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How do I know that the following variables are related (in general, any two variables)?
Below, there is an obvious relation between x and y however, the 'cor' function is giving me '0'. Is there any function in R that can detect both linear and non-linear relation?
> x <- c(-2, -1, 0, 1, 2)
> y <- c(4, 1, 0, 1, 4)
>
> cor(x,y)
[1] 0
>
Edit:
I am thinking to go for MIC/MINE algorithm despite many criticisms of this algorithm by top statisticians.
Have a look at MINE (1). They also provide a wrapper for R.
(1) Reshef, D.N, Y.A Reshef, H.K Finucane, S.R Grossman, G. McVean, P.J Turnbaugh, E.S Lander, M. Mitzenmacher, and P.C Sabeti. “Detecting Novel Associations in Large Data Sets.” Science 334, no. 6062 (2011): 1518–1524.
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Good evening,
Even though I know it will "destroy" an actual normal distribution, I need to set a maximum and a minimum to a rnorm function in R.
I'm using survival rates in vultures to calculate population trends and although I need it to fluctuate, for logic reasons, survival rates can't be over 1 or under 0.
I tried doing it with if's and else's but I think there should be a better way to do it.
Thanks!
You could sample from a large normalized rnorm draw:
rbell <- function(n) {
r <- rnorm(n * 1000)
sample((r - min(r)) / diff(range(r)), n)
}
For example:
rbell(10)
#> [1] 0.5177806 0.5713479 0.5330545 0.5987649 0.3312775 0.5508946 0.3654235 0.3897417
#> [9] 0.1925600 0.6043243
hist(rbell(1000))
This will always be curtailed to the interval (0, 1).
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I have a vector of probability and I want to sample one component from this vector by taking into account the value of its probability.
Here is an example:
x = [1/2,1/4,1/4]
The probability of taking 1 in x is 1/2, the probability for 2 and 3 is 1/4. How can I do that in R?
You can use sample function
sample(1:3, 1, prob = c(0.5, 0.25, 0.25))
It selects one value from c(1, 2, 3) with the probability of c(0.5, 0.25, 0.25) respectively.
Read ?sample for more details on how it works.
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How do I set two equations equal to each other in R to solve?
For example:
xlog(x)=8273
Find X?
Use equation in the form: x*log(x)-8273 = 0
You should have some idea of the range in which the answer lies. Then use uniroot function:
f <- function(x) (x*log(x)-8273)
uniroot(f, lower=0.1, upper=100000000)$root
[1] 1170.897
Or a more general form:
f <- function(x,y) (x*log(x)-y)
uniroot(f, y=8273, lower=0.1, upper=100000000)$root
[1] 1170.897
It turns out (with a little help from Wolfram Alpha) that this particular solution is related to the Lambert W function (which Wolfram Alpha calls the "product log" function):
library(emdbook)
exp(lambertW(8273)) ## 1170.897
The Lambert W is available in several other R packages (LambertW, spatstat, pracma, condmixt, VGAM) as well.
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Whats is the fastest way to find the maximum root of a cubic function in R?
a x^3 + b x^2 + c x + d = 0
Is there anything wrong with the base function polyroot?
Description
Find zeros of a real or complex polynomial.
An example of a cubic
polyroot(c(1,3,3,1))
# [1] -1+0i -1+0i -1-0i
Here is a function to find the maximum non-complex root of a polynomial...
maxReal <- function(params){
x <- polyroot(params)
reals <- sapply(x, function(i) isTRUE(all.equal(Im(i),0)))
max(Re(x)[reals])
}
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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.