I would like to extract the variance covariance matrix for variables b and c and have some struggles to find the right command. My original data frame has more then 100 variables. therefore to know a command to exctract that would be great
Given data:
a<-rnorm(1000, mean = 0, sd = 1)
b<-rnorm(1000, mean = 0, sd = 1)
c<-rnorm(1000, mean = 0, sd = 1)
d<-rbinom(1000, size = 1, prob = .5)
e<-rbinom(1000, size = 1, prob = .5)
f<-rbinom(1000, size = 1, prob = .5)
data<-data.frame(a,b,c,d,e,f)
test<-glm(a~b+c+d+e+f,data=data)
pe.glmCube<-test$coefficients[2:3] # point estimates
I tried the same with the variance matrix. But it seems senseless to do it that way:
vc.glmCube <- vcov(test[2:3]) # var-cov matrix
vcov(test)[c("b", "c"), c("b", "c")]
# b c
#b 1.083964e-03 -2.532682e-05
#c -2.532682e-05 9.779278e-04
Related
I am attempting to analyze a dataset using nonlinear least squares that has both measurement error and heteroscedasticity. I was able to fit the model using nls() in R and adjust for heteroscedasticity by weighting the observations by a power function of the fitted values (a technique known as the power mean method). However, when I additionally try to correct for measurement error using the simex package in R, I get the following error:
Error: measurement.error is constant 0 in column(s) 1
Which is strange because I have specified a nonzero measurement error. I have pasted example code below which reproduces this error.
library(simex)
set.seed(123456789)
x = runif(n = 1000, min = 1, max = 3.6)
x_err = x + rnorm(n = 1000, mean = 0, sd = 0.1)
y_mean = 100/(1+10^(log10(100)-x)*0.75)
y_het = y_mean + rnorm(n = 1000, mean = 0, sd = 10*x^-2)
y_het = ifelse(y_het > 0, y_het, 0)
w = (100/(1+10^(log10(100)-x_err)*0.75))^-2
nls_fit = nls(y_het ~ 100/(1+10^((log10(k)-x_err)*h)), start = list("k" = 100, "h" = 0.75), weights = 1/w)
simex(nls_fit, SIMEXvariable = "x_err", measurement.error = 0.1, asymptotic = F)
I am trying to graph a Bayesian Network (BN) with instantiated nodes using the libraries bnlearn and Rgraphviz. My workflow is as follow:
After creating a data frame with random data (the data I am actually using is obviously not random) I then discretise the data, structure learn the directed acyclic graph (DAG), fit the data to the DAG and then plot the DAG. I also plot a DAG which shows the posterior probabilities of each of the nodes.
#rm(list = ls())
library(bnlearn)
library(Rgraphviz)
# Generating random dataframe
data_clean <- data.frame(a = runif(min = 0, max = 100, n = 1000),
b = runif(min = 0, max = 100, n = 1000),
c = runif(min = 0, max = 100, n = 1000),
d = runif(min = 0, max = 100, n = 1000),
e = runif(min = 0, max = 100, n = 1000))
# Discretising the data into 3 bins
bins <- 3
data_discrete <- discretize(data_clean, breaks = bins)
# Creating factors for each bin in the data
lv <- c("low", "med", "high")
for (i in names(data_discrete)){
levels(data_discrete[, i]) = lv
}
# Structure learning the DAG from the training set
whitelist <- matrix(c("a", "b",
"b", "c",
"c", "e",
"a", "d",
"d", "e"),
ncol = 2, byrow = TRUE, dimnames = list(NULL, c("from", "to")))
bn.hc <- hc(data_discrete, whitelist = whitelist)
# Plotting the DAG
dag.hc <- graphviz.plot(bn.hc,
layout = "dot")
# Fitting the data to the structure
fitted <- bn.fit(bn.hc, data = data_discrete, method = "bayes")
# Plotting the DAG with posteriors
graphviz.chart(fitted, type = "barprob", layout = "dot")
The next thing I do is to manually change the distributions in the bn.fit object, assigned to fitted, and then plot a DAG that shows the instantiated nodes and the updated posterior probability of the response variable e.
