When plotting a function (as opposed to numerical data), how can we set the number of sample points (i.e. the number of distinct x coordinates where the function is computed)? Importantly, where can I find this information in the documentation?
Example:
plot(x -> sin(1/x), 0.001, 1)
For a useful plot in the 0–0.25 range we need many more points.
One way you can do it is:
using Gadfly;
X=1e-6:1e-6:1.0
plot(x=X, y=X .|> x -> sin(1/x), Geom.line)
or you may like this version more
using Gadfly;
X=[1/z for z=300.0:-0.05:1.0]
plot(x=X, y=X .|> x -> sin(1/x), Geom.line)
To get a docu, just do
?plot
or when you want to look at the code
methods(plot)
The number of sampling points can indeed be specified:
plot(y=[x->sin(1/x)], xmin=[0.001], xmax=[1], Stat.func(1000), Geom.line)
You can find Stat.func in the Gadfly docs here:
http://gadflyjl.org/stable/lib/statistics/#Gadfly.Stat.func.
Note you can write either Stat.func(num_samples=1000) or Stat.func(1000), since there is only one argument.
Related
I came across a Rose plot obtained with Plots.jl package in Julia:
https://goropikari.github.io/PlotsGallery.jl/src/rose.html
Two things are not clear to me. The first one is what is Julia doing on the line:
θ = 0:2pi/n:2pi
It seems that the output is (lower limit):(bin size):(upper limit) but I haven't seen this type of arithmetics previously where two ranges are divided. The second thing is that I would like to obtain a histogram polar plot as it was done with R (Making a polar histogram in ggplot2), but I haven't found the documentation for line styles or how to do it in Plots.jl. Thanks.
Note that start:step:end is a common syntax in creating ranges. Let's dissect the line:
# `pi` is a reserved variable name in Julia
julia> pi
π = 3.1415926535897...
# A simple division
julia> 2pi/1
6.283185307179586
# Simple multiplication
julia> 2pi
6.283185307179586
So the 0:2pi/n:2pi creates an object of type StepRange that starts from 0 up to 2pi with steps of size 2pi/n.
In the case of desired plot, you can use the PlotlyJS.jl package. As they provided an example here. (Scroll down until you see "Polar Bar Chart")
I tested the code myself, and it's reproducible expectedly. Unfortunately, I don't know anything about the R language.
julia> using RDatasets, DataFrames, PlotlyJS
julia> df = RDatasets.dataset("datasets", "iris");
julia> sepal = df.SepalWidth;
julia> plot(
barpolar(
r=sepal
)
)
Results in:
I found some posts and discussions about the above, but I'm not sure... could someone please check if I am doing anything wrong?
I have a set of N points of the form (x,y,z). The x and y coordinates are independent variables that I choose, and z is the output of a rather complicated (and of course non-analytical) function that uses x and y as input.
My aim is to find a set of values of (x,y) where z=z0.
I looked up this kind of problem in R-related forums, and it appears that I need to interpolate the points first, perhaps using a package like akima or fields.
However, it is less clear to me: 1) if that is necessary, or the basic R functions that do the same are sufficiently good; 2) how I should use the interpolated surface to generate a correct matrix of the desired (x,y,z=z0) points.
E.g. this post seems somewhat related to the problem I am describing, but it looks extremely complicated to me, so I am wondering whether my simpler approach is correct.
Please see below some example code (not the original one, as I said the generating function for z is very complicated).
I would appreciate if you could please comment / let me know if this approach is correct / suggest a better one if applicable.
df <- merge(data.frame(x=seq(0,50,by=5)),data.frame(y=seq(0,12,by=1)),all=TRUE)
df["z"] <- (df$y)*(df$x)^2
ta <- xtabs(z~x+y,df)
contour(ta,nlevels=20)
contour(ta,levels=c(1000))
#why are the x and y axes [0,1] instead of showing the original values?
#and how accurate is the algorithm that draws the contour?
li2 <- as.data.frame(contourLines(ta,levels=c(1000)))
#this extracts the contour data, but all (x,y) values are wrong
require(akima)
s <- interp(df$x,df$y,df$z)
contour(s,levels=c(1000))
li <- as.data.frame(contourLines(s,levels=c(1000)))
#at least now the axis values are in the right range; but are they correct?
require(fields)
image.plot(s)
fancier, but same problem - are the values correct? better than the akima ones?
Let us consider the following scenario .
for x in range (1,100)
for y in range (2,500)
#plot(f(x),g(y))
end
end
where f(x) and g(y) are some user defined functions.
The output must be the desired points on plane.
Is there any way in julia to do like what I need ?
In general I can do like this
for x in range (1,100)
for y in range (2,500)
push!(l,f(x))
push!(m,g(y))
end
end
and then plotting from the two lists l,m as x,y axes respectively.
But now I want to plot points while executing loop.
This is mostly supported in Plots... see https://github.com/tbreloff/Plots.jl/issues/30 for a little more information and some example usage.
use the display function:
for x in 1:100
p = plot(f(x),g(y))
display(p)
sleep(1)
end
(inspired by Andreas Peter on the Julia slack #helpdesk channel)
I plotted an expression curve, i.e.curve(-log((1-x)/0.9999)/x,ylim=c(0,4)).
However, I want to see the reverse relationship, i.e. y changes over x instead of x changes over y. Are there any R function can plot it automatically? Or a function that can solve the equation?
There are two obvious choices:
(i) derive the inverse function algebraically (trivial in this case),
That is, take y=-log((1-x)/0.9999) and make x the subject of the equation (which would require straightforward algebraic manipulation suitable for a question on math.SE if it's not obvious how to proceed)...
... and then use curve on the result of that, or
(ii) use plot rather than curve to plot a set of (x,y) pairs (set type="l" to get a curve), and simply interchange which is x and which is y in the call to plot.
Is there a simple way to plot the difference between two probability density functions?
I can plot the pdfs of my data sets (both are one-dimensional vectors with roughly 11000 values) on the same plot together to get an idea of the overlap/difference but it would be more useful to me if I could see a plot of the difference.
something along the lines of the following (though this obviously doesn't work):
> plot(density(data1)-density(data2))
I'm relatively new to R and have been unable to find what I'm looking for on any of the forums.
Thanks in advance
This should work:
plot(x =density(data1, from= range(c(data1, data2))[1],
to=range(c(data1, data2))[2] )$x,
y= density(data1, from= range(c(data1, data2))[1],
to=range(c(data1, data2))[2] )$y-
density(data2, from= range(c(data1, data2))[1],
to=range(c(data1, data2))[2] )$y )
The trick is to make sure the densities have the same limits. Then you can plot their differences at the same locations.My understanding of the need for the identical limits comes from having made the error of not taking that step in answering a similar question on Rhelp several years ago. Too bad I couldn't remember the right arguments.
It looks like you need to spend a little time learning how to use R (or any other language, for that matter). Help files are your friend.
From the output of ?density :
Value [i.e. the data returned by the function]
If give.Rkern is true, the number R(K), otherwise an object with class
"density" whose underlying structure is a list containing the
following components.
x the n coordinates of the points where the density is estimated.
y the estimated density values. These will be non-negative, but can
be zero [remainder of "value" deleted for brevity]
So, do:
foo<- density(data1)
bar<- density(data2)
plot(foo$y-bar$y)