x <- sin(1:20)
y <- (dplyr::lag(x)-x)-x
plot(x, y, type="l")
Yields this plot:
How do I get a smooth, cyclical trajectory? The smoothing functions I've tried all want to make a single smooth function.
Example 1 (throws an error):
lines(smooth.spline(x, y))
Example 2 (draws a function):
lo <- loess(y~x)
lines(predict(lo), col='red', lwd=2)
Example 3 (draws a function):
xspline(c(x,y), shape=1, border='blue' )
I think you're asking for both (a) smoother curves and (b) cylindrical/symmetric circles, correct?
Increase the count of points to make it smoother. Fix the aspect ratio to make it a circle (i.e., asp=1).
x <- sin((1:20000) / 1000)
y <- cos((1:20000) / 1000)
plot(x, y, type="l", asp=1)
EDIT, responding to OP's comments below.:
This uses dplyr::lag(), while increasing the number of points. Don't conceptually shift the lag though (i.e., y should be a full value of 1 from x).
(x=1, y=0), (x=1.001, y=0.001), ..., (x=20, y=19)
x <- sin((1:20000) / 1000)
y <- (dplyr::lag(x, 1000)-x)-x
plot(x, y, type="l", asp=1)
If that's not what you want, maybe sketch a picture and attach in (in like MS Paint or Inkscape).
Related
I basically want to shade an area behind a graph.
It's easy enough to do in linear scale.
x <- 1:20
y <- x^2
plot(x, y, type="l")
polygon(c(10,10,15,15),
c(-100,600,600,-100),
col=rgb(0,1,0,0.3),border=FALSE)
Produces this:
But once you put y on a log scale,
plot(x, y, type="l", log="y")
polygon(c(10,10,15,15),
c(-100,600,600,-100),
col=rgb(0,1,0,0.3),border=FALSE)
Nothing shows up.
Be careful when playing with log = "y". If your y value is negative, you get NaN. This is exactly what happened here. Try
plot(x, y, type="l", log="y")
polygon(c(10,10,15,15),
c(1e-7,600,600,1e-7), ## log(1e-7) is small enough
col=rgb(0,1,0,0.3),border=FALSE)
I have a perspective plot of a locfit model and I wish to add two things to it
Predictor variables as points in the 3D space
Color the surface according to the Z axis value
For the first, I have tried to use the trans3d function. But I get the following error even though my variables are in vector format:
Error in cbind(x, y, z, 1) %*% pmat : requires numeric/complex matrix/vector arguments
Here is a snippet of my code
library(locfit)
X <- as.matrix(loc1[,1:2])
Y <- as.matrix(loc1[,3])
zz <- locfit(Y~X,kern="bisq")
pmat <- plot(zz,type="persp",zlab="Amount",xlab="",ylab="",main="Plains",
phi = 30, theta = 30, ticktype="detailed")
x1 <- as.vector(X[,1])
x2 <- as.vector(X[,2])
Y <- as.vector(Y)
points(trans3d(x1,x2,Y,pmat))
My "loc1" data can be found here - https://www.dropbox.com/s/0kdpd5hxsywnvu2/loc1_amountfreq.txt?dl=0
TL,DR: not really in plot.locfit, but you can reconstruct it.
I don't think plot.locfit has good support for this sort of customisation. Supposedly get.data=T in your plot call will plot the original data points (point 1), and it does seem to do so, except if type="persp". So no luck there. Alternatively you can points(trans3d(...)) as you have done, except you need the perspective matrix returned by persp, and plot.locfit.3d does not return it. So again, no luck.
For colouring, typically you make a colour scale (http://r.789695.n4.nabble.com/colour-by-z-value-persp-in-raster-package-td4428254.html) and assign each z facet the colour that goes with it. However, you need the z-values of the surface (not the z-values of your original data) for this, and plot.locfit does not appear to return this either.
So to do what you want, you'll essentially be recoding plot.locfit yourself (not hard, though just cludgy).
You could put this into a function so you can reuse it.
We:
make a uniform grid of x-y points
calculate the value of the fit at each point
use these to draw the surface (with a colour scale), saving the perspective matrix so that we can
plot your original data
so:
# make a grid of x and y coords, calculate the fit at those points
n.x <- 20 # number of x points in the x-y grid
n.y <- 30 # number of y points in the x-y grid
zz <- locfit(Total ~ Mex_Freq + Cal_Freq, data=loc1, kern="bisq")
xs <- with(loc1, seq(min(Mex_Freq), max(Mex_Freq), length.out=20))
ys <- with(loc1, seq(min(Cal_Freq), max(Cal_Freq), length.out=30))
xys <- expand.grid(Mex_Freq=xs, Cal_Freq=ys)
zs <- matrix(predict(zz, xys), nrow=length(xs))
# generate a colour scale
n.cols <- 100 # number of colours
palette <- colorRampPalette(c('blue', 'green'))(n.cols) # from blue to green
# palette <- colorRampPalette(c(rgb(0,0,1,.8), rgb(0,1,0,.8)), alpha=T)(n.cols) # if you want transparency for example
# work out which colour each z-value should be in by splitting it
# up into n.cols bins
facetcol <- cut(zs, n.cols)
