search for: 2x3

Is there an element-by-element multiplication in R, like the .* operator in Matlab? eg: A (2x3) B (2x3) C=A.*B C (2x3) C = [[a11*b11 a12*b12 a13*b13]; [a21*b21 a22*b22 a23*b23]] I can't find one... Thanks -Mike Gates

Hi, all, I am wondering if there is an algorithm for calculating power of 2x3 table using Fisher exact test. There is one (algorithm 280) for 2x2 Fisher exact test but I couldn't find one for 2x3 table. If we are not lucky enough to have one, is there any other way to calculate exact power of 2x3 table? The reason why I want exact power is because some cells ar...

Dear R users, I need to to the the following. Let a= 1 2 3 4 5 6 and b= -1 -2 -3 be (2x3) matrices. -4 -5 -6 I need to combine the two matrices into a new (2x6) matrix like this: ab = ( 1 -1 2 -2 3 -3 ) 4 -4 5 -5 6 -6 How can this be done in R? ----------------------------------------------------------------- ?????? ??? - ?? ???????! www.survivor.btv.bg

Hi How do I make do.call() take "+" as a function for a list of more than two elements? Toy problem follows: f <- function(i){matrix((1:6)^i,2,3)} # Thus f() returns a matrix of size 2x3; I want to add a whole bunch of such matrices, # as in f(1) + f(2) + f(3) + f(4) # But: > do.call("+",sapply(1:4,f,simplify=FALSE)) Error in do.call("+", sapply(1:4, f, simplify = FALSE)) : operator needs one or two arguments Also, > do.call(sum,sapply(1:4,...

...ot;trace", spec="longrun",K=2))@cval[k,j]) }}} I have a problem in collecting the results of a test. The function ca.jo creates an object with various attributes, one of which is the "cval" that i can access through @cval. The attribute cval is an object of dimension 2X3. I am running recursively the test with ca.jo for 12 couples of time series, so I have an output of 12 matrices of 2X3 elements and I would like to create an object like an array of dimension (2,3,12) which contains each matrix @cval produced by ca.jo for the 12 subjects that i tested. Can anyone...

I am sorry if this is documented in apply's dcumentation or completely obvious, I could not find or work it out. Given an matricies Q: 2x3, R:1x3 and S:1x2 apply(Q, 1, '-', R) is 3x2 and apply(Q, 2, '-', S) is 2x3 Why? cheers Worik > Q [,1] [,2] [,3] [1,] 1 2 3 [2,] 10 11 12 > R [1] 1 4 7 > S [1] 1 4 > apply(Q, 1, '-', R) [,1] [,2] [1,] 0 9 [2,] -2 7 [3,] -4...

I have a 20x3 matrix as follows: > m <- replicate(3, matrix(rnorm(20),20,1)) I need to compute, say, 95th and 99th percentiles of each column such that the resulting matrix becomes 2x3 with each row representing the respective percentile. My "best effort" is to compute one column at a time as follows: > quantile(m[,1], c(0.95, 0.99)) To do the same for columns 2 and 3, I would simply change the column number accordingly. Clearly, this is not very efficient as I m...

...do.call() take "+" as a function for a list of more > > > than two elements? > > > > > > Toy problem follows: > > > > > > > > > f <- function(i){matrix((1:6)^i,2,3)} > > > > > > # Thus f() returns a matrix of size 2x3; I want to add a whole bunch > > > of such matrices, > > > # as in f(1) + f(2) + f(3) + f(4) > > > > > > # But: > > > > > > > > > > > > > do.call("+",sapply(1:4,f,simplify=FALSE)) > > > Error in do.call(...

It was a real surprise, but a student in my class found that the function eigen is buggy. He traced to the problem from his inability of getting principal component analysis to work on his data. Chong Gu Here is a matrix I generated through X'X, where X is 2x3. > jj [,1] [,2] [,3] [1,] 0.8288469 -1.269783 -0.7533517 [2,] -1.2697829 2.162132 2.0262917 [3,] -0.7533517 2.026292 4.1927134 Here are the 0.90.1 eigen results. > eigen(jj) $values [1] 3.06206547 0.20492432 -0.04955518 $vectors [,1] [,2]...

...7 > matb<-matrix(2:7,nrow=3,ncol=2) > matb [,1] [,2] [1,] 2 5 [2,] 3 6 [3,] 4 7 > which(mata==4) [1] 3 > which(matb==4) [1] 3 The function "which" returns the position "3" for both, mata and matb, but If I didn't know that mata is a 2x3 matrix and matb is a 3x2, I wouldn't know that the number 4 is at the column 2 and row 1 of the mata and at the column 1 and row 3 of the matb. Do you know any way to know the column and the row of a number in a matrix automatically? Thanks, best regards, Charles -- Um ax?! :) -- Charles...

