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I need to make a simple numeric linear interpolation in Delphi, was thinking of implementing a function, but then thought better and I think that should already be some library. I found nothing on google.
My problem is simple, I have a dataset with X and Y data, and other new dataset X data (Xb) that will be the basis for finding new Y data (Yin) interpolated.
In R, for example, have a function approx that accomplishes this easily. This function also allows Xb length is of any size.
Yin <- approx (X, Y, Xb, method = "linear")$y
There is some statistical library to do this in Delphi? Or continue to write my function (based on approx)?
Linear interpolation of 1D-data is very simple:
Find such index in X-array, that X[i] <= Xb < X[i+1]
(binary search for case of random access, linear search for case of step-by step Xb changing)
Calculate
Yb = Y[i] + (Y[i+1] - Y[i]) * (Xb - X[i]) / (X[i+1] - X[i])
Related
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What are some different packages in R that contain in built function to simulate the Zero inflated distributions, related to the popular discrete models like the Poisson, Negative Binomial, COM-Poisson, Poisson Inverse Gaussian, Poisson-Lindley except the 'iZid' package.
Have a look at the CRAN Task View on Distributions. This is a curated look at R packages that help you work with distributions. You can search the page for "inflated" to quickly find the relevant parts.
If you have an existing function that generates random deviates from a non-zero-inflated distribution, you can write a wrapper (or decorator) that creates a zero-inflated-deviate simulator. The only assumption I've made here is that the first argument of the original function is called n and specifies the number of random deviates to pick.
For example, if we want to extend rbinom to return zero-inflated binomial deviates ...
ziversion <- function(rfun) {
f <- function(n, ..., zi) {
x <- rfun(n, ...)
x <- ifelse(runif(n) < zi, 0, x)
return(x)
}
return(f)
}
rzibinom <- ziversion(rbinom)
set.seed(101)
rzibinom(10, size = 10, prob = 0.2, zi = 0.5)
## [1] 1 0 3 2 0 1 2 0 0 0
zi is the zero-inflation probability. With a little bit of effort the code could be made more efficient ...
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I'm new to R, and a formula I need to enter includes functions that are beyond the scope of my mathematics experience. In particular, I don't understand what's going on with the subscript to the gamma function. Is this an incomplete gamma function, and if so, is it upper or lower?
Anyhow, the formula is attached in the image. This is the CDF for the 4 parameter Generalized Gamma distribution, taken from a statistical software manual. How would I specify this in R? Any help is much appreciated.
This can be completely wrong, but it seems that function F in the image is a combination of an upper incomplete gamma function and other terms. The subscript seems to be a transformation of x:
y <- ((x - gamma)/beta)^k
If so it could be coded as follows.
f <- function(x, a, beta, gamma, k){
y <- ((x - gamma)/beta)^k
pgamma(y, a, lower = FALSE)*gamma(a)
}
The expression
pgamma(y, a, lower = FALSE)*gamma(a)
is a base R way to code the upper incomplete gamma function. Alternatively, package gsl, function gamma_inc could be used replacing that code line with
gsl::gamma_inc(a, y)
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I would like to estimate the parameters of a nonlinear regression model with LAD regression. In essence the LAD estimator is an M-estimator. As far as I know it is not possible to use the robustbase package to do this. How could I use R to do LAD regression? Could I use a standard package?
You could do this with the built-in optim() function
Make up some data (make sure x is positive, so that a*x^b makes sense - raising negative numbers to fractional powers is problematic):
set.seed(101)
a <- 1; b <- 2
dd <- data.frame(x=rnorm(1000,mean=7))
dd$y <- a*dd$x^b + rnorm(1000,mean=0,sd=0.1)
Define objective function:
objfun <- function(p) {
pred <- p[1]*dd$x^p[2] ## a*x^b
sum(abs(pred-dd$y)) ## least-absolute-deviation criterion
}
Test objective function:
objfun(c(0,0))
objfun(c(-1,-1))
objfun(c(1,2))
Optimize:
o1 <- optim(fn=objfun, par=c(0,0), hessian=TRUE)
You do need to specify starting values, and deal with any numerical/computational issues yourself ...
I'm not sure I know how to compute standard errors: you can use sqrt(diag(solve(o1$hessian))), but I don't know if the standard theory on which this is based still applies ...
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I was wondering if there is a way to calculate Nagelkerke R-square based upon the output produced. I know that I can calculate McFadden R-square directly. But Nagelkerke produces what we feel is a more accurate strength of the model.
I am not having luck with adding on packages to my setup, if that is the line of thought that you have.
Thanks.
This question is underdefined so I'll do my best assuming that "the output produced" is a glm object. This function should produce the appropriate pseudo-R square you want when applied to a glm object.
Nagelkerke <- function(mod) {
l_full <- exp(logLik(mod))
l_intercept <- exp(logLik( update(mod, . ~ 1) ))
N <- length(mod$y)
r_2 <- (1 - (l_intercept / l_full)^(2/N)) / (1 - l_intercept^(2/N))
return( as.numeric(r_2) )
}
Example:
model <- glm(formula = vs ~ mpg + disp, family = binomial("logit"), data = mtcars);
Nagelkerke(model);
#[1] 0.6574295
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A reviewer of a paper I submitted to a scientific journal insists that my function
f1[b_, c_, t_] := 1 - E^((c - t)/b)/2
is "mathematically equivalent" to the function
f2[b0_, b1_, t_] := 1 - b0 E^(-b1 t)
He insists
While the models might appear(superficially) to be different, the f1
model is merely a re-parameterisation of the f2 model, and this can be
seen easily using highschool mathematics.
I survived High School, but I don't see the equivalence, and FullSimplify does not yield the same results. Perhaps I am misunderstanding FullSimplify. Is there a way to authoritatively refute or confirm the assertion of the reviewer?
If c and b are constant, you can factor them out relatively easily given the property of the power operator:
e^(A + B) = e^A x e^B...
so
e^((c - t)/b) = e^(c/b - t/b) = e^(c/b) x e^(-t/b) = b0 x e^(-t/b)
The latter expression is commonly used to simplify linear differential equation.