Using fmsb to calculate incidence rate ratios.
Example:
rr <- rateratio(853, 7703, 284, 3508)
rr
Cases Person-time
Exposed 853 284
Unexposed 7703 3508
Total 8556 3792
Incidence rate ratio estimate and its significance probability
data: 853 7703 284 3508
p-value <2.2e-16
95 percent confidence interval:<br/>
1.274426 1.468068
sample estimates:
[1] 1.367824
Then when I try to return the values to store later in a table:
p <- rr$p.value
p
[1] 0
How can I change the format so that it returns the p-value with more digits? Preferably 3 or 4 or in scientific notation.
fmsb::rateratio() does an imprecise computation of the tail value of the Normal distribution, resulting in the value underflowing to zero. You can fix this yourself (it would be public-spirited to contact the maintainer (maintainer("fmsb")) to suggest this improvement to the package).
Save the source code of the function to a file:
dump("rateratio",file="my_rateratio.R")
Edit the file in a text editor (or RStudio or whatever) and change line 14, which is
p.v <- 2 * (1 - pnorm(abs(.CHI)))
to
p.v <- 2 * pnorm(abs(.CHI), lower=FALSE)
(which is an equivalent, but more accurate computation: see ?pnorm or here). Then read the file back in:
source("my_rateratio.R")
(this creates a new version of rateratio() in your global workspace, masking the version from fmsb).
Now if you re-run the code you'll see that rr$p.value is 2.893638e-18.
I am using cuhre from R2Cuba 1.1-0 for integrating the following function
fn <- function(x) {
pnorm((-2-sum(sqrt(vecRho)*x))/sqrt(1-sum(vecRho)))*prod(dnorm(x))
}
where vecRho is a vector of 6 numbers between 0 and 0.1, i.e.
vecRho<-runif(6,0,0.1)
By definition, the integrand fn is between 0 and 1. The integration is expected to be positive. However, using cuhre the result becomes negative when the length of vecRho exceeds 5.
NDIM<-length(vecRho)
cuhre(NDIM, 1, fn,
flags = list(verbose =0),
lower = rep(-10,NDIM),
upper = rep(10,NDIM))$value
[1] -0.4738284
Moreover, when the length of vecRho >=6 the absolute value of the integration increases as the length of vecRho increases.
Is there something I can do to fix this? Thanks!
ok, got it, you have 6D integral with 6 gaussian kernels inside. I know good answer for 1D integration with gaussian kernel. It is called Gauss-Hermite quadrature, and there is an R package for that. If you go this route, you'll have to
make curry functions, but it might worth it.
Some sample code:
library(gaussquad)
n.quad <- 128 # integration order
# get the particular (weights,abscissas) as data frame
# with 2 observables and n.quad observations
rule <- ghermite.h.quadrature.rules(n.quad, mu = 0.0)[[n.quad]]
# test function - integrate 1 over exp(-x^2) from -Inf to Inf
# should get sqrt(pi) as an answer
f <- function(x) {
1.0
}
q <- ghermite.h.quadrature(f, rule)
print(q - sqrt(pi))
I'm looking into GoF (goodness of fit) testing, and wanted to see if the quantiles of a vector of data followed the expected frequency of a normal distribution N(0, 1), and before running the chi square test, I generated these frequencies for the normal distribution:
< -2 SD's (standard deviations), between -2 and -1 SD's, between -1 and 0 SD's, between 0 and 1 SD's, between 1 and 2 SD's, and more than 2 SD's.
To do so I took the long route:
(Normal_distr <- c(pnorm(-2), pnorm(-1) - pnorm(-2), pnorm(0) - pnorm(-1),
pnorm(1) - pnorm(0), pnorm(2) - pnorm(1), pnorm(2, lower.tail = F)))
[1] 0.02275013 0.13590512 0.34134475 0.34134475 0.13590512 0.02275013
I see that the symmetry allows me to cut down the length of the code, but isn't there an easier way... something (I don't think this will work, but the idea of...) like pnorm(-2:-1) returning an identical value to pnorm(-1) - pnorm(-2) = 0.13590512?
Question: Is there an R function that calculates the area under the normal curve between quantiles so that we can pass a vector such as c(-3:3) through it, as opposed to subtracting pnorm()'s of adjacent standard deviations or other quantiles?
I'n not sure if there is a specific function to do this, but you can do it pretty simply like so:
#Get difference between adjacent quantiles
diff(pnorm(-2:-1))
[1] 0.1359051
#Get area under normal curve from -2 to 2 sd's:
sum(diff(pnorm(-2:2)))
[1] 0.9544997
Im attempting fitting a powerlaw distribution to a data set, using the method outlined by Aaron Clauset, Cosma Rohilla Shalizi and M.E.J. Newman in their paper "Power-Law Distributions in Empirical Data".
