Displaying 20 results from an estimated 1000 matches similar to: "expand.gird with constraints?"
2011 Aug 03
1
create a list under constraints
Hi, R users,
Here is an example.
k <- c(1,2,3,4,5)
i <- c(0,1,3,2,1)
if k=1, then i=0
if k=2, then i=0, 1
if k=3, then i=0, 1, 2, 3
if k=4, then i=0, 1, 2
if k=5, then i=0, 1
so i'd like to create a list like below.
> list
k i
1 1 0
2 2 0
3 2 1
4 3 0
5 3 1
6 3 2
7 3 3
8 4 0
9 4 1
10 4 2
11 5 0
12 5 1
I tried expand.grid, but I can't.
Any suggestion will be
2014 Jan 11
1
Fortran BLAS giving bad results
Hello r-devel,
When compiling Fortran code containing BLAS functions and calling it using
dyn.load, I am getting incorrect results. A small example with which I can
reproduce the problem is below.
I am running on OSX Mavericks (upgraded R, Xcode, etc per instructions on
this list), but I do not think it is related since the check on my
"blupsurv" package on r-forge seems to be showing
2011 Aug 11
5
generate two sets of random numbers that are correlated
Dear R users
I'd like to generate two sets of random numbers with a fixed correlation
coefficient, say .4, using R.
Any suggestion will be greatly appreciated.
Regards,
Kathryn Lord
--
View this message in context: http://r.789695.n4.nabble.com/generate-two-sets-of-random-numbers-that-are-correlated-tp3735695p3735695.html
Sent from the R help mailing list archive at Nabble.com.
2011 Oct 13
5
Counting the number of integers at one swoop
Dear R users,
I'd like to count the number of integers in a vector y.
Here is an example.
y <- c(0,1,1,3,3,3,5,5,6)
In fact, I know how to count the number of specific number in y.
sum(y==0) -> 1
sum(y==1) -> 2
sum(y==2) -> 0
sum(y==3) -> 3
sum(y==4) -> 0
sum(y==5) -> 2
sum(y==6) -> 1
However, in one computation I want to get this vector [1,2,0,3,0,2,1].
Thank
2012 Jun 25
4
do.call or something instead of for
Dear R users,
I'd like to compute X like below.
X_{i,t} = 0.1*t + 2*X_{i,t-1} + W_{i,t}
where W_{i,t} are from Uniform(0,2) and X_{i,0} = 1+5*W_{i,0}
Of course, I can do this with "for" statement, but I don't think it's good
idea because "i" and "t" are too big.
So, my question is that
Is there any better idea to avoid "for" statement
2009 Jul 28
3
character vector -> numeric matrix ??
Dear R users...
I'd like to change this character vector, "zz",
zz <- c("12","56","89")
to the following numeric matrix.
[,1] [,2]
[1,] 1 2
[2,] 5 6
[3,] 8 9
Actually, "zz" vector has a long length.
Any comments will be greatly appreciated.
Kathryn Lord
--
View this message in context:
2008 Oct 08
3
Re move repeated values
Dear R users,
I'd like to make this data
rem.y = c(-1,0,2,4,5)
from
y = c(-1,-1,0,2,2,2,2,4,4,5,5,5,5,5).
That is, I need to remove repeated values.
Here is my code, but I don't think it is efficient. How could I improve
this?
#------------------------------------------------------------------------
y = c(-1,-1,0,2,2,2,2,4,4,5,5,5,5,5)
n=length(y)
for (i in 1:n) #
2010 Apr 16
3
VERY SIMPLE QUESTION
Dear R users,
I am looking for more efficient way to compute the followings
--------------------------------------------------------------------------
a <- matrix(c(1,1,1,1,2,2,2,2),4,2)
b <- matrix(c(1,2,3,4),4,1)
Eventually, I want to get this matrix, `c`.
c <- matrix(c(1/1,1/2,1/3,1/4,2/1,2/2,2/3,2/4),4,2)
--------------------------------------------------------------------------
2009 Apr 25
2
Changing gird marks in ggplot2
Hi,
When I zoom into a graph created in ggplot2 with the
coord_cartesian(ylim=c(0,5)) option, I have no values labelled on my y-axis.
For this graph ggplot2 only puts labels the y-axis at intervals of 10 (i.e.
0, 10, 20, ...). However, the major portion of the graph I am interested in
is located between the values of 0 and 5 on the y-axis (thus why I am
zoooming). How can I coerce ggplot2 into
2008 Nov 17
2
re sults from "do.call" function
Dear R users...
I made this by help of one of R users.
