[R] How to generate a matrix of Beta or Binomial distribution
Hi folks,
I have a question about how to efficiently produce random numbers from Beta
and Binomial distributions.
For Beta distribution, suppose we have two shape vectors shape1 and shape2.
I hope to generate a 1 x 2 matrix X whose i th rwo is a sample from
reta(2,shape1[i]mshape2[i]). Of course this can be done via loops:
for(i in 1:1)
{
X[i,]=rbeta(2,shape1[i],shape2[i])
}
However, the above code is time consuming. It would be better to directly
generate X without any loops. I tried the following code
X- rbeta(2,shape1,shape2)
but it looks not right. So I was wondering if there is an R code doing this.
For Binomial distribution, I have a similar question. I hope to generate a
1 x n matrix Y whose i th row is a sample from rbinom(n,2,p[i]), where p
is a vector of binomial probabilities. We can also do it by loops
for(i in 1:1)
{
Y[i,]=rbinom(n,2,p[i])
}
But it would be nice to do it without using any loops. Is this possible in
R?
Many thanks for great help and suggestions!
Jeff
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Re: [R] How to generate a matrix of Beta or Binomial distribution
Dear Michael, Thanks very much for your explicit explanation! That makes much sense. Best wishes, Jeff -- View this message in context: http://r.789695.n4.nabble.com/How-to-generate-a-matrix-of-Beta-or-Binomial-distribution-tp4641422p4641472.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to add a variable to a graph title
Thank you for this useful code! -- View this message in context: http://r.789695.n4.nabble.com/how-to-add-a-variable-to-a-graph-title-tp4636364p4636405.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] how to add a variable to a graph title
Hi folks, To simplify my question, consider an example: x=runif(1,0,1) plot(1:5,1:5) Now I want to add a title to the above plot showing the vaue of x, so if the generated x is 0.3, graphically the tile should be like Figure 1: x=0.3. How do I achieve this in R? Thanks. Jeff -- View this message in context: http://r.789695.n4.nabble.com/how-to-add-a-variable-to-a-graph-title-tp4636364.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Local linear regression
locpoly() might be useful -- View this message in context: http://r.789695.n4.nabble.com/Local-linear-regression-tp4512566p4634166.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to find eigenfunctions and eigenvalues of a fourth order ODE
Dear all, I am having trouble with the following problem. Suppose we have the fourth order ODE with boundary conditions: http://r.789695.n4.nabble.com/file/n4591748/problem.jpg problem.jpg where q(t) is a known function. Note here the lambda parameter is changing, so essentially we have a series of ODEs. lambda is called an eigenvalue, the solution y is called an eigenfunction associated with lambda. The question is to find the eigenvalues lambda and eigenfunctions y. In matlab, this can be done. I am wondering if R has similar functions to achieve this goal. Any suggestions will be highly appreciated. Best regards, Jeff -- View this message in context: http://r.789695.n4.nabble.com/How-to-find-eigenfunctions-and-eigenvalues-of-a-fourth-order-ODE-tp4591748p4591748.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] efficiently finding the integrals of a sequence of functions
Thanks,Hans!
I agree that this is a good way of solving this problem.
Here is another way. Instead of defining a vector of uni-dimensional
functions and trying to integrating
each component (a uni-dimensional function), we can do something below
my.integrand-function(x,k)
{
return(f[x,k]) ## use matrix to represent the sequence of uni-dimensional
functions
}
my.integral-function(k) ## k=1,...,5000 denotes the function labels
{
return(integrate(my.integrand,lower=...,upper=...,k)$value)
}
When calculating the integrals, just perform
sapply(1:5000, my.integral)
This is a way of avoiding loops but the computing time needs to be carefully
examined.
