如何通过Python设置F曲线的插值？ （搅拌机）

Question

I have a Scale F-Curve of an Object and what I need Is to set its interpolation as CUBIC for example.

What is the simpliest and the fastest way to do it?

Answer 1

It's a long winding path to the fcurves ;), but once you get there it is quick to work with.

Starting with the active object, you want to go to the fcurves

fc = bpy.context.active_object.animation_data.action.fcurves

Other fcurves can be found in similar paths eg for material nodes it is

fc = mat.node_tree.animation_data.action.fcurves

fcurves is a list of all curves, it is usually easiest to use find to get the curve you want (the index values 0,1,2 match x,y,z) unless you want to loop through and change them all.

loc_x_curve = fc.find('scale', index=0)

Then each curve is a list of keyframe items that have their own interpolation setting.

for k in loc_x_curve.keyframe_points:
    # k.co[0] is the frame number
    # k.co[1] is the keyed value
    k.interpolation = 'CUBIC'
    k.easing = 'EASE_IN'

Answer 2

Try using SciPy, for example , the following would work:

>>> from scipy.interpolate import interp1d
>>> x = np.linspace(0, 10, num=11, endpoint=True)
>>> y = np.cos(-x**2/9.0)
>>> f = interp1d(x, y)
>>> f2 = interp1d(x, y, kind='cubic')
>>> xnew = np.linspace(0, 10, num=41, endpoint=True)
>>> import matplotlib.pyplot as plt
>>> plt.plot(x, y, 'o', xnew, f(xnew), '-', xnew, f2(xnew), '--')
>>> plt.legend(['data', 'linear', 'cubic'], loc='best')
>>> plt.show()