Here is what I'd like to do :
So, I have my function but I would like to change a parameter (Temperature,in case you're wondering) previously defined and have the 4 functions on the same plot. I'm looking for a option like replot in Gnuplot for instance. Here is what I've done so far using just Plots :
function L(v)
defined the function here
end;
plot(L,0,2000,title="Spectral Radiance",lw=2, xlabel = L"photon \ energy \ (cm^{-1})",
ylabel = L"W . m^{-2} . sr^{-1} . wavenumber^{-1}", label="300K")
So the Temperature has been previously defined for T = 2000 K, but now how can I change it to 3500,4500 and 5600 and display the results in the same plot ? Many thanks for any tips !
As #crstnbr says, in Plots.jl you can do:
using Plots
function L(v)
v .* rand(100)
end;
p = plot()
for v in 1:4
plot!(L(v), label="factor $v")
end
display(p)
Related
I have a function in Julia that produces a plot using the Plots package plot() command. I'd like to set some optional arguments for the plot by passing arguments into my outer function. For example, I'd like to set title and axis labels without needing to program a bunch of if statements to check what parameters I'm trying to pass in. A MWE of what I'd like to do is as follows:
function outer(data; plot_options...)
x = data.x
y = data.y
plot(x,y, plot_options...)
end
So that if I call something like outer(data, title="My Title", lw=2) I produce a plot with title set to "My Title" and a linewidth of 2. Trying the naive thing that I programmed above results in an error.
function outer(data; plot_options...)
x = data.x
y = data.y
plot(x,y; plot_options...)
end
missed a semicolon?
Julia has the delightful ability to generate plots constructed from Unicode symbols which are printed directly to the command line in a very straightforward way. For example, the following code generates a Unicode plot of a sine function directly to the command line:
using Plots
unicodeplots();
x = [0:0.1:2*pi;];
y = sin.(x);
plot(x,y)
I would like to try to find a way to create an animated plot of this form directly on the command line. Ideally, I would like to generate a single plot in Unicode that is ``updated" in such a way that it appears animated.
However, although printing hundreds of distinct frames to the command line is naturally less appealing, such a solution is acceptable if it ``looks" like an animation. Another less acceptable solution is to print such Unicode plots into a gif in a way that is consistent for all platforms; attempts to do any of this involving jury-rigging #animate and #gif have largely failed, since either function cannot even print Unicode plots to a file in the Windows form of Julia.
UPDATE: Here is an example of code that generates an "animation" in the command line that is not really acceptable, which simply plots each distinct frame followed by "spacing" in the command line provided by a special Unicode character (tip provided by niczky12):
using Plots
unicodeplots();
n = 100;
x = [0:0.1:4*pi;];
for i = 1:30
y = sin.(x .+ (i/2));
plot(x, y, show=true, xlims = (0,4*pi), ylims = (-1,1))
sleep(0.01)
println("\33[2J")
end
A slight improvement might be this:
let
print("\33[2J")
for i in 1:30
println("\33[H")
y = sin.(x .+ (i/2));
plot(x, y, show=true, xlims = (0,4*pi), ylims = (-1,1))
sleep(0.01)
end
end
I want to plot the function
4(x)^2 = ((y)^2/(1-y));
how can I plot this?
--> 4*(x) = ((y^2)*(1-y)^-1)^0.5;
4*(x) = ((y^2)*(1-y)^-1)^0.5;
^^
Error: syntax error, unexpected =, expecting end of file
Since Scilab 6.1.0, plotimplicit() does it:
plotimplicit "4*x^2 = y^2/(1-y)"
xgrid()
Can't do more simple. Result:
Well, you have to first create a function and for that you have to express one variable in terms of the other.
function x = f(y)
x = (((y^2)*(1-y)^-1)^0.5)/4;
endfunciton
Then you need to generate the input data (i.e, the points at which you want to evaluate the function)
ydata = linspace(1, 10)
Now you push your input point through the function to get your output points
xdata = f(ydata)
Then, you can plot the pairs of x and y using:
plot(xdata, ydata)
Or even easier, without the intermediate step of generating the output data, you can simply do:
plot(f(ydata), ydata)
BTW. I find it strange that the function you are trying to plot is x in terms of y, usually, x is the input variable, but I hope you know what you are trying to accomplish.