I've found code to compare to my own, but im a bit mystified where some of it comes from, the story thus far is,
to identify a suitable xmin for the powerlaw fit, we take each possible xmin fit a powerlaw to that data and then compute the corresponding exponet (a) then the KS statistic (D) for the fit and the observed data, then find the xmin that corresponds to the minimum of D. The KS statistic if computed as follows,
cx <- c(0:(n-1))/n # n is the sample size for the data >= xmin
cf <- 1-(xmin/z)^a # the cdf for a powerlaw z = x[x>=xmin]
D <- max(abs(cf-cx))
what i dont get is where cx comes for, surely we should be comparing the distance between the empirical distributions and the calculated distribution. something along the lines of:
cx = ecdf(sort(z))
cf <- 1-(xmin/z)^a
D <- max(abs(cf-cx(z)))
I think im just missing something very basic but please do correct me!
The answer is that they are (almost) the same. The easiest way to see this is to generate some data:
z = sort(runif(5,xmin, 10*xmin))
n = length(x)
Then examine the values of the two CDFs
R> (cx1 = c(0:(n-1))/n)
[1] 0.0 0.2 0.4 0.6 0.8
R> (cx2 = ecdf(sort(z)))
[1] 0.2 0.4 0.6 0.8 1.0
Notice that they are almost the same - essentially the cx1 gives the CDF for greater than or equal to whilst cx2 is greater than.
The advantage of the top approach is that it is very efficient and quick to calculate. The disadvantage is that if your data isn't truly continuous, i.e. z=c(1,1,2), cx1 is wrong. But then you shouldn't be fitting your data to a CTN distribution if this were the case.
As an assignment I had to develop and algorithm and generate a samples for a given geometric distribution with PMF
Using the inverse transform method, I came up with the following expression for generating the values:
Where U represents a value, or n values depending on the size of the sample, drawn from a Unif(0,1) distribution and p is 0.3 as stated in the PMF above.
I have the algorithm, the implementation in R and I already generated QQ Plots to visually assess the adjustment of the empirical values to the theoretical ones (generated with R), i.e., if the generated sample follows indeed the geometric distribution.
Now I wanted to submit the generated sample to a goodness of fit test, namely the Chi-square, yet I'm having trouble doing this in R.
[I think this was moved a little hastily, in spite of your response to whuber's question, since I think before solving the 'how do I write this algorithm in R' problem, it's probably more important to deal with the 'what you're doing is not the best approach to your problem' issue (which certainly belongs where you posted it). Since it's here, I will deal with the 'doing it in R' aspect, but I would urge to you go back an ask about the second question (as a new post).]
Firstly the chi-square test is a little different depending on whether you test
H0: the data come from a geometric distribution with parameter p
or
H0: the data come from a geometric distribution with parameter 0.3
If you want the second, it's quite straightforward. First, with the geometric, if you want to use the chi-square approximation to the distribution of the test statistic, you will need to group adjacent cells in the tail. The 'usual' rule - much too conservative - suggests that you need an expected count in every bin of at least 5.
I'll assume you have a nice large sample size. In that case, you'll have many bins with substantial expected counts and you don't need to worry so much about keeping it so high, but you will still need to choose how you will bin the tail (whether you just choose a single cut-off above which all values are grouped, for example).
I'll proceed as if n were say 1000 (though if you're testing your geometric random number generation, that's pretty low).
First, compute your expected counts:
dgeom(0:20,.3)*1000
[1] 300.0000000 210.0000000 147.0000000 102.9000000 72.0300000 50.4210000
[7] 35.2947000 24.7062900 17.2944030 12.1060821 8.4742575 5.9319802
[13] 4.1523862 2.9066703 2.0346692 1.4242685 0.9969879 0.6978915
[19] 0.4885241 0.3419669 0.2393768
Warning, dgeom and friends goes from x=0, not x=1; while you can shift the inputs and outputs to the R functions, it's much easier if you subtract 1 from all your geometric values and test that. I will proceed as if your sample has had 1 subtracted so that it goes from 0.
I'll cut that off at the 15th term (x=14), and group 15+ into its own group (a single group in this case). If you wanted to follow the 'greater than five' rule of thumb, you'd cut it off after the 12th term (x=11). In some cases (such as smaller p), you might want to split the tail across several bins rather than one.
> expec <- dgeom(0:14,.3)*1000
> expec <- c(expec, 1000-sum(expec))
> expec
[1] 300.000000 210.000000 147.000000 102.900000 72.030000 50.421000
[7] 35.294700 24.706290 17.294403 12.106082 8.474257 5.931980
[13] 4.152386 2.906670 2.034669 4.747562
The last cell is the "15+" category. We also need the probabilities.