I want to add an outer subtitle to my 2x3-plot, but it's distance to the lowest plots is very high: 6 lines or more?! If I choose oma=c(5,0,2,0), it lies out of the plotting region and disapprears. Obviously I make something wrong here. Help appreciated, Thomas x=seq(from=1, to=3.5, length=100) par(mfrow = c(2, 3), oma=c(6,0,2,0)) ##...

talking with a client about various options for backup. They are an advertising agency so most of the space is used by various binary files JPG/TIFF/Movies. We have been using LTO-2 but if we jump up their storage to say 2Tb, that's going to be $5000 for LTO-3 with magazines. This affords us 3 week rotations rather simply. If we add hard drive storage for backups instead of buying LTO-3 tape

I have a 2 x 3 matrix called snp and I want to compute the following probability: choose(sum(snp[,1]), snp[1,1]) * choose(sum(snp[,2]), snp[1,2]) * choose(sum(snp[,3]), snp[1,3])/choose(sum(snp), sum(snp[1,])) but I keep getting Infs and NaNs. Is there a function that can do this in R? -- Thanks, Jim. [[alternative HTML version deleted]]

...a number of different RxC tests based on how many genotypes (e.g. 2 genotypes vs 4 gives 8 cells, 2 by 2 gives you a 4 cell test etc.). Instead of writing a series of if then statements for the chi-square test on the obs-exp numbers for each possible gene pair (i've started to do this....2x2, 2x3, 3x2, 2x4, 3x4,etc) is there a generic way to pass the number of unique elements for each gene (which would define the number of genotypes for that gene) into a calculation of the obs-exp and then the chisquare test? Here is some example code to illustrate what I am trying to work with. ###exampl...

Hello everyone, I have problem rebalancing Gluster volume. Gluster version is 3.7.3. My 1x3 replicated volume become full, so I've added three more bricks to make it 2x3 and wanted to rebalance. But every time I start rebalancing, it fails immediately. Rebooting Gluster nodes doesn't help. # gluster volume rebalance gsae_artifactory_cluster_storage start volume rebalance: gsae_artifactory_cluster_storage: success: Rebalance on gsae_artifactory_cluster_storage...

...42, Szymon Miotk <szymon.miotk at gmail.com> wrote: >> >> Hello everyone, >> >> >> I have problem rebalancing Gluster volume. >> Gluster version is 3.7.3. >> My 1x3 replicated volume become full, so I've added three more bricks >> to make it 2x3 and wanted to rebalance. >> But every time I start rebalancing, it fails immediately. >> Rebooting Gluster nodes doesn't help. >> >> # gluster volume rebalance gsae_artifactory_cluster_storage start >> volume rebalance: gsae_artifactory_cluster_storage: success: R...

Hi everybody, I'm trying to construct contrasts for an ANOVA to determine if there is a significant trend in the means of my groups. In the following example, based on the type of 2x3 ANOVA I'm trying to perform, does the linear polynomial contrast generated by contr.poly allow me to test for a linear trend across groups? doi=data.frame( Group=c( rep(1, 5), rep(2, 5), rep(3, 5), rep(1, 5), rep(2, 5), rep(3, 5)), Mean_1=c( rnorm(5, mean=3, sd=1), rnorm(5, mea...

On 7 July 2017 at 15:42, Szymon Miotk <szymon.miotk at gmail.com> wrote: > Hello everyone, > > > I have problem rebalancing Gluster volume. > Gluster version is 3.7.3. > My 1x3 replicated volume become full, so I've added three more bricks > to make it 2x3 and wanted to rebalance. > But every time I start rebalancing, it fails immediately. > Rebooting Gluster nodes doesn't help. > > # gluster volume rebalance gsae_artifactory_cluster_storage start > volume rebalance: gsae_artifactory_cluster_storage: success: Rebalance > on gsa...

...gt; >> >> Hello everyone, >> >> >> >> >> >> I have problem rebalancing Gluster volume. >> >> Gluster version is 3.7.3. >> >> My 1x3 replicated volume become full, so I've added three more bricks >> >> to make it 2x3 and wanted to rebalance. >> >> But every time I start rebalancing, it fails immediately. >> >> Rebooting Gluster nodes doesn't help. >> >> >> >> # gluster volume rebalance gsae_artifactory_cluster_storage start >> >> volume rebalance:...

...il.com> wrote: > >> > >> Hello everyone, > >> > >> > >> I have problem rebalancing Gluster volume. > >> Gluster version is 3.7.3. > >> My 1x3 replicated volume become full, so I've added three more bricks > >> to make it 2x3 and wanted to rebalance. > >> But every time I start rebalancing, it fails immediately. > >> Rebooting Gluster nodes doesn't help. > >> > >> # gluster volume rebalance gsae_artifactory_cluster_storage start > >> volume rebalance: gsae_artifactory_cl...