Hi:
The idea is as follows:
* string your four matrices into vectors, cbinding them so that the
columns correspond to what you want as the (1,1), (1, 2), (2, 1) and (2, 2)
elements, respectively, of the matrix/table to be used for the Fisher test;
* Operate row-wise on the constructed matr
Hi,
I would recommend reformatting the data as a 2x2x1000 array and using apply.
Jonathan
On Mon, Jan 3, 2011 at 7:57 AM, zhaoxing731 wrote:
> Hello
>
> I have 4 1000*1000 matrix A,B,C,D. I want to use the corresponding element of
> the 4 matrices. Using the "for loop" as follow:
>
> E<-o
> f
org wrote on 01/03/2011 09:57:14 AM:
> [image removed]
>
> [R] matrices call a function element-wise
>
> zhaoxing731
>
> to:
>
> R-help
>
> 01/03/2011 01:46 PM
>
> Sent by:
>
> r-help-boun...@r-project.org
>
> Hello
>
> I have 4 100
Hello
I have 4 1000*1000 matrix A,B,C,D. I want to use the corresponding element of
the 4 matrices. Using the "for loop" as follow:
E<-o
for (i in 1:1000)
{for (j in 1:1000)
{
E<-fisher.test(matrix(c(A[i][j],B[i][j],C[i][j],D[i][j]),2))#call
fisher.test for every element
}
}
Why not just use matrix(runif(9),nrow=3) ?
--
View this message in context:
http://n4.nabble.com/Matrices-with-randomly-generated-entries-tp1018777p1024617.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org ma
On Jan 20, 2010, at 4:55 PM, jshort wrote:
I'm attempting to generate matrices where the entries are randomly
generated
numbers of specified distribution.
The following code was an attempt to create a 3 by 3 matrix, where my
entries where randomly generated from a uniform (0,1) distributio
I'm attempting to generate matrices where the entries are randomly generated
numbers of specified distribution.
The following code was an attempt to create a 3 by 3 matrix, where my
entries where randomly generated from a uniform (0,1) distribution.
x = matrix(0,ncol = 3, byrow = T)
for(i in 1:
On Mon, 2009-08-10 at 14:02 -0400, mmv.listservs wrote:
> Your example works because you have only 5 really long labels. I tried
> changing 5 to 474 because that is how many I have and the plot looks
> illegible. Should I increase the margin or the height of the plotting
> window? Thanks for your h
Your example works because you have only 5 really long labels. I tried
changing 5 to 474 because that is how many I have and the plot looks
illegible. Should I increase the margin or the height of the plotting
window? Thanks for your help again.
## dummy data
set.seed(123)
dummy <- data.frame(A =
On Mon, 2009-08-10 at 13:41 -0400, mmv.listservs wrote:
> I should give an example
Thanks for that - I just two seconds ago sent a reply to the list
complaining because you didn't provide one. However, the code below
doesn't work.
>
> for(p in 1:100)
> {
> str <- paste("Task", p, sep=" ")
>
On Mon, 2009-08-10 at 13:33 -0400, mmv.listservs wrote:
> Gavin and Stefan,
>
> Both the subset commands and the flag were exactly what I needed. On another
> note, I'm dealing with variables that are categorical and have long names
> like "Task XYZ", "Task ABC" "Task CCC"
>
> When I try to plot
I should give an example
for(p in 1:100)
{
str <- paste("Task", p, sep=" ")
task_name[p] <- str
}
## first set the random seed so we get the same results
set.seed(123)
## now produce some dummy data
dummy <- data.frame(A = sample(LETTERS[1:4], 100, replace = TRUE),
B =
Gavin and Stefan,
Both the subset commands and the flag were exactly what I needed. On another
note, I'm dealing with variables that are categorical and have long names
like "Task XYZ", "Task ABC" "Task CCC"
When I try to plot against the probability it doesn't show me the Task name
anymore. How
On Mon, 2009-08-10 at 11:17 -0400, mmv.listservs wrote:
> yy<-poisson2[poisson2$Reboot.Id=="Reboot
> 2",poisson2$Task.Status=="F",,drop=FALSE]
The above doesn't make any sense and can't be working or doing what you
think it is doing.
Lets dissect this command:
yy <- poisson2[poisson2$Reboot.Id==
yy<-poisson2[poisson2$Reboot.Id=="Reboot
2",poisson2$Task.Status=="F",,drop=FALSE]
doesn't work either? Any other ideas?
On Mon, Aug 10, 2009 at 11:01 AM, mmv.listservs wrote:
> How do you access all the column attributes associated with a column reboot
> instance?
>
> The variables
>
> poisson
How do you access all the column attributes associated with a column reboot
instance?
The variables
poisson2 ~ a matrix with 10,000 rows and 8 column attributes.
Things I tried:
This command only returns a vector for one of the column attributes
x1_prob <- poisson2$Probability[poisson2$Reboot.
I suspect you probably need to provide an example that illustrates why
this seems difficult:
> x <- 1:10
> f1 <- function(x) exp(x)
> g1 <- function (x) log(x)
> h1 <- function (x) x^(1/2)
# any of these functions would return a 10 element vector if given x
as an argument
> f1(x)*g1(x)*h1(x
Hi,
I'm a new R user and would appreciate your help regarding the following: Can
I create a matrix whose elements are n functions of a vector x? In my
problem I have 3 vectors (a,b,c) with elements a=[f1(x) f2(x)fn(x)];
b=[g1(x) g2(x)gn(x)]; c=[h1(x) h2(x)hn(x)]. I need to create a fin
18 matches
Mail list logo
