With ScatteredInterpolation.jl we can get from the three vectors, x, y, z, a matrix zg, consisting in interpolated values at the points of a meshgrid associated to linear grids, xg and yg.
As long as we don’t know what interpolation method is used by
plotly.js, we cannot obtain exactly the same contours. Among different methods of interpolation from ScatteredInterpolation.jl, the radial basis function method with Multiquadratic basis function defines a Plots.contourf somehow similar to PlotlyJS.contour.
using ScatteredInterpolation, Plots, ColorSchemes
let
x = [1, 2, 3, 1]
y = [1, 1, 2, 2]
z = [1.0, 2, 3, 4]
points = stack(zip(x,y))
itp = interpolate(Multiquadratic(), points, z);
xm, xM = extrema(x)
ym, yM = extrema(y)
n = 50
m = 25
xg = range(xm, xM, n)
yg = range(ym, yM, m)
X = [s for t in yg, s in xg] #size(X) is (m,n)
Y = [t for t in yg, s in xg] # size(Y) is also (m,n)
gridP = stack(vec([[x, y] for (x,y) in zip(X, Y)])) #2 x m*n - matrix; each column is a grid point
interpolated = evaluate(itp, gridP)
zg = reshape(interpolated, m, n)
p = Plots.contourf(xg, yg, zg; levels=6, c=cgrad(ColorSchemes.plasma))
end
