Using R, it is trivial to calculate the quantiles for given probabilities in a sampled distribution:
x <- rnorm(1000, mean=4, sd=2)
quantile(x, .9) # results in 6.705755
However, I can't find an easy way to do the inverse—calculate the probability for a given quantile in the sample x. The closest I've come is to use pnorm() with the same mean and standard deviation I used when creating the sample:
pnorm(5, mean=4, sd=2) # results in 0.6914625
However, because this is calculating the probability from the full normal distribution, and not the sample x, it's not entirely accurate.
Is there a function that essentially does the inverse of quantile()? Something that essentially lets me do the same thing as pnorm() but with a sample? Something like this:
backwards_quantile(x, 5)
I've found the ecdf() function, but can't figure out a way to make it result in a single probability instead of a full equation object.
ecdf returns a function: you need to apply it.
f <- ecdf(x)
f( quantile(x,.91) )
# Equivalently:
ecdf(x)( quantile(x,.91) )
Just for convenience, this function helps:
quantInv <- function(distr, value) ecdf(distr)(value)
set.seed(1)
x <- rnorm(1000, mean=4, sd=2)
quantInv(x, c(4, 5, 6.705755))
[1] 0.518 0.685 0.904
You more or less have the answer yourself. When you want to write
backwards_quantile(x, 5)
just write
ecdf(x)(5)
This corresponds to the inverse of quantile() with type=1. However, if you want other types (I favour the NIST standard, corresponding to Excel's Percentile.exc, which is type=6), you have more work to do.
In these latter cases, consider which use you are going to put it to. If all you want is to plot it, for instance, then consider
yVals<-seq(0,1,0.01)
plot(quantile(x,yVals,type=6))
But if you want the inverse for a single value, like 5, then you need to write a solving function to find the P that makes
quantile(x,P,type=6) = 5
For instance this, which uses binary search between the extreme values of x:
inverse_quantile<-function(x,y,d=0.01,type=1) {
A<-min(x)
B<-max(x)
k<-(log((B-A)/d)/log(2))+1
P=0.5
for (i in 1:k) {
P=P+ifelse((quantile(x,P,type=type)<y),2^{-i-1},-2^{-i-1})
}
P
}
So if you wanted the type 4 quantile of your set x for the number 5, with precision 0.00001, then you would write
inverse_quantile<-function(x,5,d=0.00001,type=4)
Related
I apply the sensitivity package in R. In particular, I want to use sobolroalhs as it uses a sampling procedure for inputs that allow for evaluations of models with a large number of parameters. The function samples uniformly [0,1] for all inputs. It is stated that desired distributions need to be obtained as follows
####################
# Test case: dealing with non-uniform distributions
x <- sobolroalhs(model = NULL, factors = 3, N = 1000, order =1, nboot=0)
# X1 follows a log-normal distribution:
x$X[,1] <- qlnorm(x$X[,1])
# X2 follows a standard normal distribution:
x$X[,2] <- qnorm(x$X[,2])
# X3 follows a gamma distribution:
x$X[,3] <- qgamma(x$X[,3],shape=0.5)
# toy example
toy <- function(x){rowSums(x)}
y <- toy(x$X)
tell(x, y)
print(x)
plot(x)
I have non-zero mean and standard deviations for some input parameter that I want to sample out of a normal distribution. For others, I want to uniformly sample between a defined range (e.g. [0.03,0.07] instead [0,1]). I tried using built in R functions such as
SA$X[,1] <- rnorm(1000, mean = 579, sd = 21)
but I am afraid this procedure messes up the sampling design of the package and resulted in odd results for the sensitivity indices. Hence, I think I need to adhere for the uniform draw of the sobolroalhs function in which and use the sampled value between [0, 1] when drawing out of the desired distribution (I think as density draw?). Does this make sense to anyone and/or does anyone know how I could sample out of the right distributions following the syntax from the package description?
You can specify mean and sd in qnorm. So modify lines like this:
x$X[,2] <- qnorm(x$X[,2])
to something like this:
x$X[,2] <- qnorm(x$X[,2], mean = 579, sd = 21)
Similarly, you could use the min and max parameters of qunif to get values in a given range.
