From 文档:
c(ndarray, shape (4, n-1, ...)) 每段多项式的系数。尾部尺寸与以下尺寸相匹配y
,
不包括轴。例如,如果y
是 1-d,那么c[k, i]
是一个
系数为(x-x[i])**(3-k)
在之间的段上x[i]
and
x[i+1]
.
So in your example, the coefficients for the first segment [x1, x2] would be in column 0:
-
y1 would be
s.c[3, 0]
-
b1 would be
s.c[2, 0]
-
c1 would be
s.c[1, 0]
-
d1 would be
s.c[0, 0]
.
Then for the second segment [x2, x3] you would have s.c[3, 1]
, s.c[2, 1]
, s.c[1, 1]
and s.c[0, 1]
for y2, b2, c2, d2, and so on and so forth.
例如:
x = np.array([1, 2, 4, 5]) # sort data points by increasing x value
y = np.array([2, 1, 4, 3])
arr = np.arange(np.amin(x), np.amax(x), 0.01)
s = interpolate.CubicSpline(x, y)
fig, ax = plt.subplots(1, 1)
ax.hold(True)
ax.plot(x, y, 'bo', label='Data Point')
ax.plot(arr, s(arr), 'k-', label='Cubic Spline', lw=1)
for i in range(x.shape[0] - 1):
segment_x = np.linspace(x[i], x[i + 1], 100)
# A (4, 100) array, where the rows contain (x-x[i])**3, (x-x[i])**2 etc.
exp_x = (segment_x - x[i])[None, :] ** np.arange(4)[::-1, None]
# Sum over the rows of exp_x weighted by coefficients in the ith column of s.c
segment_y = s.c[:, i].dot(exp_x)
ax.plot(segment_x, segment_y, label='Segment {}'.format(i), ls='--', lw=3)
ax.legend()
plt.show()
![enter image description here](https://i.stack.imgur.com/Bk7rm.png)