Now we don't yet have a sample; I'll just generate one:
y <- rgeom(1000,0.3)
but now we want a table of observed counts:
(x <- table(factor(y,levels=0:14),exclude=NULL))
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 <NA>
292 203 150 96 79 59 47 25 16 10 6 7 0 2 5 3
Now you could compute the chi-square directly and then calculate the p-value:
> (chisqstat <- sum((x-expec)^2/expec))
[1] 17.76835
(pval <- pchisq(chisqstat,15,lower.tail=FALSE))
[1] 0.2750401
but you can also get R to do it:
> chisq.test(x,p=expec/1000)
Chi-squared test for given probabilities
data: x
X-squared = 17.7683, df = 15, p-value = 0.275
Warning message:
In chisq.test(x, p = expec/1000) :
Chi-squared approximation may be incorrect
Now the case for unspecified p is similar, but (to my knowledge) you can no longer get chisq.test to do it directly, you have to do it the first way, but you have to estimate the parameter from the data (by maximum likelihood or minimum chi-square), and then test as above but you have one fewer degree of freedom for estimating the parameter.
See the example of doing a chi-square for a Poisson with estimated parameter here; the geometric follows the much same approach as above, with the adjustments as at the link (dealing with the unknown parameter, including the loss of 1 degree of freedom).
Let us assume you've got your randomly-generated variates in a vector x. You can do the following:
x <- rgeom(1000,0.2)
x_tbl <- table(x)
x_val <- as.numeric(names(x_tbl))
x_df <- data.frame(count=as.numeric(x_tbl), value=x_val)
# Expand to fill in "gaps" in the values caused by 0 counts
all_x_val <- data.frame(value = 0:max(x_val))
x_df <- merge(all_x_val, x_df, by="value", all.x=TRUE)
x_df$count[is.na(x_df$count)] <- 0
# Get theoretical probabilities
x_df$eprob <- dgeom(x_df$val, 0.2)
# Chi-square test: once with asymptotic dist'n,
# once with bootstrap evaluation of chi-sq test statistic
chisq.test(x=x_df$count, p=x_df$eprob, rescale.p=TRUE)
chisq.test(x=x_df$count, p=x_df$eprob, rescale.p=TRUE,
simulate.p.value=TRUE, B=10000)
There's a "goodfit" function described as "Goodness-of-fit Tests for Discrete Data" in package "vcd".
G.fit <- goodfit(x, type = "nbinomial", par = list(size = 1))
I was going to use the code you had posted in an earlier question, but it now appears that you have deleted that code. I find that offensive. Are you using this forum to gather homework answers and then defacing it to remove the evidence? (Deleted questions can still be seen by those of us with sufficient rep, and the interface prevents deletion of question with upvoted answers so you should not be able to delete this one.)
Generate a QQ Plot for testing a geometrically distributed sample
--- question---
I have a sample of n elements generated in R with
sim.geometric <- function(nvals)
{
p <- 0.3
u <- runif(nvals)
ceiling(log(u)/log(1-p))
}
for which i want to test its distribution, specifically if it indeed follows a geometric distribution. I want to generate a QQ PLot but have no idea how to.
--------reposted answer----------
A QQ-plot should be a straight line when compared to a "true" sample drawn from a geometric distribution with the same probability parameter. One gives two vectors to the functions which essentially compares their inverse ECDF's at each quantile. (Your attempt is not particularly successful:)
sim.res <- sim.geometric(100)
sim.rgeom <- rgeom(100, 0.3)
qqplot(sim.res, sim.rgeom)
Here I follow the lead of the authors of qqplot's help page (which results in flipping that upper curve around the line of identity):
png("QQ.png")
qqplot(qgeom(ppoints(100),prob=0.3), sim.res,
main = expression("Q-Q plot for" ~~ {G}[n == 100]))
dev.off()
---image not included---
You can add a "line of good fit" by plotting a line through through the 25th and 75th percentile points for each distribution. (I added a jittering feature to this to get a better idea where the "probability mass" was located:)
sim.res <- sim.geometric(500)
qqplot(jitter(qgeom(ppoints(500),prob=0.3)), jitter(sim.res),
main = expression("Q-Q plot for" ~~ {G}[n == 100]), ylim=c(0,max( qgeom(ppoints(500),prob=0.3),sim.res )),
xlim=c(0,max( qgeom(ppoints(500),prob=0.3),sim.res )))
qqline(sim.res, distribution = function(p) qgeom(p, 0.3),
prob = c(0.25, 0.75), col = "red")