Easy plotting of vector valued function in Plots.jl

Specific Domains Modelling & Simulations
,
1

Hi,

I’d like to plot a function from R -> R^N in Plots.jl. My code is:

using Plots 
N = 9
v₁ = ones(N)
v₂ = [1, -8, ones(N-2)...]
z₀ = 1.0

function z(t; v₁ = v₁, v₂ = v₂, z₀ = z₀)
	(z₀ / 9.0) .* (v₁ - exp(-9.0*t)*v₂)
end

The function works as expected, but I’d like to be able to do something like:

plot(t -> z(t), 0.0:0.01:1.0)

but that errors. The best I can do is:

data = [[z(t)[i] for t ∈ 0.0:0.001:1.0] for i ∈ 1:N]
plot(data)

which works (and looks great) but I wondered if there is a better way to interact with the Plots API, or if there is another package that connects Plots to vector valued functions?

2

Try this:

t= 0.0:0.01:1.0
plot(t, reduce(hcat, z.(t))')
1 Like
3

A similar solution to reduce(hcat, z.(t))' is to use the SplitApplyCombine.invert function, which does what your data construct does. (I wish there were a general unzip function in base). For your problem, the use would be:

using SplitApplyCombine, Plots
plot(t, invert(z.(t)))

As an aside, this is also of use for the more traditional plot of f:R → R^n with n in 2 or 3, that being a space curve, which can be plotted with a pattern like:

t = 0:pi/10:2pi
plot(invert(sincos.(t))...)
1 Like
4

Hi both,

Thanks for these - just what I was looking for!

5

As in Julia we are spoilt for choice, we shouldn’t deprive ourselves :

using TensorCast
plot(t, @cast _[j][i] := z.(t)[i][j])