# draw surface, with colours (col=...)
pmat <- persp(x=xs, y=ys, zs, theta=30, phi=30, ticktype='detailed', main="plains", xlab="", ylab="", zlab="Amount", col=palette[facetcol])
# draw your original data
with(loc1, points(trans3d(Mex_Freq,Cal_Freq,Total,pmat), pch=20))
Note - doesn't look that pretty! might want to adjust say your colour scale colours, or the transparency of the facets, etc. Re: adding legend, there are some other questions that deal with that.
(PS: what a shame ggplot doesn't do 3D scatter plots.)
I'm trying to calculate a Bezier-like spline curve that passes through a sequence of x-y coordinates. An example would be like the following output from the cscvn function in Matlab (example link):
I believe the (no longer maintained) grid package used to do this (grid.xspline function?), but I haven't been able to install an archived version of the package, and don't find any examples exactly along the lines of what I would like.
The bezier package also looks promising, but it is very slow and I also can't get it quite right:
library(bezier)
set.seed(1)
n <- 10
x <- runif(n)
y <- runif(n)
p <- cbind(x,y)
xlim <- c(min(x) - 0.1*diff(range(x)), c(max(x) + 0.1*diff(range(x))))
ylim <- c(min(y) - 0.1*diff(range(y)), c(max(y) + 0.1*diff(range(y))))
plot(p, xlim=xlim, ylim=ylim)
text(p, labels=seq(n), pos=3)
bp <- pointsOnBezier(cbind(x,y), n=100)
lines(bp$points)
arrows(bp$points[nrow(bp$points)-1,1], bp$points[nrow(bp$points)-1,2],
bp$points[nrow(bp$points),1], bp$points[nrow(bp$points),2]
)
As you can see, it doesn't pass through any points except the end values.
I would greatly appreciate some guidance here!
There is no need to use grid really. You can access xspline from the graphics package.
Following from your code and the shape from #mrflick:
set.seed(1)
n <- 10
x <- runif(n)
y <- runif(n)
p <- cbind(x,y)
xlim <- c(min(x) - 0.1*diff(range(x)), c(max(x) + 0.1*diff(range(x))))
ylim <- c(min(y) - 0.1*diff(range(y)), c(max(y) + 0.1*diff(range(y))))
plot(p, xlim=xlim, ylim=ylim)
text(p, labels=seq(n), pos=3)
You just need one extra line:
xspline(x, y, shape = c(0,rep(-1, 10-2),0), border="red")
It may not the be the best approach, bit grid certainly isn't inactive. It's included as a default package with the R installation. It's the underlying graphics engine for plotting libraries like lattice and ggplot. You shouldn't need to install it, you should just be able to load it. Here's how I might translate your code to use grid.xpline
set.seed(1)
n <- 10
x <- runif(n)
y <- runif(n)
xlim <- c(min(x) - 0.1*diff(range(x)), c(max(x) + 0.1*diff(range(x))))
ylim <- c(min(y) - 0.1*diff(range(y)), c(max(y) + 0.1*diff(range(y))))
library(grid)
grid.newpage()
pushViewport(viewport(xscale=xlim, yscale=ylim))
grid.points(x, y, pch=16, size=unit(2, "mm"),
default.units="native")
grid.text(seq(n), x,y, just=c("center","bottom"),
default.units="native")
grid.xspline(x, y, shape=c(0,rep(-1, 10-2),0), open=TRUE,
default.units="native")
popViewport()
which results in
note that grid is pretty low-level so it's not super easy to work with, but it does allow you far more control of what and where you plot.
And if you want to extract the points along the curve rather than draw it, look at the ?xsplinePoints help page.
Thanks to all that helped with this. I'm summarizing the lessons learned plus a few other aspects.
Catmull-Rom spline vs. cubic B-spline
Negative shape values in the xspline function return a Catmull-Rom type spline, with spline passing through the x-y points. Positive values return a cubic B type spline. Zero values return a sharp corner. If a single shape value is given, this is used for all points. The shape of end points is always treated like a sharp corner (shape=0), and other values do not influence the resulting spline at the end points:
# Catmull-Rom spline vs. cubic B-spline
plot(p, xlim=extendrange(x, f=0.2), ylim=extendrange(y, f=0.2))
text(p, labels=seq(n), pos=3)
# Catmull-Rom spline (-1)
xspline(p, shape = -1, border="red", lwd=2)
# Catmull-Rom spline (-0.5)
xspline(p, shape = -0.5, border="orange", lwd=2)
# cubic B-spline (0.5)
xspline(p, shape = 0.5, border="green", lwd=2)
# cubic B-spline (1)
xspline(p, shape = 1, border="blue", lwd=2)
legend("bottomright", ncol=2, legend=c(-1,-0.5), title="Catmull-Rom spline", col=c("red", "orange"), lty=1)
legend("topleft", ncol=2, legend=c(1, 0.5), title="cubic B-spline", col=c("blue", "green"), lty=1)
Extracting results from xspline for external plotting
This took some searching, but the trick is to apply the argument draw=FALSE to xspline.