Jeff
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[R] efficiently finding the integrals of a sequence of functions
Hi folks, I am having a question about efficiently finding the integrals of a list of functions. To be specific, here is a simple example showing my question. Suppose we have a function f defined by f-function(x,y,z) c(x,y^2,z^3) Thus, f is actually corresponding to three uni-dimensional functions f_1(x)=x, f_2(y)=y^2 and f_3(z)=z^3. What I am looking for are the integrals of these three functions f_1,f_2,f_3 over some interval, say, (0,1). More specifically, the integrals \int_0^1 f_1(x) dx, \int_0^1 f_2(y) dy and \int_0^1 f_3(z) dz. For this simple example, of course we can do these three integrals one by one using integrate (f_1, lower=0, upper=1) However, in practice, I have a sequence of 5000 uni-dimensional functions and hope to find all of their integrals (over some regions), so using loops to do this one by one is not efficient. A possible idea is to convert the sequence of functions to a list of objects, and use sapply() which allow us to find the integrals in a fashion of vectorizing. But I don't know how to convert the above example function f to a list. In my research problem, I can only define the 5000 functions in a vectorizing way like f-function(x1,x2,...,x5000) c(f1(x1),f2(x2),...,f5000(x5000)) So how to convert it to a list will be a crucial step to efficiently get the integrals. Thanks for all the suggestions! Jeff -- View this message in context: http://r.789695.n4.nabble.com/efficiently-finding-the-integrals-of-a-sequence-of-functions-tp4179452p4179452.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Overlaying density plot on forest plot
Frank, Have you tried the R function overlay(), it is exactly applying to your question. Jeff -- View this message in context: http://r.789695.n4.nabble.com/Overlaying-density-plot-on-forest-plot-tp4179654p4180430.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Regression Models
So your question is about fitting a regression model for all the subsets of predictors? Then there would be 2^13 submodesl? Probably leaps() does what you want. This function does a all-subset regresion. -- View this message in context: http://r.789695.n4.nabble.com/Regression-Models-tp4173278p4180447.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] A question about R image function
Hi Sarah, Thanks a lot! You are right, my data is not over a regular grid cell locations. One advantage of image() is that it can produce continuous color change for the data value. When the data value is over a large range, this will make it more convenient. Thanks! Jeff -- View this message in context: http://r.789695.n4.nabble.com/A-question-about-R-image-function-tp3875581p4102588.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Using GIS data in R
Hi Rolf, I have a similar question. I want to test whether a point with certain coordinates is inside a state, say Texas. It seems that inside.owin() only works for testing if a point lies in a regular region, say a polygon. Since Texas has irregular boundary, how do we achieve this? Or there is some alternative way we can use. Thanks! Jeff -- View this message in context: http://r.789695.n4.nabble.com/Using-GIS-data-in-R-tp1748266p4089242.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] how to use quadrature to integrate some complicated functions
Hello to all,
I am having trouble with intregrating a complicated uni-dimensional function
of the following form
Phi(x-a_1)*Phi(x-a_2)*...*Phi(x-a_{n-1})*phi(x-a_n).
Here n is about 5000, Phi is the cumulative distribution function of
standard normal,
phi is the density function of standard normal, and x ranges over
(-infty,infty).
My idea is to to use quadrature to handle this integral. But since Phi has
not cloaed form,
I don't know how to do this effeciently. I appreciate very much if someone
has any ideas about it.
Thanks!
Jeff
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[R] ploting pie charts centered at given points
Hi to all, Suppose we have a group of points on the plane, at each point, we want to draw a pie chart centered at that point. I have found that pie.labels() function might be useful. This function allows us to achieve my goal, but before using the function, we have to use the function plot(), say, the following example code: pieval-c(2,4,6,8) plot(1:5,type=n,axes=FALSE) box() bisect.angles-floating.pie(3,3,pieval) pie.labels(3,3,bisect.angles,c(two,four,six,eight)) I am wondering why use 1:5 in plot? Can I use for loop language to draw many pie charts on a plane? For this case, using 1:5 may not work. I also tried other numbers such like 1:10, 1:3, all failed. I stronglly appreciate your time and help! Jeff -- View this message in context: http://r.789695.n4.nabble.com/ploting-pie-charts-centered-at-given-points-tp3896771p3896771.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] A question about R image function
Dear folks, I have a question about the image() function in R. I found the following link talking about this but the replies didn't help with my situations. http://r.789695.n4.nabble.com/question-on-image-function-td839275.html#a839276 To be simple, I will keep using the example in the above link. Suppose the data are like xy mcpvalue 0.4603578 0.6247629 1.001 0.4603715 0.62477881.001 0.4603852 0.6247948 1.001 0.4110561 0.5664841 0.995 So we have four points with coordinates given by the four pairs of (x,y) values. Each point is associated with a mcpvalue. How do we use image() to plot mcpvalue as a two-dimensional plot? I hope that the points are positioned by the coordinates and the color of each point in that plot is changing with the value of mcpvalue. Using image() does not work as the coordinates of the points are not ascending. Thanks a lot! Jeff -- View this message in context: http://r.789695.n4.nabble.com/A-question-about-R-image-function-tp3875581p3875581.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