# Manually instantiating
fitted_evidence <- fitted
cpt.a = matrix(c(1, 0, 0), ncol = 3, dimnames = list(NULL, lv))
cpt.c = c(1, 0, 0,
0, 1, 0,
0, 0, 1)
dim(cpt.c) <- c(3, 3)
dimnames(cpt.c) <- list("c" = lv, "b" = lv)
cpt.b = c(1, 0, 0,
0, 1, 0,
0, 0, 1)
dim(cpt.b) <- c(3, 3)
dimnames(cpt.b) <- list("b" = lv, "a" = lv)
cpt.d = c(0, 0, 1,
0, 1, 0,
1, 0, 0)
dim(cpt.d) <- c(3, 3)
dimnames(cpt.d) <- list("d" = lv, "a" = lv)
fitted_evidence$a <- cpt.a
fitted_evidence$b <- cpt.b
fitted_evidence$c <- cpt.c
fitted_evidence$d <- cpt.d
# Plotting the DAG with instantiation and posterior for response
graphviz.chart(fitted_evidence, type = "barprob", layout = "dot")
This is the result I get but my actual BN is much larger with many more arcs and it would be impractical to manually change the bn.fit object.
I would like to find out if there is a way to plot a DAG with instantiation without changing the bn.fit object manually? Is there a workaround or function that I am missing?
I think/hope I have read the documentation for bnlearn thoroughly. I appreciate any feedback and would be happy to change anything in the question if I have not conveyed my thoughts clearly enough.
Thank you.
How about using cpdist to draw samples from the posterior given the evidence. You can then estimate the updated parameters using bn.fit using the cpdist samples. Then plot as before.
An example:
set.seed(69184390) # for sampling
# Your evidence vector
ev <- list(a = "low", b="low", c="low", d="high")
# draw samples
updated_dat <- cpdist(fitted, nodes=bnlearn::nodes(fitted), evidence=ev, method="lw", n=1e6)
# refit : you'll get warnings over missing levels
updated_fit <- bn.fit(bn.hc, data = updated_dat)
# plot
par(mar=rep(0,4))
graphviz.chart(updated_fit, type = "barprob", layout = "dot")
Note I used bnlearn::nodes as nodes is masked by a dependency of Rgraphviz. I tend to load bnlearn last.
I want to repeatedly sample values based on a certain condition. For example I want to create a sample of 100 values.
With probability of 0.7 it will be sampled from one distribution, and from another probability, otherwise.
Here is a way to do what I want:
set.seed(20)
A<-vector()
for (i in 1:100){
A[i]<-ifelse(runif(1,0,1)>0.7,rnorm(1, mean = 100, sd = 20),runif(1, min = 0, max = 1))
}
I am sure there are other more elegant ways, without using for loop.
Any suggestions?
You can sample an indiactor, which defines what distribution you draw from.
ind <- sample(0:1, size = 100, prob = c(0.3, 0.7), replace = TRUE)
A <- ind * rnorm(100, mean = 100, sd = 20) + (1 - ind) * runif(100, min = 0, max = 1)
In this case you don't use a for-loop but you need to sample more random variables.
If the percentage of times is not random, you can draw the right amount of each distribution then shuffle the result :
n <- 100
A <- sample(c(rnorm(0.7*n, mean = 100, sd = 20), runif(0.3*n, min = 0, max = 1)))
I was debugging my simulation and I find that when I run rnorm(), my random normal values don't look random to me at all. ccc is the mean sd vector that is given parametrically. How can I get really random normal realizations? Since my original simulation is quite long, I don't want to go into Gibbs sampling... Should you know why I get non-random looking realizations of normal random variables?
