week. I recently found
your site and have been able to search the archives of posts. This has
given me some great information that has allowed me to craft an initial
design to an inquiry I would like to make into the breakdown of McNemar's
test. I have read an intro to R manual and the posting guides and hope I am
not violating
2013 Feb 13
3
date and matrices
Hi Elisa,
Try this:
date1<-format(seq.Date(as.Date("1991.1.1",format="%Y.%m.%d"),as.Date("1996.12.31",format="%Y.%m.%d"),by="day"),"%Y.%m.%d")
?length(date1)
#[1] 2192
mat1<-matrix(c(.314,.314,.273,.273,.236,.236,.236,.236,.273,.314,.403,.314),ncol=1)
res1<-
2010 Mar 02
2
turn character string into unevaluated R object
Hi,
How to turn a character string into an unevaluated R object? I want to load some files in a directory into data matrix R objects. I could do this with read.table and assign (see below). Then, I want to turn the character string representing a file name (the evaluated expression of i) into an unevaluated R object. Basically, I want to create matrices whose names are the same as the related file
2005 Apr 29
1
na.action
Hi,
I had the following code:
testp <- rcorr(t(datcm1),type = "pearson")
mat1 <- testp[[1]][,] > 0.6
mat2 <- testp[[3]][,] < 0.05
mat3 <- mat1 + mat2
The resulting mat3 (smaller version) matrix looks like:
NA 0 0 0
0 NA 0 NA
0 0 NA 2
0 0 2 NA
To get to the number of times a '2' appears in the rows, I was
2011 Apr 24
2
random roundoff?
On my CentOS 5 box, in a C++ program that does much arithmetic,
including numerous matrix multiplications, I have a situation in
in which the result depends on the nature of nearby I/O. Thus,
with all arithmetic done with type double, and where values
are mostly in the range [-1.0e0,+1.0e0] or nearby, I do:
cerr << "some stuff" << endl;
mat3 = matmult(mat1,mat2);
I
2011 May 11
3
Reordering inputs
Hello All,
I have 2 matrices consisting of the same inputs, but having different
outputs. I created a heatmap for both of them; the point is to compare them
side by side. The best way to organize the inputs is to make sure that the
order of the inputs are the same for both heatmaps. How would I go about
making sure that the order of inputs of both heatmaps are the same?
As it is right now, I can
2013 Apr 24
2
Distance matrices Combinations
Dear UseRs,
MY PROBLEM IS A SMALL PIECE OF A REAL BIG AND A COMPLICATED PROBLEM. IF I DELIBERATE IN A VERY SIMPLE WAY THEN ALL I
WANT IS TO PUT ALL THE POSSIBLE COMBINATIONS OF 75 DISTANCE MATRICES (BY TAKING 4 MATRICES, MORE COMMONLY 75C4), in the following equation.
t<-as.matrix((MAT1)^2+(MAT2)^2+(MAT3)^2+(MAT4)^2+,upper=T,diag=T))
Then "1215450" values of "t"(one for
2012 Oct 07
1
variances of random effects in coxme
Dear R users,
I'm using the function coxme of the package coxme in order to build Cox
models with complex random effects. Unfortunately, I sometimes get
surprising estimations of the variances of the random effects.
I ran models with different fixed covariates but always with the same 3
random effects defined by the argument
varlist=coxmeMlist(list(mat1,mat2,mat3), rescale = F, pdcheck = F,
2010 Feb 12
3
Code working but too slow, any idea for how to speed it up ?(no loop in it)
Hello my friends,
here is a code I wrote with no loops on matrix that is taking too long (2
seconds and I call him 720 times --> 12 minutes):
mat1 and mat2 are both matrix with 103 columns and 164 rows.
sequence <- matrix(seq(1 : ncol(mat1)))
returns <- apply(sequence, 1, function, mat1= mat1, mat2 = mat2, day = 1)
function<- function(mat1, mat2, colNb, day){
2005 Nov 03
4
merging dataframes
Dear List,
I often have to merge two or more data frames containing unique row
names but with some columns (names) common to the two data frames and
some columns not common. This toy example will explain the kind of setup
I am talking about:
mat1 <- as.data.frame(matrix(rnorm(20), nrow = 5))
mat2 <- as.data.frame(matrix(rnorm(20), nrow = 4))
rownames(mat1) <- paste("site",
2010 Aug 25
3
approxfun-problems (yleft and yright ignored)
Dear all,
I have run into a problem when running some code implemented in the
Bioconductor panp-package (applied to my own expression data), whereby gene
expression values of known true negative probesets (x) are interpolated onto
present/absent p-values (y) between 0 and 1 using the *approxfun -
function*{stats}; when I have used R version 2.8, everything had
worked fine,
however, after updating
2005 Oct 06
1
Compare two distance matrices
Hi all,
I am trying to compare two distance matrices with R. I would like to
create a XY plot of these matrices and do some linear regression on
it. But, I am a bit new to R, so i have a few questions (I searched in
the documentation with no success).
The first problem is loading a distance matrix into R. This matrix is
the output of a the Phylip program Protdist and lookes like this:
5
2007 Nov 12
1
update matrix with subset of it where only row names match
I guess this has a simple solution:
I have matrix 'mat1' which has row and column names, e.g.:
A B C
row1 0 0 0
row2 0 0 0
....
rown 0 0 0
I have a another matrix 'mat2', essentially a subset of 'mat1' where the
rownames are all in 'mat1' e.g.:
B
row3 5
row8 6
row54 7
I want to insert the values of matrix mat2 for column B (in reality it
could be some or
2003 Oct 31
2
Summing elements in a list
Hi,
Suppose that I have a list where each component is a list of two
matrices. I also have a vector of weights. How can I collapse my
list of lists into a single list of two matrices where each matrix
in the result is the weighted sum of the corresponding matrices.
I could use a loop but this is a nested calculation so I was hoping
there is a more efficient way to do this. To help clarify,