Need help with array calculation

New to Julia
1

I would like to ask for help about array calculation

I want to plot contourline of f in which:

    f =  q + p*ln(p )

To do that,

    n = 100
    p = range(0, stop = 200, length =n )
    q = range(0, stop = 200, length = n)
    p_grid = repeat(p, 1, n)
    q_grid = repeat(q', n, 1)

Then, how to generate f_grid? ( f = q + p*ln(p ))
If possible, I can use PyPlot to plot:

  contour(p_grid, q_grid, f_grid, levels=0);

( in case of f2 : f2 = q²+ p*(p - 1)
I could create: f2_grid = (q_grid.^2 + p_grid.^2 - p_grid)

2

First, let’s define f as an actual function:

julia> f = (x, y) -> x + y * log(y)
#11 (generic function with 1 method)

You can broadcast this function over p and q to create a grid:

julia> f.(p, q')
100×100 Array{Float64,2}:
...

But you actually don’t need to! In fact, Plots.jl already knows how to plot the contour of a function if you just pass it the input vectors p and q and the function itself:

julia> using Plots

julia> n = 100
100

julia> q = range(0, stop = 200, length = n)
0.0:2.0202020202020203:200.0

julia> p = range(0, stop = 200, length =n )
0.0:2.0202020202020203:200.0

julia> f = (x, y) -> x + y * log(y)
#11 (generic function with 1 method)

julia> contour(p, q, f)

Note that we also don’t need to generate p_grid or q_grid because Plots.jl is smart enough to work directly from the vectors.

4 Likes
3

Because of the way broadcasting works you very rarely need to define grids like p_grid and q_grid. So if you find yourself using repeat on a vector, there’s a good chance you’re doing something unnecessary.

In fact the same is true for both Matlab and numpy too, so this technique is quite outdated these days.

2 Likes
4

Fantastic! Thank you very much for you quick reply.
I have only used PyPlot. This may makes me switch to Plots!