Dear Michael,
Thanks very much for your explicit explanation! That makes much sense.
Best wishes,
Jeff
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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 co
Thank you for this useful code!
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h
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
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View this me
locpoly() might be useful
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https://stat.ethz.ch/m
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 seri
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]) ## u
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.
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Frank,
Have you tried the R function overlay(), it is exactly applying to your
question.
Jeff
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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 func
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
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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
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 nor
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
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 lin
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