_________________________________________________________________
X=matrix(seq(1,4), 2 , 2)
B=matrix(c(0.6,1.0,2.5,1.5) , 2 , 2)
func <- function(i,y0,j) { y0*exp(X[i,]%*%B[,j]) }
list1 <- expand.grid( i=c(1,2) , y0=c(1,2) , j=c(1,2) )
results <- do.call( func , list1 )
2011 Aug 13
3
optimization problems
Dear R users
I am trying to use OPTIMX(OPTIM) for nonlinear optimization.
There is no error in my code but the results are so weird (see below).
When I ran via OPTIM, the results are that
Initial values are that theta0 = 0.6 1.6 0.6 1.6 0.7. (In fact true vales
are 0.5,1.0,0.8,1.2, 0.6.)
--------------------------------------------------------------------------------------------
>
2011 Aug 29
3
gradient function in OPTIMX
Dear R users
When I use OPTIM with BFGS, I've got a significant result without an error
message. However, when I use OPTIMX with BFGS( or spg), I've got the
following an error message.
----------------------------------------------------------------------------------------------------
> optimx(par=theta0, fn=obj.fy, gr=gr.fy, method="BFGS",
>
2009 Jul 27
2
Splitting matrix into several small matrices
Dear R users...
I need to split this matrix(or dataframe), for example,
z <- matrix(c(13,1,1,1,1,12,0,0,0,0,8,1,0,1,1,8,0,1,0,0,
10,1,1,1,1,3,0,1,0,0,3,1,0,1,1,6,1,1,1,1),8,5,byrow = T)
> z
[,1] [,2] [,3] [,4] [,5]
[1,] 13 1 1 1 1
[2,] 12 0 0 0 0
[3,] 8 1 0 1 1
[4,] 9 0 1 0 0
[5,] 10 1 1 1 1
2011 Jan 19
1
expand.grid
Hello list.
I feel like an idiot.
There exists a method called expand.grid which, from the documentation,
appears to do just what I want, but then it doesn''t, and I can''t get it to
behave.
Given a dataframe
dfr<-data.frame(c1=c("a", "b", NA, "a", "a"), c2=c("d", NA, "d", "e", "e"),
2009 Aug 22
1
computation of matrices in list of list
Dear R users,
I have the list as follows;
#------------------------------------------------------
> z
[[1]]
[[1]][[1]]
matrix(A)
[[1]][[2]]
matrix(B)
[[1]][[3]]
matrix(C)
[[2]]
[[2]][[1]]
matrix(D)
[[2]][[2]]
matrix(E)
[[2]][[3]]
matrix(F)
#---------------------------------------------
I'd like to compute
matrix(A)+matrix(B)+matrix(C)+matrix(D)+matrix(E)+matrix(F)
In
2009 Aug 27
1
ignore an error and go back to ....
Dear R users,
is there way to ignore an error and go back to 1st line?
I mean,
#-------------------------------
while (or repeat) ------------
{
1
2
.
.
.
6
}
#-----------------------------
For example, if I have an error in the 6th line, then I'd like to go back to
the 1st line.
I've already tried "try", but it didn't work.
Any suggestion will be greatly
2008 Aug 29
1
more efficient double summation...
Dear R users...
I made the R-code for this double summation computation
http://www.nabble.com/file/p19213599/doublesum.jpg
-------------------------------------------------
Here is my code..
sum(sapply(1:m, function(k){sum(sapply(1:m,
function(j){x[k]*x[j]*dnorm((mu[j]+mu[k])/sqrt(sig[k]+sig[j]))/sqrt(sig[k]+sig[j])}))}))
-------------------------------------------------
In fact, this is
2011 Aug 02
1
My R code is not efficient
Dear R users,
I have two n*1 integer vectors, y1 and y2, where n is very very large.
I'd like to compute
elbp = 4^(y1) * 5^(y2) * sum_{i=0}^{max(y1, y2)} [{ (y1-i)! * (i)! *
(y2-i)! }^(-1)];
that is, I need to compute "elbp" for each (y1, y2) pair.
So I made R code like below, but I don't think it's efficient
Would you plz tell me how to avoid this "for"
2009 Aug 20
1
how to compute this summation...
Dear R users,
I try to compute this summation,
http://www.nabble.com/file/p25054272/dd.jpg
where
f(y|x) = Negative Binomial(y, mu=exp(x' beta), size=1/alp)
http://www.nabble.com/file/p25054272/aa.jpg
http://www.nabble.com/file/p25054272/cc.jpg
In fact, I tried to use "do.call" function to compute each u(y,x) before the
summation, but I got an error, "Error in X[i, ]
2010 Apr 01
2
pdf files in loops
I need to make a bunch of PDF files of histograms. I tried
gatelist = unique(mdf$ArrivalGate)
for( gate in gatelist) {
outfile = paste("../", airport, "/", airport, "taxiHistogram", gate,
".pdf", sep="")
pdf(file = outfile, width = 10, height=8, par(lwd=1))
title=paste("Taxi time for Arrival Gate", gate, "by