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I'm confused by the coefficients produced by the output of lm
Here's a copy of the data I'm working with
(postprocessed.csv)
"","time","value"
"1",1,2.61066016308988
"2",2,3.41246054742996
"3",3,3.8608767964033
"4",4,4.28686048552237
"5",5,4.4923132964825
"6",6,4.50557049744317
"7",7,4.50944447661246
"8",8,4.51097373134893
"9",9,4.48788748823809
"10",10,4.34603985656981
"11",11,4.28677073671406
"12",12,4.20065901625172
"13",13,4.02514194962519
"14",14,3.91360194972916
"15",15,3.85865748409081
"16",16,3.81318053258601
"17",17,3.70380706527433
"18",18,3.61552922363713
"19",19,3.61405310598722
"20",20,3.64591327503384
"21",21,3.70234435835577
"22",22,3.73503970503372
"23",23,3.81003078640584
"24",24,3.88201196162666
"25",25,3.89872518158949
"26",26,3.97432743542362
"27",27,4.2523675144599
"28",28,4.34654855854847
"29",29,4.49276038902684
"30",30,4.67830892029687
"31",31,4.91896819673664
"32",32,5.04350767355202
"33",33,5.09073406942046
"34",34,5.18510849382162
"35",35,5.18353176529036
"36",36,5.2210776270173
"37",37,5.22643491929207
"38",38,5.11137006553725
"39",39,5.01052467981257
"40",40,5.0361056705898
"41",41,5.18149486951409
"42",42,5.36334869132276
"43",43,5.43053620818444
"44",44,5.60001072279525
I have fitted a 4th order polynomial to this data using the following script:
library(ggplot2)
library(matrixStats)
library(forecast)
df_input <- read.csv("postprocessed.csv")
x <- df_input$time
y <- df_input$value
df <- data.frame(x, y)
poly4model <- lm(y~poly(x, degree=4), data=df)
v <- seq(30, 40)
vv <- poly4model$coefficients[1] +
poly4model$coefficients[2] * v +
poly4model$coefficients[3] * (v ^ 2) +
poly4model$coefficients[4] * (v ^ 3) +
poly4model$coefficients[5] * (v ^ 4)
pdf("postprocessed.pdf")
plot(df)
lines(v, vv, col="red", pch=20, lw=3)
dev.off()
I initially tried using the predict function to do this, but couldn't get that to work, so resorted to implementing this "workaround" using some new vectors v and vv to store the data for the line in the region I am trying to plot.
Ultimatly, I am trying to do this:
Fit a 4th order polynomial to the data
Plot the 4th order polynomial over the range of data in one color
Plot the 4th order polynomial over the range from the last value to the last value + 10 (prediction) in a different color
At the moment I am fairly sure using v and vv to do this is not "the best way", however I would have thought it should work. What is happening is that I get very large values.
Here is a screenshot from Desmos. I copied and pasted the same coefficients as shown by typing poly4model$coefficients into the console. However, something must have gone wrong because this function is nothing like the data.
I think I've provided enough info to be able to run this short script. However I will add the pdf as well.
It is easiest to use the predict function to create your line. To do that, you pass the model and a data frame with the desired independent variables to the predict function.
x <- df_input$time
y <- df_input$value
df <- data.frame(x, y)
poly4model <- lm(y~poly(x, degree=4), data=df)
v <- seq(30, 40)
#Notice the column in the dataframe is the same variable name
# as the variable in the model!
predict(poly4model, data.frame(x=v))
plot(df)
lines(v, predict(poly4model, data.frame(x=seq(30, 40))), col="red", pch=20, lw=3)
NOTE
The function poly "Returns or evaluates orthogonal polynomials of degree 1 to degree over the specified set of points x: these are all orthogonal to the constant polynomial of degree 0." To return the "normal" polynomial coefficients one needs to use the "raw=TRUE" option in the function.
poly4model <- lm(y~poly(x, degree=4, raw=TRUE), data=df)
Now your equation above will work.
I have frequency values changing with the time (x axis units), as presented on the picture below. After some normalization these values may be seen as data points of a density function for some distribution.
Q: Assuming that these frequency points are from Weibull distribution T, how can I fit best Weibull density function to the points so as to infer the distribution T parameters from it?
sample <- c(7787,3056,2359,1759,1819,1189,1077,1080,985,622,648,518,
611,1037,727,489,432,371,1125,69,595,624)
plot(1:length(sample), sample, type = "l")
points(1:length(sample), sample)
Update.
To prevent from being misunderstood, I would like to add little more explanation. By saying I have frequency values changing with the time (x axis units) I mean I have data which says that I have:
7787 realizations of value 1
3056 realizations of value 2
2359 realizations of value 3 ... etc.