Of course, it's also possible to transform standard normals or uniforms to the ones you want using things like X <- 579 + 21*Z or Y <- 0.03 + 0.04*U, where Z is a standard normal and U is standard uniform, but for some distributions those transformations aren't so simple and using the q* functions can be easier.
Forgive me if this has been asked before (I feel it must have, but could not find precisely what I am looking for).
Have can I draw one element of a vector of whole numbers (from 1 through, say, 10) using a probability function that specifies different chances of the elements. If I want equal propabilities I use runif() to get a number between 1 and 10:
ceiling(runif(1,1,10))
How do I similarly sample from e.g. the exponential distribution to get a number between 1 and 10 (such that 1 is much more likely than 10), or a logistic probability function (if I want a sigmoid increasing probability from 1 through 10).
The only "solution" I can come up with is first to draw e6 numbers from the say sigmoid distribution and then scale min and max to 1 and 10 - but this looks clumpsy.
UPDATE:
This awkward solution (and I dont feel it very "correct") would go like this
#Draw enough from a distribution, here exponential
x <- rexp(1e3)
#Scale probs to e.g. 1-10
scaler <- function(vector, min, max){
(((vector - min(vector)) * (max - min))/(max(vector) - min(vector))) + min
}
x_scale <- scaler(x,1,10)
#And sample once (and round it)
round(sample(x_scale,1))
Are there not better solutions around ?
I believe sample() is what you are looking for, as #HubertL mentioned in the comments. You can specify an increasing function (e.g. logit()) and pass the vector you want to sample from v as an input. You can then use the output of that function as a vector of probabilities p. See the code below.
logit <- function(x) {
return(exp(x)/(exp(x)+1))
}
v <- c(seq(1,10,1))
p <- logit(seq(1,10,1))
sample(v, 1, prob = p, replace = TRUE)
I'm trying to get a 95% confidence interval around some predicted values, but am not capable of achieving this.
Basically, I estimated a growth curve like this:
set.seed(123)
dat=data.frame(size=rnorm(50,10,3),age=rnorm(50,5,2))
S <- function(t,ts,C,K) ((C*K)/(2*pi))*sin(2*pi*(t-ts))
sommers <- function(t,Linf,K,t0,ts,C)
Linf*(1-exp(-K*(t-t0)-S(t,ts,C,K)+S(t0,ts,C,K)))
model <- nls(size~sommers(age,Linf,K,t0,ts,C),data=dat,
start=list(Linf=10,K=4.7,t0=2.2,C=0.9,ts=0.1))
I have independent size measurements, for which I would like to predict the age. Therefore, the inverse of the function, which is not very straightforward, I calculated like this:
model.out=coef(model)
S.out <- function(t)
((model.out[[4]]*model.out[[2]])/(2*pi))*sin(2*pi*(t-model.out[[5]]))
sommers.out <- function(t)
model.out[[1]]*(1-exp(-model.out[[2]]*(t-model.out[[3]])-S.out(t)+S.out(model.out[[3]])))
inverse = function (f, lower = -100, upper = 100) {
function (y) uniroot((function (x) f(x) - y), lower = lower, upper = upper)[1]
}
sommers.inverse = inverse(sommers.out, 0, 25)
x= sommers.inverse(10) #this works with my complete dataset, but not with this fake one
Although this works fine, I need to know the confidence interval (95%) around this estimate (x). For linear models there is for example "predict(... confidence=)". I could also bootstrap the function somehow to get the quantiles associated with the parameters (didn't find how), to then use the extremes of those to calculate the maximum and minimum values predictable. But that doesn't really look like the good way of doing this....
Any help would be greatly appreciated.
EDIT after answer:
So this worked (explained in the book of Ben Bolker, see answer):
vmat = mvrnorm(1000, mu = coef(mfit), Sigma = vcov(mfit))
dist = numeric(1000)
for (i in 1:1000) {dist[i] = sommers_inverse(9.938,vmat[i,])}
quantile(dist, c(0.025, 0.975))
On the rather bad fake data I gave, this works of course rather horrible. But on the real data (which I have a problem recreating), this is ok!
