How to plot a vector field in Julia?

问题

I want to plot a vector field in Julia. I could not find an example here. Here there are some examples using plotly, however, they do not work for me. I would like to plot the vector field by plotlyjs or plotly.

Here is an example code in Julia:

using Plots
pyplot()
x = collect(linspace(0,10,100));
X = repmat(x,1,length(x));
Y = repmat(x',length(x),1);
U = cos.(X.*Y);
V = sin.(X.*Y);
streamplot(X,Y,U,V)

Here is the Matlab example:

[x,y] = meshgrid(0:0.2:2,0:0.2:2);
u = cos(x).*y;
v = sin(x).*y;

figure
quiver(x,y,u,v)

回答1:

The short answer: use quiver from Plots.jl.

quiver(x, y, quiver=(u, v))

---

In the following, I'll attempt to fully recreate the example you showed in Matlab.

First, we'll import Plots and enable the plotly backend.

using Plots
plotly()

We need to define a function similar to Matlab's meshgrid. Since Plots will operate on our arrays of points regardless of their dimensionality, I chose to simply use repeat and use the "flattened" outputs.

 meshgrid(x, y) = (repeat(x, outer=length(y)), repeat(y, inner=length(x)))

Now, we can create x, y, u, and v using the same logic as the Matlab code. For the sake of brevity, we can use the @. macro to vectorize all calls in the given expression.

x, y = meshgrid(0:0.2:2, 0:0.2:2)
u = @. cos(x) * y
v = @. sin(x) * y

From here, we can simply use the quiver function from Plots, passing u and v as a 2-tuple to the keyword argument quiver.

quiver(x, y, quiver=(u, v))

The result is close to the Matlab output, but it seems that Plots.jl scales the arrows to be longer than they are in Matlab. This is easily fixable, though; we can simply broadcast-multiply u and v by a scale constant.

scale = 0.2
u = @. scale * cos(x) * y
v = @. scale * sin(x) * y

来源：https://stackoverflow.com/questions/51466537/how-to-plot-a-vector-field-in-julia