Some way towards my goal (incorrect one, as I think) would be to create a set of these realizations:
# Loop to simulate values
set.values <- c()
for(i in 1:length(sample)){
set.values <<- c(set.values, rep(i, times = sample[i]))
}
hist(set.values)
lines(1:length(sample), sample)
points(1:length(sample), sample)
and use fitdistr on the set.values:
f2 <- fitdistr(set.values, 'weibull')
f2
Why I think it is incorrect way and why I am looking for a better solution in R?
in the distribution fitting approach presented above it is assumed that set.values is a complete set of my realisations from the distribution T
in my original question I know the points from the first part of the density curve - I do not know its tail and I want to estimate the tail (and the whole density function)
Here is a better attempt, like before it uses optim to find the best value constrained to a set of values in a box (defined by the lower and upper vectors in the optim call). Notice it scales x and y as part of the optimization in addition to the Weibull distribution shape parameter, so we have 3 parameters to optimize over.
Unfortunately when using all the points it pretty much always finds something on the edges of the constraining box which indicates to me that maybe Weibull is maybe not a good fit for all of the data. The problem is the two points - they ares just too large. You see the attempted fit to all data in the first plot.
If I drop those first two points and just fit the rest, we get a much better fit. You see this in the second plot. I think this is a good fit, it is in any case a local minimum in the interior of the constraining box.
library(optimx)
sample <- c(60953,7787,3056,2359,1759,1819,1189,1077,1080,985,622,648,518,
611,1037,727,489,432,371,1125,69,595,624)
t.sample <- 0:22
s.fit <- sample[3:23]
t.fit <- t.sample[3:23]
wx <- function(param) {
res <- param[2]*dweibull(t.fit*param[3],shape=param[1])
return(res)
}
minwx <- function(param){
v <- s.fit-wx(param)
sqrt(sum(v*v))
}
p0 <- c(1,200,1/20)
paramopt <- optim(p0,minwx,gr=NULL,lower=c(0.1,100,0.01),upper=c(1.1,5000,1))
popt <- paramopt$par
popt
rms <- paramopt$value
tit <- sprintf("Weibull - Shape:%.3f xscale:%.1f yscale:%.5f rms:%.1f",popt[1],popt[2],popt[3],rms)
plot(t.sample[2:23], sample[2:23], type = "p",col="darkred")
lines(t.fit, wx(popt),col="blue")
title(main=tit)
You can directly calculate the maximum likelihood parameters, as described here.
# Defining the error of the implicit function
k.diff <- function(k, vec){
x2 <- seq(length(vec))
abs(k^-1+weighted.mean(log(x2), w = sample)-weighted.mean(log(x2),
w = x2^k*sample))
}
# Setting the error to "quite zero", fulfilling the equation
k <- optimize(k.diff, vec=sample, interval=c(0.1,5), tol=10^-7)$min
# Calculate lambda, given k
l <- weighted.mean(seq(length(sample))^k, w = sample)
# Plot
plot(density(rep(seq(length(sample)),sample)))
x <- 1:25
lines(x, dweibull(x, shape=k, scale= l))
Assuming the data are from a Weibull distribution, you can get an estimate of the shape and scale parameter like this:
sample <- c(7787,3056,2359,1759,1819,1189,1077,1080,985,622,648,518,
611,1037,727,489,432,371,1125,69,595,624)
f<-fitdistr(sample, 'weibull')
f
If you are not sure whether it is distributed Weibull, I would recommend using the ks.test. This tests whether your data is from a hypothesised distribution. Given your knowledge of the nature of the data, you could test for a few selected distributions and see which one works best.
For your example this would look like this:
ks = ks.test(sample, "pweibull", shape=f$estimate[1], scale=f$estimate[2])
ks
The p-value is insignificant, hence you do not reject the hypothesis that the data is from a Weibull distribution.
Update: The histograms of either the Weibull or exponential look like a good match to your data. I think the exponential distribution gives you a better fit. Pareto distribution is another option.
f<-fitdistr(sample, 'weibull')
z<-rweibull(10000, shape= f$estimate[1],scale= f$estimate[2])
hist(z)
f<-fitdistr(sample, 'exponential')
z = rexp(10000, f$estimate[1])
hist(z)
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?
This question already has answers here:
How do I best simulate an arbitrary univariate random variate using its probability function?
(4 answers)
Closed 9 years ago.
How can I generate random sample data from the quantiles of the unknown density f(x) for x between 0 and 4 in R?
f = function(x) ((x-1)^2) * exp(-(x^3/3-2*x^2/2+x))
If I understand you correctly (??) you want to generate random samples with the distribution whose density function is given by f(x). One way to do this is to generate a random sample from a uniform distribution, U[0,1], and then transform this sample to your density. This is done using the inverse cdf of f, a methodology which has been described before, here.
So, let
f(x) = your density function,
F(x) = cdf of f(x), and
F.inv(y) = inverse cdf of f(x).
