我在 2D 网格上有布尔资料,想用来matplotlib
在资料所在的区域True
和False
.
然而,这些区域之间的分离在实际资料中并不平滑。给定这些资料,如何计算平滑计数?
这是一个最小的例子:
import numpy as np
import matplotlib.pyplot as plt
# generate some non-smooth example data
MESHSIZE = 10
REFINEMENT = 4*MESHSIZE
x = np.linspace(-MESHSIZE, MESHSIZE, REFINEMENT)
xv, yv = np.meshgrid(x, x)
xvf = xv.reshape(-1)
yvf = yv.reshape(-1)
def choppy_circle(x, y):
inner = x.astype(int)**2 y.astype(int)**2 < 10.0
return inner
# consider this the *actual* data given to me as-is
my_x = xvf
my_y = yvf
my_z = choppy_circle(xvf, yvf)
# need to visualize the contour that separates areas where
# my_z is True/False
plt.tricontour(my_x, my_y, my_z, levels=np.array([1.0-1e-3]))
plt.scatter(xv, yv, s=0.1)
plt.show()
这会产生以下情节,该情节忠实于资料,但不是我想要的:
如何使用 中给出的资料my_x
,my_y
并围绕其中的my_z
域构建平滑轮廓?my_z
True
像这样的东西:
uj5u.com热心网友回复:
您可以将样条曲线拟合到轮廓。并通过选择样条曲线的平滑自变量使其尽可能平滑。
首先,您获得边界点
import functools
import itertools
mask = my_z.reshape(40,40)
mask &= functools.reduce(np.logical_or,[~np.roll(np.roll(mask, shift_x, 0),shift_y,1)
for shift_x,shift_y in itertools.product((-1,0,1),repeat=2)])
x,y = my_x[mask.reshape(-1)],my_y[mask.reshape(-1)]
plt.scatter(x,y)
现在我们通过相应复数的自变量对您的点进行排序。如果您不知道我的意思是该点与原点和点 (1,0) 形成的角度。并为其拟合样条。
import scipy.interpolate as interpolate
import matplotlib.pyplot as plt
arr = np.array(sorted(zip(x,y), key=lambda x: cmath.phase(x[0] 1j*x[1])))
s=1
tck, u = interpolate.splprep([arr[:,0],arr[:,1]],per=1, s=s)
x_i, y_i = interpolate.splev(np.linspace(0, 1, 10**4), tck)
ax = plt.gca()
ax.plot(x_i, y_i)
ax.scatter(arr[:,0],arr[:,1])
ax.set_title(f"{s=}")
ax.set_aspect('equal')
结果看起来会有所不同,具体取决于s
. 我为您绘制了一些:
uj5u.com热心网友回复:
您可以使用shapely
获取任意形状的质心和边界框,然后绘制一个圆:
# […] same as previously
# get points
cs = plt.tricontour(my_x, my_y, my_z, levels=np.array([1.0-1e-3]))
v = cs.collections[0].get_paths()[0].vertices
from shapely.geometry import Polygon
# find centroid coordinates and bounding box
p = Polygon(v)
x,y =p.centroid.coords[0]
minx, miny, maxx, maxy = p.bounds
# plot circle
# depending on the data, one could also plot an ellipse or rectangle
r = max((maxx-minx)/2, (maxy-miny)/2)
circle = plt.Circle((x, y), r, color='r', fill=False)
plt.gca().add_patch(circle)
输出:
uj5u.com热心网友回复:
提取此答案中提出的轮廓资料并使用@user2640045 提出的样条插值允许对任意轮廓执行此操作:
# my_x, my_y, my_z as above...
# get contour data
cs = plt.tricontour(my_x, my_y, my_z, levels=np.array([1.0-1e-3]))
print(type(cs))
# process each contour
for contour in cs.collections[0].get_paths():
vert = contour.vertices
vert_x = vert[:, 0]
vert_y = vert[:, 1]
# plot contour
plt.plot(vert_x, vert_y)
# interpolate contour points
s = 20
tck, u = interpolate.splprep([vert_x, vert_y], per=1, s=s)
x_interp, y_interp = interpolate.splev(np.linspace(0, 1, 10**3), tck)
# plot interpolated contour
plt.plot(x_interp, y_interp)
# plot grid
plt.scatter(xv, yv, s=0.1)
# display plot
plt.show()
重要的一点是回圈头
for contour in cs.collections[0].get_paths():
其中得到每条等高线的xy资料。
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