Unless I'm mistaken, you're going to have to use either regular (parametric) bootstrapping or a method called either "population predictive intervals" (e.g., see section 5 of chapter 7 of Bolker 2008), which assumes that the sampling distributions of your parameters are multivariate Normal. However, I think you may have bigger problems, unless I've somehow messed up your model in adapting it ...
Generate data (note that random data may actually bad for testing your model - see below ...)
set.seed(123)
dat <- data.frame(size=rnorm(50,10,3),age=rnorm(50,5,2))
S <- function(t,ts,C,K) ((C*K)/(2*pi))*sin(2*pi*(t-ts))
sommers <- function(t,Linf,K,t0,ts,C)
Linf*(1-exp(-K*(t-t0)-S(t,ts,C,K)+S(t0,ts,C,K)))
Plot the data and the initial curve estimate:
plot(size~age,data=dat,ylim=c(0,16))
agevec <- seq(0,10,length=1001)
lines(agevec,sommers(agevec,Linf=10,K=4.7,t0=2.2,ts=0.1,C=0.9))
I had trouble with nls so I used minpack.lm::nls.lm, which is slightly more robust. (There are other options here, e.g. calculating the derivatives and providing the gradient function, or using AD Model Builder or Template Model Builder, or using the nls2 package.)
For nls.lm we need a function that returns the residuals:
sommers_fn <- function(par,dat) {
with(c(as.list(par),dat),size-sommers(age,Linf,K,t0,ts,C))
}
library(minpack.lm)
mfit <- nls.lm(fn=sommers_fn,
par=list(Linf=10,K=4.7,t0=2.2,C=0.9,ts=0.1),
dat=dat)
coef(mfit)
## Linf K t0 C ts
## 10.6540185 0.3466328 2.1675244 136.7164179 0.3627371
Here's our problem:
plot(size~age,data=dat,ylim=c(0,16))
lines(agevec,sommers(agevec,Linf=10,K=4.7,t0=2.2,ts=0.1,C=0.9))
with(as.list(coef(mfit)), {
lines(agevec,sommers(agevec,Linf,K,t0,ts,C),col=2)
abline(v=t0,lty=2)
abline(h=c(0,Linf),lty=2)
})
With this kind of fit, the results of the inverse function are going to be extremely unstable, as the inverse function is many-to-one, with the number of inverse values depending sensitively on the parameter values ...
sommers_pred <- function(x,pars) {
with(as.list(pars),sommers(x,Linf,K,t0,ts,C))
}
sommers_pred(6,coef(mfit)) ## s(6)=9.93
sommers_inverse <- function (y, pars, lower = -100, upper = 100) {
uniroot(function(x) sommers_pred(x,pars) -y, c(lower, upper))$root
}
sommers_inverse(9.938, coef(mfit)) ## 0.28
If I pick my interval very carefully I can get back the correct answer ...
sommers_inverse(9.938, coef(mfit), 5.5, 6.2)
Maybe your model will be better behaved with more realistic data. I hope so ...
Using R, it is trivial to calculate the quantiles for given probabilities in a sampled distribution:
x <- rnorm(1000, mean=4, sd=2)
quantile(x, .9) # results in 6.705755
However, I can't find an easy way to do the inverse—calculate the probability for a given quantile in the sample x. The closest I've come is to use pnorm() with the same mean and standard deviation I used when creating the sample:
pnorm(5, mean=4, sd=2) # results in 0.6914625
However, because this is calculating the probability from the full normal distribution, and not the sample x, it's not entirely accurate.
Is there a function that essentially does the inverse of quantile()? Something that essentially lets me do the same thing as pnorm() but with a sample? Something like this:
backwards_quantile(x, 5)
I've found the ecdf() function, but can't figure out a way to make it result in a single probability instead of a full equation object.
ecdf returns a function: you need to apply it.
f <- ecdf(x)
f( quantile(x,.91) )
# Equivalently:
ecdf(x)( quantile(x,.91) )
Just for convenience, this function helps:
quantInv <- function(distr, value) ecdf(distr)(value)
set.seed(1)
x <- rnorm(1000, mean=4, sd=2)
quantInv(x, c(4, 5, 6.705755))
[1] 0.518 0.685 0.904
You more or less have the answer yourself. When you want to write
backwards_quantile(x, 5)
just write
ecdf(x)(5)
This corresponds to the inverse of quantile() with type=1. However, if you want other types (I favour the NIST standard, corresponding to Excel's Percentile.exc, which is type=6), you have more work to do.