In R code:
f <- function(x) {((x-1)^2) * exp(-(x^3/3-2*x^2/2+x))}
F <- function(x) {integrate(f,0,x)$value}
F <- Vectorize(F)
F.inv <- function(y){uniroot(function(x){F(x)-y},interval=c(0,10))$root}
F.inv <- Vectorize(F.inv)
x <- seq(0,5,length.out=1000)
y <- seq(0,1,length.out=1000)
par(mfrow=c(1,3))
plot(x,f(x),type="l",main="f(x)")
plot(x,F(x),type="l",main="CDF of f(x)")
plot(y,F.inv(y),type="l",main="Inverse CDF of f(x)")
In the code above, since f(x) is only defined on [0,Inf], we calculate F(x) as the integral of f(x) from 0 to x. Then we invert that using the uniroot(...) function on F-y. The use of Vectorize(...) is needed because, unlike almost all R functions, integrate(...) and uniroot(...) do not operate on vectors. You should look up the help files on these functions for more information.
Now we just generate a random sample X drawn from U[0,1] and transform it with Z = F.inv(X)
X <- runif(1000,0,1) # random sample from U[0,1]
Z <- F.inv(X)
Finally, we demonstrate that Z is indeed distributed as f(x).
par(mfrow=c(1,2))
plot(x,f(x),type="l",main="Density function")
hist(Z, breaks=20, xlim=c(0,5))
Rejection sampling is easy enough:
drawF <- function(n) {
f <- function(x) ((x-1)^2) * exp(-(x^3/3-2*x^2/2+x))
x <- runif(n, 0 ,4)
z <- runif(n)
subset(x, z < f(x)) # Rejection
}
Not the most efficient but it gets the job done.
Use sample . Generate a vector of probablities from your existing function f, normalized properly. From the help page:
sample(x, size, replace = FALSE, prob = NULL)
Arguments
x Either a vector of one or more elements from which to choose, or a positive integer. See ‘Details.’
n a positive number, the number of items to choose from. See ‘Details.’
size a non-negative integer giving the number of items to choose.
replace Should sampling be with replacement?
prob A vector of probability weights for obtaining the elements of the vector being sampled.
Let's say I have a set of numbers that I suspect come from the same distribution.
set.seed(20130613)
x <- rcauchy(10)
I would like a function that randomly generates a number from that same unknown distribution. One approach I have thought of is to create a density object and then get the CDF from that and take the inverse CDF of a random uniform variable (see Wikipedia).
den <- density(x)
#' Generate n random numbers from density() object
#'
#' #param n The total random numbers to generate
#' #param den The density object from which to generate random numbers
rden <- function(n, den)
{
diffs <- diff(den$x)
# Making sure we have equal increments
stopifnot(all(abs(diff(den$x) - mean(diff(den$x))) < 1e-9))
total <- sum(den$y)
den$y <- den$y / total
ydistr <- cumsum(den$y)
yunif <- runif(n)
indices <- sapply(yunif, function(y) min(which(ydistr > y)))
x <- den$x[indices]
return(x)
}
rden(1, den)
## [1] -0.1854121
My questions are the following:
Is there a better (or built into R) way to generate a random number from a density object?
Are there any other ideas on how to generate a random number from a set of numbers (besides sample)?
To generate data from a density estimate you just randomly choose one of the original data points and add a random "error" piece based on the kernel from the density estimate, for the default of "Gaussian" this just means choose a random element from the original vector and add a random normal with mean 0 and sd equal to the bandwidth used:
den <- density(x)
N <- 1000
newx <- sample(x, N, replace=TRUE) + rnorm(N, 0, den$bw)
Another option is to fit a density using the logspline function from the logspline package (uses a different method of estimating a density), then use the rlogspline function in that package to generate new data from the estimated density.
If all you need is to draw values from your existing pool of numbers, then sample is the way to go.
If you want to draw from the presumed underlying distribution, then use density , and fit that to your presumed distribution to get the necessary coefficients (mean, sd, etc.), and use the appropriate R distribution function.
Beyond that, I'd take a look at Chapter7.3 ("rejection method") of Numerical Recipes in C for ways to "selectively" sample according to any distribution. The code is simple enough to be easily translated into R .
My bet is someone already has done so and will post a better answer than this.
Greg Snow's answer was helpful to me, and I realized that the output of the density function has all the data needed to create random numbers from the input distribution. Building on his example, you can do the following to get random values using the density output.
x <- rnorm(100) # or any numeric starting vector you desire
dens <- density(x)
N <- 1000
newx <- sample(x = dens$x, N, prob = dens$y, replace=TRUE) + rnorm(N, 0, dens$bw)
You can even create a simple random number generating function
rdensity <- function(n, dens) {
return(sample(x = dens$x, n, prob = dens$y, replace=TRUE) + rnorm(n, 0, dens$bw))
}