标签：scatter-plot python matplotlib polygon polygons
我有一堆带有两组数据的交叉图,并且一直在寻找一种matploltib方法,用平滑的多边形轮廓突出显示它们的绘图区域.
目前我只使用Adobe Illustrator并修改已保存的情节,但这并不理想.例：
我很感激任何指针/链接到示例.
干杯
解决方法:
在这里,你有一个例子.我写了主要的想法,但显然,你可以做得更好.
简短说明：
1)你需要计算凸壳(http://en.wikipedia.org/wiki/Convex_hull)
2)使用船体,您可以对其进行缩放以保留所有数据.
3)您必须插入结果曲线.
第一部分是在http://wiki.scipy.org/Cookbook/Finding_Convex_Hull完成的.第二部分是微不足道的.第三个是非常通用的,你可以执行任何方法,有很多不同的方法来做同样的事情.我采用了@ Jaime的方法(Smooth spline representation of an arbitrary contour, f(length) –> x,y),我认为这是一种非常好的方法.
我希望它可以帮到你……
#Taken from http://wiki.scipy.org/Cookbook/Finding_Convex_Hull
import numpy as n, pylab as p, time
def _angle_to_point(point, centre):
'''calculate angle in 2-D between points and x axis'''
delta = point - centre
res = n.arctan(delta[1] / delta[0])
if delta[0] < 0:
res += n.pi
return res
def _draw_triangle(p1, p2, p3, **kwargs):
tmp = n.vstack((p1,p2,p3))
x,y = [x[0] for x in zip(tmp.transpose())]
p.fill(x,y, **kwargs)
def area_of_triangle(p1, p2, p3):
'''calculate area of any triangle given co-ordinates of the corners'''
return n.linalg.norm(n.cross((p2 - p1), (p3 - p1)))/2.
def convex_hull(points, graphic=False, smidgen=0.0075):
'''
Calculate subset of points that make a convex hull around points
Recursively eliminates points that lie inside two neighbouring points until only convex hull is remaining.
:Parameters:
points : ndarray (2 x m)
array of points for which to find hull
graphic : bool
use pylab to show progress?
smidgen : float
offset for graphic number labels - useful values depend on your data range
:Returns:
hull_points : ndarray (2 x n)
convex hull surrounding points
'''
if graphic:
p.clf()
p.plot(points[0], points[1], 'ro')
n_pts = points.shape[1]
assert(n_pts > 5)
centre = points.mean(1)
if graphic: p.plot((centre[0],),(centre[1],),'bo')
angles = n.apply_along_axis(_angle_to_point, 0, points, centre)
pts_ord = points[:,angles.argsort()]
if graphic:
for i in xrange(n_pts):
p.text(pts_ord[0,i] + smidgen, pts_ord[1,i] + smidgen, \
'%d' % i)
pts = [x[0] for x in zip(pts_ord.transpose())]
prev_pts = len(pts) + 1
k = 0
while prev_pts > n_pts:
prev_pts = n_pts
n_pts = len(pts)
if graphic: p.gca().patches = []
i = -2
while i < (n_pts - 2):
Aij = area_of_triangle(centre, pts[i], pts[(i + 1) % n_pts])
Ajk = area_of_triangle(centre, pts[(i + 1) % n_pts], \
pts[(i + 2) % n_pts])
Aik = area_of_triangle(centre, pts[i], pts[(i + 2) % n_pts])
if graphic:
_draw_triangle(centre, pts[i], pts[(i + 1) % n_pts], \
facecolor='blue', alpha = 0.2)
_draw_triangle(centre, pts[(i + 1) % n_pts], \
pts[(i + 2) % n_pts], \
facecolor='green', alpha = 0.2)
_draw_triangle(centre, pts[i], pts[(i + 2) % n_pts], \
facecolor='red', alpha = 0.2)
if Aij + Ajk < Aik:
if graphic: p.plot((pts[i + 1][0],),(pts[i + 1][1],),'go')
del pts[i+1]
i += 1
n_pts = len(pts)
k += 1
return n.asarray(pts)
if __name__ == "__main__":
import scipy.interpolate as interpolate
# fig = p.figure(figsize=(10,10))
theta = 2*n.pi*n.random.rand(1000)
r = n.random.rand(1000)**0.5
x,y = r*p.cos(theta),r*p.sin(theta)
points = n.ndarray((2,len(x)))
points[0,:],points[1,:] = x,y
scale = 1.03
hull_pts = scale*convex_hull(points)
p.plot(x,y,'ko')
x,y = [],[]
convex = scale*hull_pts
for point in convex:
x.append(point[0])
y.append(point[1])
x.append(convex[0][0])
y.append(convex[0][1])
x,y = n.array(x),n.array(y)
#Taken from https://stackoverflow.com/questions/14344099/numpy-scipy-smooth-spline-representation-of-an-arbitrary-contour-flength
nt = n.linspace(0, 1, 100)
t = n.zeros(x.shape)
t[1:] = n.sqrt((x[1:] - x[:-1])**2 + (y[1:] - y[:-1])**2)
t = n.cumsum(t)
t /= t[-1]
x2 = interpolate.spline(t, x, nt)
y2 = interpolate.spline(t, y, nt)
p.plot(x2, y2,'r--',linewidth=2)
p.show()
有一些有用的论文,例如：
http://repositorium.sdum.uminho.pt/bitstream/1822/6429/1/ConcaveHull_ACM_MYS.pdf
此外,您可以尝试：http://en.wikipedia.org/wiki/Convex_hull000013000000″ rel=”noreferrer”>http://resources.arcgis.com/en/help/main/10.1/index.html#//0http://en.wikipedia.org/wiki/Convex_hull000013000000
标签：scatter-plot,python,matplotlib,polygon,polygons 来源： https://codeday.me/bug/20191006/1858337.html
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