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
Dear Michael,
Thanks very much for your explicit explanation! That makes much sense.
Best wishes,
Jeff
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Thank you for this useful code!
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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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locpoly() might be useful
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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
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]) ##
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
Frank,
Have you tried the R function overlay(), it is exactly applying to your
question.
Jeff
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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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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
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(),
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
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