Re: [R] How to find local minimum between distributions using mixtools?
I presumed there was some reason why the 'optimise' function did not suit you.
optimize(function (x) a*dnorm(x, x1, s1) + (1-a)*dnorm(x, x2, s2),
range(c(x1, x2)))
should do the trick.
On Thu, 14 Oct 2021 at 23:42, Luigi Marongiu wrote:
>
> Thank you, I hoped there might be an automatic method more than
> function analysis...
>
> On Thu, Oct 14, 2021 at 9:06 AM Richard O'Keefe wrote:
> >
> > Do you really want the minimum?
> > It sounds as though your model is a*N(x1,s1) + (1-a)*N(x2,s2) where
> > you use mixtools to estimate
> > the parameters. Finding the derivative of that is fairly
> > straightforward calculus, and solving for the
> > derivative being zero gives you extrema (you want the one between x1 and
> > x2).
> >
> > In practice I might start with a quick and dirty hack like
> > x <- seq(from min(x1,x2), to=max(x1,x2), length=399)
> > p <- a*dnorm(x, x1, s1) + (1-a)*dnorm(x, x2, s2)
> > m <- min(p)
> > x[x == m]
> >
> > On Wed, 13 Oct 2021 at 22:12, Luigi Marongiu
> > wrote:
> > >
> > > Hello,
> > > I have two peaks distributions (and some noise values in between), and
> > > I used mixtools to characterize the two overlapping data. How can I
> > > find the minimum between the peak distributions?
> > > The sigma value of the second population is close to that value, but I
> > > am not sure if this is really the cut-off point between the two
> > > distributions (by eye, I would say the value is 0.125 against 0.1 of
> > > the sigma value). Is there a better approach?
> > > Thanks
> > >
> > > ```
> > > set.seed(10)
> > > negative_mr <- runif(50, 0, 0.099)
> > > negative_fcn <- runif(50, 1, 40)
> > > positive_mr <- c(runif(30, 0.2, 0.5), runif(20, 0.4, 0.5))
> > > positive_fcn <- c(runif(30, 25, 40), runif(20, 10, 25))
> > > uncertain_mr <- runif(10, 0.099, 0.2)
> > > uncertain_fcn <- runif(10, 2, 40)
> > > df <- data.frame(Y=c(negative_mr, uncertain_mr, positive_mr),
> > > X=c(negative_fcn, uncertain_fcn, positive_fcn),
> > > GROUP=c(rep("negative", length(negative_mr)),
> > > rep("uncertain", length(uncertain_mr)),
> > > rep("positive", length(positive_mr
> > > library(mixtools)
> > > model = normalmixEM((x = df$Y))
> > > summary(model)
> > > # plot
> > > plot(model, which=2)
> > > Cut_off <- model$sigma[2]
> > > Cut_offE <- 0.125
> > > abline(v=Cut_off, col="blue", lwd=2)
> > > abline(v=Cut_offE, col="blue", lwd=2, lty=2)
> > > plot(df$Y[df$GROUP=="negative"]~df$X[df$GROUP=="negative"],
> > > pch=1, ylim=c(0,0.5), xlim=c(0,41), cex=1.5,
> > > xlab=expression(bold("X")),
> > > ylab=expression(bold("Y")))
> > > points(df$Y[df$GROUP=="positive"]~df$X[df$GROUP=="positive"],
> > > pch=16, cex=1.5)
> > > points(df$Y[df$GROUP=="uncertain"]~df$X[df$GROUP=="uncertain"],
> > > pch=16, cex=1.5, col="grey")
> > > abline(h=Cut_off, col="blue", lwd=2)
> > > abline(h=Cut_offE, col="blue", lwd=2, lty=2)
> > > ```
> > >
> > >
> > > --
> > > Best regards,
> > > Luigi
> > >
> > > __
> > > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> > > https://stat.ethz.ch/mailman/listinfo/r-help
> > > PLEASE do read the posting guide
> > > http://www.R-project.org/posting-guide.html
> > > and provide commented, minimal, self-contained, reproducible code.
>
>
>
> --
> Best regards,
> Luigi
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to find local minimum between distributions using mixtools?