In these latter cases, consider which use you are going to put it to. If all you want is to plot it, for instance, then consider
yVals<-seq(0,1,0.01)
plot(quantile(x,yVals,type=6))
But if you want the inverse for a single value, like 5, then you need to write a solving function to find the P that makes
quantile(x,P,type=6) = 5
For instance this, which uses binary search between the extreme values of x:
inverse_quantile<-function(x,y,d=0.01,type=1) {
A<-min(x)
B<-max(x)
k<-(log((B-A)/d)/log(2))+1
P=0.5
for (i in 1:k) {
P=P+ifelse((quantile(x,P,type=type)<y),2^{-i-1},-2^{-i-1})
}
P
}
So if you wanted the type 4 quantile of your set x for the number 5, with precision 0.00001, then you would write
inverse_quantile<-function(x,5,d=0.00001,type=4)
i'm comparing different measures of distance and similarity for vector profiles (Subtest results) in R, most of them are easy to compute and/or exist in dist().
Unfortunately, one that might be interesting and is to difficult for me to calculate myself is Cattel's Rp. I can not find it in R.
Does anybody know if this exists already?
Or can you help me to write a function?
The formula (Cattell 1994) of Rp is this:
(2k-d^2)/(2k + d^2)
where:
k is the median for chi square on a sample of size n;
d is the sum of the (weighted=m) difference between the two profiles,
sth like: sum(m(x(i)-y(i)));
one thing i don't know is, how to get the chi square median in there
Thank you
What i get without defining the k is:
Rp.Cattell <- function(x,y){z <- (2k-(sum(x-y))^2)/(2k+(sum(x-y))^2);return(z)}
Vector examples are:
x <- c(-1.2357,-1.1999,-1.4727,-0.3915,-0.2547,-0.4758)
y <- c(0.7785,0.9357,0.7165,-0.6067,-0.4668,-0.5925)
They are measures by the same device, but related to different bodyparts. They don't need to be standartised or weighted, i would say.
This page gives a general formula for k, and then gives a more thorough method using SAS/IML which pretty much gives the same results. So I used the general formula, added calculation of degrees of freedom, which leads to this:
Rp.Cattell <- function(x,y) {
dof <- (2-1) * (length(y)-1)
k <- (1-2/(9*dof))^3
z <- (2*k-sum(sum(x-y))^2)/(2*k+sum(sum(x-y))^2)
return(z)
}
x <- c(-1.2357,-1.1999,-1.4727,-0.3915,-0.2547,-0.4758)
y <- c(0.7785,0.9357,0.7165,-0.6067,-0.4668,-0.5925)
Rp.Cattell(x, y)
# [1] -0.9012083
Does this figure appear to make sense?
Trying to verify the function, I found out now that the median of chisquare is the chisquare value for 50% probability - relating to random. So the function should be:
Rp.Cattell <- function(x,y){
dof <- (2-1) * (length(y)-1)
k <- qchisq(.50, df=dof)
z <- (2k-(sum(x-y))^2)/(2k+(sum(x-y))^2);
return(z)}
It is necessary though to standardize the Values before, so the results are distributed correctly.
So:
library ("stringr")
# they are centered already
x <- as.vector(scale(c(-1.2357,-1.1999,-1.4727,-0.3915,-0.2547,-0.4758),center=F, scale=T))
y <- as.vector(scale(c(0.7785,0.9357,0.7165,-0.6067,-0.4668,-0.5925),center=F, scale=T))
Rp.Cattell(x, y) -0.584423
This sounds reasonable now - or not?
I consider calculation of z is incorrect.
You need to calculate the sum of the squared differences. Not the square of the sum of differences. Besides product operator is missing in 2k.
It should be
z <- (2*k-sum((x-y)^2))/(2*k+sum((x-y)^2))
Do you agree?