# Extract xy values
plot(p, xlim=extendrange(x, f=0.1), ylim=extendrange(y, f=0.1))
text(p, labels=seq(n), pos=3)
spl <- xspline(x, y, shape = -0.5, draw=FALSE)
lines(spl)
arrows(x0=(spl$x[length(spl$x)-0.01*length(spl$x)]), y0=(spl$y[length(spl$y)-0.01*length(spl$y)]),
x1=(spl$x[length(spl$x)]), y1=(spl$y[length(spl$y)])
)
I have managed to create a scatterplot with two datasets on a single plot. One set of data has an X axis that ranges from 0 -40 (Green), while the other only ranges from 0 -15 (Red).
I used this code to add trend lines to the red and green data separately (using par(new)).
plot( x1,y1, col="red", axes=FALSE, xlab="",ylab="",ylim= range(0:1), xlim= range(0:40))
f <- function(x1,a,b,d) {(a*x1^2) + (b*x1) + d}
fit <- nls(y1 ~ f(x1,a,b,d), start = c(a=1, b=1, d=1))
co <- coef(fit)
curve(f(x, a=co[1], b=co[2], d=co[3]), add = TRUE, col="red", lwd=1)
My issue is I can't seem to find a way to stop the red trend line at 15 on the x axis. I "googled" around and nothing seemed to come up for my issue. Lots on excel trend lines! I tired adding an end= statement to fit<- and that did not work either.
Please help,
I hope I have posted enough information. Thanks in advance.
Try using ggplot. Following example uses mtcars data:
library(ggplot2)
ggplot(mtcars, aes(qsec, wt, color=factor(vs)))+geom_point()+ stat_smooth(se=F)
You can do this in base graphics with the from and to arguments of the curve function (see the help for curve for more details). For example:
# Your function
f <- function(x1,a,b,d) {(a*x1^2) + (b*x1) + d}
# Plot the function from x=-100 to x=100
curve(f(x, a=-2, b=3, d=0), from=-100, to=100, col="red", lwd=1, lty=1)
# Same curve going from x=-100 to x=0 (and shifted down by 1000 units so it's
# easy to see)
curve(f(x, a=-2, b=3, d=-1000), from=-100, to=0,
add=TRUE, col="blue", lwd=2, lty=1)
If you want to set the curve x-limits programmatically, you can do something like this (assuming your data frame is called df and your x-variable is called x):
curve(f(x, a=-2, b=3, d=0), from=range(df$x)[1], to=range(df$x)[2],
add=TRUE, col="red", lwd=1, lty=1)
I have fit a LOESS local regression to some data and I want to be able to find the X value associated with a given Y value.
plot(cars, main = "Stopping Distance versus Speed")
car_loess <- loess(cars$dist~cars$speed,span=.5)
lines(1:50, predict(car_loess,data.frame(speed=1:50)))
I was hoping that I could use teh inverse.predict function from the chemCal package, but that does not work for LOESS objects.
Does anyone have any idea how I might be able to do this calibrationa in a better way than predicticting Y values from a long vector of X values and looking through the resulting fitted Y for the Y value of interest and taking its corresponding X value?
Practically speaking in the above example, let's say I wanted to find the speed at which the stopping distance is 15.
Thanks!
The predicted line that you added to the plot is not quite right. Use code like this instead:
# plot the loess line
lines(cars$speed, car_loess$fitted, col="red")
You can use the approx() function to get a linear approximation from the loess line at a give y value. It works just fine for the example that you give:
# define a given y value at which you wish to approximate x from the loess line
givenY <- 15
estX <- approx(x=car_loess$fitted, y=car_loess$x, xout=givenY)$y
# add corresponding lines to the plot
abline(h=givenY, lty=2)
abline(v=estX, lty=2)
But, with a loess fit, there may be more than one x for a given y. The approach I am suggesting does not provide you with ALL of the x values for the given y. For example ...
# example with non-monotonic x-y relation
y <- c(1:20, 19:1, 2:20)
x <- seq(y)
plot(x, y)
fit <- loess(y ~ x)
# plot the loess line
lines(x, fit$fitted, col="red")
# define a given y value at which you wish to approximate x from the loess line
givenY <- 15
estX <- approx(x=fit$fitted, y=fit$x, xout=givenY)$y
# add corresponding lines to the plot
abline(h=givenY, lty=2)
abline(v=estX, lty=2)