Reference: https://www.scilab.org/tutorials/getting-started/plotting
Take care that y must be in [-inf 1[
y=linspace(-10 ,1.00001,1000);
x = sqrt(y^2./(1-y))/4;
clf; plot(y,x),plot(y,-x)
If x is a solution -x is also solution
I would like to create a stacked area chart, similar to this for example, in Julia using Plots.
I know / suppose that you can do this if you directly use the Gadfly or PyPlot backends in Julia, but I was wondering if there was a recipe for this. If not, how can you contribute to the Plots Recipes? Would be a useful addition.
There's a recipe for something similar in
https://docs.juliaplots.org/latest/examples/pgfplots/#portfolio-composition-maps
For some reason the thumbnail looks broken now though (but the code works).
The exact plot in the matlab example can be produced by
plot(cumsum(Y, dims = 2)[:,end:-1:1], fill = 0, lc = :black)
As a recipe that would look like
#userplot AreaChart
#recipe function f(a::AreaChart)
fillto --> 0
linecolor --> :black
seriestype --> :path
cumsum(a.args[1], dims = 2)[:,end:-1:1]
end
If you want to contribute a recipe to Plots you can open a pull request on Plots, or, eg. on StatsPlots - there's a good description of contributing here: https://docs.juliaplots.org/latest/contributing/
It's a bit of reading, but very generally useful as an introduction to contributing to Julia packages.
You can read this thread in the Julia discourse forum where the question is developed in deep.
One solution posted there using Plots is :
# a simple "recipe" for Plots.jl to get stacked area plots
# usage: stackedarea(xvector, datamatrix, plotsoptions)
#recipe function f(pc::StackedArea)
x, y = pc.args
n = length(x)
y = cumsum(y, dims=2)
seriestype := :shape
# create a filled polygon for each item
for c=1:size(y,2)
sx = vcat(x, reverse(x))
sy = vcat(y[:,c], c==1 ? zeros(n) : reverse(y[:,c-1]))
#series (sx, sy)
end
end
a = [1,1,1,1.5,2,3]
b = [0.5,0.6,0.4,0.3,0.3,0.2]
c = [2,1.8,2.2,3.3,2.5,1.8]
sNames = ["a","b","c"]
x = [2001,2002,2003,2004,2005,2006]
plotly()
stackedarea(x, [a b c], labels=reshape(sNames, (1,3)))
(by user NiclasMattsson)
Other ways presented there include using the VegaLite.jl package.
I am using a package(LightGraphs.jl) in Julia, and it has a predefined histogram method that creates the degree distribution of a network g.
deg_hist = degree_histogram(g)
I want to make a plot of this but i am new to plotting in Julia. The object returned is a StatsBase.Histogram which has the following as its inner fields:
StatsBase.Histogram{Int64,1,Tuple{FloatRange{Float64}}}
edges: 0.0:500.0:6000.0
weights: [79143,57,32,17,13,4,4,3,3,2,1,1]
closed: right
Can you help me how I can make use of this object to plot the histogram?
I thought this was already implemented, but I just added the recipe to StatPlots. If you check out master, you'll be able to do:
julia> using StatPlots, LightGraphs
julia> g = Graph(100,200);
julia> plot(degree_histogram(g))
For reference, the associated recipe that I added to StatPlots:
#recipe function f(h::StatsBase.Histogram)
seriestype := :histogram
h.edges[1], h.weights
end
Use the histogram fields .edges and .weights to plot it e.g.
using PyPlot, StatsBase
a = rand(1000); # generate something to plot
test_hist = fit(Histogram, a)
# line plot
plot(test_hist.edges[1][2:end], test_hist.weights)
# bar plot
bar(0:length(test_hist.weights)-1, test_hist.weights)
xticks(0:length(test_hist.weights), test_hist.edges[1])
or you could create/extend a plotting function adding a method like so:
function myplot(x::StatsBase.Histogram)
... # your code here
end
Then you will be able to call your plotting functions directly on the histogram object.