mcmc | R package mcmc | Development Tools library
kandi X-RAY | mcmc Summary
kandi X-RAY | mcmc Summary
The main branch is morph (not master). We will probably never go back to master being the main branch. This is the source tree for the R contributed package mcmc. The version for users is at CRAN (This package suggests packages Iso and xtable. So if don't have need to add to R. In R do. Since one of the vignettes takes a long time, probably want. (instead of without --no-vignettes) except for one last check before commit.
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QUESTION
I have data that looks like this:
...ANSWER
Answered 2021-Jun-01 at 22:07We can use lapply
QUESTION
I’m trying to use PyMC3 Minibatch ADVI for Bayesian Regression. The pm.fit function throws the following error and I’m not sure how to fix it.
It says that the ‘str’ object has no attribute ‘type’. What is any ‘str’ object from the error message here? I’ve mapped float tensors for more_replacements to the best of what I know.
...ANSWER
Answered 2021-May-31 at 17:34The blog post you are working from shows
QUESTION
I have a multivariate model with this (approximate) form:
...ANSWER
Answered 2021-May-21 at 01:08The gelman.diag
requires a mcmc.list
. If we are running models with different set of parameters, extract the 'Sol' and place it in a list
(Below, it is the same model)
QUESTION
I am trying to use a custom VJP (vector-Jacobian product) function as a model for a HMC-NUTS in numpyro. I was able to make a single variable function that works for HMC-NUTS as follows:
...ANSWER
Answered 2021-Feb-11 at 03:08def model(x,y):
sigma = numpyro.sample('sigma', dist.Exponential(1.))
x0 = numpyro.sample('x0', dist.Uniform(-1.,1.))
A = numpyro.sample('A', dist.Exponential(1.))
hv=vmap(h,(0,None),0)
mu=hv(x-x0,A)
numpyro.sample('y', dist.Normal(mu, sigma), obs=y)
QUESTION
I used mcmc_trace
function from the bayesplot
package to plot traceplot with mcmc list, which is a ggplot item so it can be further edited by ggplot
function.
Follows is the plot that produced by the function. I needed to change the title k1...k[20] to subject 1... subject 20. Are there any approaches I can achieve this with ggplot function?
Follows is a simple reproducible model.
...ANSWER
Answered 2021-Apr-02 at 20:00Use colnames<-
to modify the column names. Since the object is a 1-element list containing a matrix-like object, you need to use [[1]]
; if you have multiple chains you'll need to lapply()
(or use a for
loop) to apply the solution to every chain (i.e., every element in the list).
QUESTION
I used r2jags
to run jags and do the sampling. Then, I used coda
package to check convergence. As I have many subjects and many parameters, I would need to manually type them as the codes below. Are there any easier approaches that I can for example, choose gamma
for every subject? Or, just select gamma
from subject 1 to 10 without typing them manually?
ANSWER
Answered 2021-Mar-31 at 20:03A reproducible example would be nice but presumably
QUESTION
I am trying to run an exponential random graph model (ergm) on a weighted network (network_ex). This network shows the interactions between individuals in four different groups. Interactions between groups cannot occur so a blockdiagonal constraint needs to be included in the model. However when I include the blockdiagonal constraint the ergm function (that runs otherwise) returns an error (see below) and my R session either runs indefinitely or abort directly.
Is there a way to run an ergm based on a weighted block diagonal matrix?
...ANSWER
Answered 2021-Mar-24 at 02:10That does appear to be a bug. I've fixed it in the latest development version. These can be obtained by going to https://github.com/statnet/ergm and https://github.com/statnet/ergm.count and either using install_github() or downloading the binaries found in the README. Note, also that, the blockdiag() constraint has been provisionally moved to https://github.com/statnet/tergm .
QUESTION
ANSWER
Answered 2021-Mar-14 at 11:09Something like this perhaps?
QUESTION
I use the str to print out information for a calculate variable
...ANSWER
Answered 2021-Mar-07 at 18:00The attr
implies there is an attribute named 'mcpar' which is a numeric
vector
QUESTION
I have data that is a 50:50 mix of a normal distribution and a constant value:
...ANSWER
Answered 2021-Mar-07 at 03:45The issue here is that the likelihood of coming from each model involves probability density for the Gaussian and mass for the discrete, which are not commensurate. Specifically, the computation for comparing where a zero observation came from, will involve likelihoods
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