Plotting function of (x,y) with 3D plot

Question

I have a function that takes one 2D vector and returns pdf (a scalar) of that point.

As an illustration, myPDF( np.array[4,4] ) returns a single pdf value at <4,4>. This function cannot take a list of vectors, and it only accepts one 2D vector as an argument.

Now I want to plot this pdf using plot_surface like below: This is what I've tried

x = np.arange(-10,10)
y = np.arange(-10,10)
z = [ myPDF( np.array([xx,yy])) for xx in x for yy in y]
xy = [ [xx, yy] for xx in x for yy in y]

And here I am stuck. I have to somehow reshape z and xy in some proper forms to use them with plot_surface, but I do not now how to do so. I also tried meshgrid, but did not know how to convert different shapes of data into legitimate forms for plot_surface.

x, y = np.mgrid[-5:5,-5:5]

If I use mgrid as above, it gives me 10x10 matrix. Then, I have to generate corresponding z values in 10x10 matrix using my function, but could not find a way to decompose x and y, generate z values and reformat z values in a proper form.

The major problem here is that I have little knowledge in manipulating data, so could not find a good way of doing this task. Thus, I'd like to learn a proper way of generating a dataset for plot_surface using my pdf function.

What's a good way to generate data points to plot like this using my pdf function? Thanks for helping this newB!

"I tried arange, meshgrid and so on, but could not find a way to make a proper data points" - what was the problem with these methods? Show us your code. — ali_m
– ali_m, Commented Sep 26, 2016 at 18:19

Eric · Accepted Answer · 2016-09-27 00:07:13Z

1

Use apply_along_axis

xy = np.stack(np.mgrid[-5:5,-5:5], axis=-1)      # xy[0,0] = x, y
z = np.apply_along_axis(myPDF, axis=-1, arr=xy)  # z[0,0] = myPDF(xy[0,0])

answered Sep 27, 2016 at 0:07

Eric

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3 Comments

noclew Over a year ago

By the way, why axis is -1?

Eric Over a year ago

Doesn't actually matter here, but axis=- makes "x or y" the last axis, which makes (xij, yij) == (xy[i,j,0], xy[i,j,1]) == xy [i,j]. You could a pick a different axis, but then then 1 and 0 would not be last.

noclew Over a year ago

Thanks! last axis1

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