Thank you, I hoped there might be an automatic method more than
function analysis...
On Thu, Oct 14, 2021 at 9:06 AM Richard O'Keefe wrote:
>
> Do you really want the minimum?
> It sounds as though your model is a*N(x1,s1) + (1-a)*N(x2,s2) where
> you use mixtools to estimate
> the parameters. Finding the derivative of that is fairly
> straightforward calculus, and solving for the
> derivative being zero gives you extrema (you want the one between x1 and x2).
>
> In practice I might start with a quick and dirty hack like
> x <- seq(from min(x1,x2), to=max(x1,x2), length=399)
> p <- a*dnorm(x, x1, s1) + (1-a)*dnorm(x, x2, s2)
> m <- min(p)
> x[x == m]
>
> On Wed, 13 Oct 2021 at 22:12, Luigi Marongiu wrote:
> >
> > Hello,
> > I have two peaks distributions (and some noise values in between), and
> > I used mixtools to characterize the two overlapping data. How can I
> > find the minimum between the peak distributions?
> > The sigma value of the second population is close to that value, but I
> > am not sure if this is really the cut-off point between the two
> > distributions (by eye, I would say the value is 0.125 against 0.1 of
> > the sigma value). Is there a better approach?
> > Thanks
> >
> > ```
> > set.seed(10)
> > negative_mr <- runif(50, 0, 0.099)
> > negative_fcn <- runif(50, 1, 40)
> > positive_mr <- c(runif(30, 0.2, 0.5), runif(20, 0.4, 0.5))
> > positive_fcn <- c(runif(30, 25, 40), runif(20, 10, 25))
> > uncertain_mr <- runif(10, 0.099, 0.2)
> > uncertain_fcn <- runif(10, 2, 40)
> > df <- data.frame(Y=c(negative_mr, uncertain_mr, positive_mr),
> > X=c(negative_fcn, uncertain_fcn, positive_fcn),
> > GROUP=c(rep("negative", length(negative_mr)),
> > rep("uncertain", length(uncertain_mr)),
> > rep("positive", length(positive_mr
> > library(mixtools)
> > model = normalmixEM((x = df$Y))
> > summary(model)
> > # plot
> > plot(model, which=2)
> > Cut_off <- model$sigma[2]
> > Cut_offE <- 0.125
> > abline(v=Cut_off, col="blue", lwd=2)
> > abline(v=Cut_offE, col="blue", lwd=2, lty=2)
> > plot(df$Y[df$GROUP=="negative"]~df$X[df$GROUP=="negative"],
> > pch=1, ylim=c(0,0.5), xlim=c(0,41), cex=1.5,
> > xlab=expression(bold("X")),
> > ylab=expression(bold("Y")))
> > points(df$Y[df$GROUP=="positive"]~df$X[df$GROUP=="positive"],
> > pch=16, cex=1.5)
> > points(df$Y[df$GROUP=="uncertain"]~df$X[df$GROUP=="uncertain"],
> > pch=16, cex=1.5, col="grey")
> > abline(h=Cut_off, col="blue", lwd=2)
> > abline(h=Cut_offE, col="blue", lwd=2, lty=2)
> > ```
> >
> >
> > --
> > Best regards,
> > Luigi
> >
> > __
> > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
--
Best regards,
Luigi
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to find local minimum between distributions using mixtools?
Do you really want the minimum?
It sounds as though your model is a*N(x1,s1) + (1-a)*N(x2,s2) where
you use mixtools to estimate
the parameters. Finding the derivative of that is fairly
straightforward calculus, and solving for the
derivative being zero gives you extrema (you want the one between x1 and x2).
In practice I might start with a quick and dirty hack like
x <- seq(from min(x1,x2), to=max(x1,x2), length=399)
p <- a*dnorm(x, x1, s1) + (1-a)*dnorm(x, x2, s2)
m <- min(p)
x[x == m]
On Wed, 13 Oct 2021 at 22:12, Luigi Marongiu wrote:
>
> Hello,
> I have two peaks distributions (and some noise values in between), and
> I used mixtools to characterize the two overlapping data. How can I
> find the minimum between the peak distributions?
