You can use the GR framework, e.g. srand(0)
xd = -2 + 4 * rand(100) yd = -2 + 4 * rand(100) zd = [Float64(xd[i] * exp(-xd[i]^2 - yd[i]^2)) for i = 1:100] using GR contour(xd, yd, zd) <https://lh3.googleusercontent.com/-K-nsCh7jQcE/V4CjycA0NBI/AAAAAAAAADI/R2mkyGUKfroWY8mV-bHecTAVZKb13-EMQCLcB/s1600/Screen%2BShot%2B2016-07-09%2Bat%2B09.10.17.png> If you only need the mesh data, use the gridit function, e.g. (for a 200 x 200 grid): x, y, z = GR.gridit(xd, yd, zd, 200, 200) On Saturday, July 9, 2016 at 12:57:51 AM UTC+2, Kaj Wiik wrote: > > Is there a Julia version of irregularly spaced data gridding that does > zi = griddata(x,y,z,xi,yi), i.e. all arguments are 1d vectors? It seems > that Julia interp and contour packages require x, y, z[x,y]. > > > https://scipy.github.io/old-wiki/pages/Cookbook/Matplotlib/Gridding_irregularly_spaced_data.html > > Thanks, > Kaj > >
