我想知道找到两组轮廓线之间所有交点（舍入误差）的最佳方法。哪种方法最好？这是示例：
import matplotlib.pyplot as plt
import numpy as np
x = np.linspace(-1,1,500)
X,Y = np.meshgrid(x,x)
Z1 = np.abs(np.sin(2*X**2+Y))
Z2 = np.abs(np.cos(2*Y**2+X**2))
plt.contour(Z1,colors='k')
plt.contour(Z2,colors='r')
plt.show()
我想要一些类似于：
intersection_points = intersect(contour1,contour2)
print intersection_points
[(x1,y1),...,(xn,yn)]
import collections
import matplotlib.pyplot as plt
import numpy as np
import scipy.spatial as spatial
import scipy.spatial.distance as dist
import scipy.cluster.hierarchy as hier
def intersection(points1, points2, eps):
tree = spatial.KDTree(points1)
distances, indices = tree.query(points2, k=1, distance_upper_bound=eps)
intersection_points = tree.data[indices[np.isfinite(distances)]]
return intersection_points
def cluster(points, cluster_size):
dists = dist.pdist(points, metric='sqeuclidean')
linkage_matrix = hier.linkage(dists, 'average')
groups = hier.fcluster(linkage_matrix, cluster_size, criterion='distance')
return np.array([points[cluster].mean(axis=0)
for cluster in clusterlists(groups)])
def contour_points(contour, steps=1):
return np.row_stack([path.interpolated(steps).vertices
for linecol in contour.collections
for path in linecol.get_paths()])
def clusterlists(T):
'''
http://stackoverflow.com/a/2913071/190597 (denis)
T = [2, 1, 1, 1, 2, 2, 2, 2, 2, 1]
Returns [[0, 4, 5, 6, 7, 8], [1, 2, 3, 9]]
'''
groups = collections.defaultdict(list)
for i, elt in enumerate(T):
groups[elt].append(i)
return sorted(groups.values(), key=len, reverse=True)
# every intersection point must be within eps of a point on the other
# contour path
eps = 1.0
# cluster together intersection points so that the original points in each flat
# cluster have a cophenetic_distance < cluster_size
cluster_size = 100
x = np.linspace(-1, 1, 500)
X, Y = np.meshgrid(x, x)
Z1 = np.abs(np.sin(2 * X ** 2 + Y))
Z2 = np.abs(np.cos(2 * Y ** 2 + X ** 2))
contour1 = plt.contour(Z1, colors='k')
contour2 = plt.contour(Z2, colors='r')
points1 = contour_points(contour1)
points2 = contour_points(contour2)
intersection_points = intersection(points1, points2, eps)
intersection_points = cluster(intersection_points, cluster_size)
plt.scatter(intersection_points[:, 0], intersection_points[:, 1], s=20)
plt.show()
产量
根据@unutbu 的答案和直接从numpy-and-line-intersections中提取的交集算法，我想出了这个。由于在循环中找到交叉点和循环内循环的蛮力方式，它很慢。可能有一种方法可以使循环更快，但我不确定相交算法。
import matplotlib.pyplot as plt
import numpy as np
def find_intersections(A, B):
#this function stolen from https://stackoverflow.com/questions/3252194/numpy-and-line-intersections#answer-9110966
# min, max and all for arrays
amin = lambda x1, x2: np.where(x1<x2, x1, x2)
amax = lambda x1, x2: np.where(x1>x2, x1, x2)
aall = lambda abools: np.dstack(abools).all(axis=2)
slope = lambda line: (lambda d: d[:,1]/d[:,0])(np.diff(line, axis=0))
x11, x21 = np.meshgrid(A[:-1, 0], B[:-1, 0])
x12, x22 = np.meshgrid(A[1:, 0], B[1:, 0])
y11, y21 = np.meshgrid(A[:-1, 1], B[:-1, 1])
y12, y22 = np.meshgrid(A[1:, 1], B[1:, 1])
m1, m2 = np.meshgrid(slope(A), slope(B))
m1inv, m2inv = 1/m1, 1/m2
yi = (m1*(x21-x11-m2inv*y21) + y11)/(1 - m1*m2inv)
xi = (yi - y21)*m2inv + x21
xconds = (amin(x11, x12) < xi, xi <= amax(x11, x12),
amin(x21, x22) < xi, xi <= amax(x21, x22) )
yconds = (amin(y11, y12) < yi, yi <= amax(y11, y12),
amin(y21, y22) < yi, yi <= amax(y21, y22) )
return xi[aall(xconds)], yi[aall(yconds)]
x = np.linspace(-1,1,500)
X,Y = np.meshgrid(x,x)
Z1 = np.abs(np.sin(2*X**2+Y))
Z2 = np.abs(np.cos(2*Y**2+X**2))
contour1 = plt.contour(Z1,colors='k')