> ccc
# [1] 144.66667 52.52671
> rnorm(20, ccc)
# [1] 144.72325 52.31605 144.44628 53.07380 144.64438 53.87741 144.91300 54.06928 144.76440
# [10] 52.09181 144.61817 52.17339 145.01374 53.38597 145.51335 52.37353 143.02516 52.49332
# [19] 144.27616 54.22477
> rnorm(20, ccc)
# [1] 143.88539 52.42435 145.24666 50.94785 146.10255 51.59644 144.04244 51.78682 144.70936
# [10] 53.51048 143.63903 51.25484 143.83508 52.94973 145.53776 51.93892 144.14925 52.35716
# [19] 144.08803 53.34002
It's a basic concept to set parameters in a function. Take rnorm() for example:
Its structure is rnorm(n, mean = 0, sd = 1). Obviously, mean and sd are two different parameters, so you need to put respective values to them. Here is a confusing situation where you may get stuck:
arg <- c(5, 10)
rnorm(1000, arg)
This actually means rnorm(n = 1000, mean = c(5, 10), sd = 1). The standard deviation is set to 1 because the position of arg represents the parameter mean and you don't set sd additionally. Therefore, rnorm() will take the default value 1 to sd. However, what does mean = c(5, 10) mean? Let's check:
x <- rnorm(1000, arg)
hist(x, breaks = 50, prob = TRUE)
# lines(density(x), col = 2, lwd = 2)
mean = c(5, 10) and sd = 1 will recycle to length 1000, i.e.
rnorm(n = 1000, mean = c(5, 10, 5, 10, ...), sd = c(1, 1, 1, 1, ...))
and hence the final sample x is actually a blend of 500 N(5, 1) samples and 500 N(10, 1) samples which are drawn alternately, i.e.
c(rnorm(1, 5, 1), rnorm(1, 10, 1), rnorm(1, 5, 1), rnorm(1, 10, 1), ...)
As for your question, it should be:
arg <- c(5, 10)
rnorm(1000, arg[1], arg[2])
and this means rnorm(n = 1000, mean = 5, sd = 10). Check it again, and you will get a normal distribution with mean = 5 and sd = 10.
x <- rnorm(1000, arg[1], arg[2])
hist(x, breaks = 50, prob = T)
# curve(dnorm(x, arg[1], arg[2]), col = 2, lwd = 2, add = T)
I am trying to create a data frame in R, with a set of variables that are normally distributed. Firstly, we only create the data frame with the following variables:
RootCause <- rnorm(500, 0, 9)
OtherThing <- rnorm(500, 0, 9)
Errors <- rnorm(500, 0, 4)
df <- data.frame(RootCuase, OtherThing, Errors)
In the second part, we're asked to redo the above, but with a defined correlation between RootCause and OtherThing of 0.5. I have tried reading through a couple of pages and articles explaining correlation commands in R, but I am afraid I am struggling with comprehending it.
Easy answer
Draw another random variable OmittedVar and add it to the other variables:
n <- 1000
OmittedVar <- rnorm(n, 0, 9)
RootCause <- rnorm(n, 0, 9) + OmittedVar
OtherThing <- rnorm(n, 0, 9) + OmittedVar
Errors <- rnorm(n, 0, 4)
cor(RootCause, OtherThing)
[1] 0.4942716
Other answer: use multivariate normal function from MASS package:
But you have to define the variance/covariance matrix that gives you the correlation you like (the Sigma argument here):
d <- MASS::mvrnorm(n = n, mu = c(0, 0), Sigma = matrix(c(9, 4.5, 4.5, 9), nrow = 2, ncol = 2), tol = 1e-6, empirical = FALSE, EISPACK = FALSE)
cor(d[,1], d[,2])
[1] 0.5114698
Note:
Getting a correlation other than 0.5 depends on the process; if you want to change it from 0.5, you'll change the details (from adding 1 * OmittedVar in the first strat or changing Sigma in the second strat). But you'll have to look up details on variance rulse of the normal distribution.