> The sigma value of the second population is close to that value, but I
> am not sure if this is really the cut-off point between the two
> distributions (by eye, I would say the value is 0.125 against 0.1 of
> the sigma value). Is there a better approach?
> Thanks
>
> ```
> set.seed(10)
> negative_mr <- runif(50, 0, 0.099)
> negative_fcn <- runif(50, 1, 40)
> positive_mr <- c(runif(30, 0.2, 0.5), runif(20, 0.4, 0.5))
> positive_fcn <- c(runif(30, 25, 40), runif(20, 10, 25))
> uncertain_mr <- runif(10, 0.099, 0.2)
> uncertain_fcn <- runif(10, 2, 40)
> df <- data.frame(Y=c(negative_mr, uncertain_mr, positive_mr),
> X=c(negative_fcn, uncertain_fcn, positive_fcn),
> GROUP=c(rep("negative", length(negative_mr)),
> rep("uncertain", length(uncertain_mr)),
> rep("positive", length(positive_mr
> library(mixtools)
> model = normalmixEM((x = df$Y))
> summary(model)
> # plot
> plot(model, which=2)
> Cut_off <- model$sigma[2]
> Cut_offE <- 0.125
> abline(v=Cut_off, col="blue", lwd=2)
> abline(v=Cut_offE, col="blue", lwd=2, lty=2)
> plot(df$Y[df$GROUP=="negative"]~df$X[df$GROUP=="negative"],
> pch=1, ylim=c(0,0.5), xlim=c(0,41), cex=1.5,
> xlab=expression(bold("X")),
> ylab=expression(bold("Y")))
> points(df$Y[df$GROUP=="positive"]~df$X[df$GROUP=="positive"],
> pch=16, cex=1.5)
> points(df$Y[df$GROUP=="uncertain"]~df$X[df$GROUP=="uncertain"],
> pch=16, cex=1.5, col="grey")
> abline(h=Cut_off, col="blue", lwd=2)
> abline(h=Cut_offE, col="blue", lwd=2, lty=2)
> ```
>
>
> --
> Best regards,
> Luigi
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] How to find local minimum between distributions using mixtools?
Hello,
I have two peaks distributions (and some noise values in between), and
I used mixtools to characterize the two overlapping data. How can I
find the minimum between the peak distributions?
The sigma value of the second population is close to that value, but I
am not sure if this is really the cut-off point between the two
distributions (by eye, I would say the value is 0.125 against 0.1 of
the sigma value). Is there a better approach?
Thanks
```
set.seed(10)
negative_mr <- runif(50, 0, 0.099)
negative_fcn <- runif(50, 1, 40)
positive_mr <- c(runif(30, 0.2, 0.5), runif(20, 0.4, 0.5))
positive_fcn <- c(runif(30, 25, 40), runif(20, 10, 25))
uncertain_mr <- runif(10, 0.099, 0.2)
uncertain_fcn <- runif(10, 2, 40)
df <- data.frame(Y=c(negative_mr, uncertain_mr, positive_mr),
X=c(negative_fcn, uncertain_fcn, positive_fcn),
GROUP=c(rep("negative", length(negative_mr)),
rep("uncertain", length(uncertain_mr)),
rep("positive", length(positive_mr
library(mixtools)
model = normalmixEM((x = df$Y))
summary(model)
# plot
plot(model, which=2)
Cut_off <- model$sigma[2]
Cut_offE <- 0.125
abline(v=Cut_off, col="blue", lwd=2)
abline(v=Cut_offE, col="blue", lwd=2, lty=2)
plot(df$Y[df$GROUP=="negative"]~df$X[df$GROUP=="negative"],
pch=1, ylim=c(0,0.5), xlim=c(0,41), cex=1.5,
xlab=expression(bold("X")),
ylab=expression(bold("Y")))
points(df$Y[df$GROUP=="positive"]~df$X[df$GROUP=="positive"],
pch=16, cex=1.5)
points(df$Y[df$GROUP=="uncertain"]~df$X[df$GROUP=="uncertain"],
pch=16, cex=1.5, col="grey")
abline(h=Cut_off, col="blue", lwd=2)
abline(h=Cut_offE, col="blue", lwd=2, lty=2)
```
--
Best regards,
Luigi
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