contour2 = plt.contour(Z2,colors='r')
xi = np.array([])
yi = np.array([])
for linecol in contour1.collections:
for path in linecol.get_paths():
for linecol2 in contour2.collections:
for path2 in linecol2.get_paths():
xinter, yinter = find_intersections(path.vertices, path2.vertices)
xi = np.append(xi, xinter)
yi = np.append(yi, yinter)
plt.scatter(xi, yi, s=20)
plt.show()
编辑： find_intersections上面不处理垂直和水平线也不重叠线段。下面的linelineintersect函数确实处理了这些情况，但整个过程仍然很慢。我已经添加了一个计数，因此您对要走多长时间有所了解。我还更改contour1 = plt.contour(Z1,colors='k')为contour1 = plt.contour(X,Y,Z1,colors='k')，轴和您的交叉点位于网格的实际 X 和 Y 坐标中，而不是网格点的计数。
我认为问题在于您只有很多数据。一个等高线集中有 24 条线，共 12818 个线段，另一个等高线集中有 29 条线，共 11411 个线段。有很多线段组合需要检查（696 次调用linelineintersect）。通过使用更粗略的网格来计算函数，您可以更快地获得结果（100 x 100 比原来的 500 x 500 快得多）。你也许可以做某种线扫描算法，但我不知道如何实现这些东西。线交点问题在计算机游戏中有很多应用，被称为碰撞检测——必须有一些图形库可以快速确定所有交点。
import numpy as np
from numpy.lib.stride_tricks import as_strided
import matplotlib.pyplot as plt
import itertools
def unique(a, atol=1e-08):
"""Find unique rows in 2d array
Parameters
----------
a : 2d ndarray, float
array to find unique rows in
atol : float, optional
tolerance to check uniqueness
Returns
-------
out : 2d ndarray, float
array of unique values
Notes
-----
Adapted to include tolerance from code at https://stackoverflow.com/questions/8560440/removing-duplicate-columns-and-rows-from-a-numpy-2d-array#answer-8564438
"""
if np.issubdtype(a.dtype, float):
order = np.lexsort(a.T)
a = a[order]
diff = np.diff(a, axis=0)
np.abs(diff,out = diff)
ui = np.ones(len(a), 'bool')
ui[1:] = (diff > atol).any(axis=1)
return a[ui]
else:
order = np.lexsort(a.T)
a = a[order]
diff = np.diff(a, axis=0)
ui = np.ones(len(a), 'bool')
ui[1:] = (diff != 0).any(axis=1)
return a[ui]
def linelineintersect(a, b, atol=1e-08):
"""Find all intersection points of two lines defined by series of x,y pairs
Intersection points are unordered
Colinear lines that overlap intersect at any end points that fall within the overlap
Parameters
----------
a, b : ndarray
2 column ndarray of x,y values defining a two dimensional line. 1st
column is x values, 2nd column is x values.
Notes
-----
An example of 3 segment line is: a = numpy.array([[0,0],[5.0,3.0],[4.0,1]])
Function faster when there are no overlapping line segments
"""
def x_y_from_line(line):
"""1st and 2nd column of array"""
return (line[:, 0],line[:, 1])
def meshgrid_as_strided(x, y, mask=None):
"""numpy.meshgrid without copying data (using as_strided)"""
if mask is None:
return (as_strided(x, strides=(0, x.strides[0]), shape=(y.size, x.size)),
as_strided(y, strides=(y.strides[0],0), shape=(y.size, x.size)))
else:
return (np.ma.array(as_strided(x, strides=(0, x.strides[0]), shape=(y.size, x.size)), mask=mask),
np.ma.array(as_strided(y, strides=(y.strides[0],0), shape=(y.size, x.size)), mask=mask))
#In the following the indices i, j represents the pairing of the ith segment of b and the jth segment of a
#e.g. if ignore[i,j]==True then the ith segment of b and the jth segment of a cannot intersect
ignore = np.zeros([b.shape[0]-1, a.shape[0]-1], dtype=bool)
x11, x21 = meshgrid_as_strided(a[:-1, 0], b[:-1, 0], mask=ignore)
x12, x22 = meshgrid_as_strided(a[1: , 0], b[1: , 0], mask=ignore)
y11, y21 = meshgrid_as_strided(a[:-1, 1], b[:-1, 1], mask=ignore)
y12, y22 = meshgrid_as_strided(a[1: , 1], b[1: , 1], mask=ignore)
#ignore segements with non-overlappiong bounding boxes
ignore[np.ma.maximum(x11, x12) < np.ma.minimum(x21, x22)] = True
ignore[np.ma.minimum(x11, x12) > np.ma.maximum(x21, x22)] = True
ignore[np.ma.maximum(y11, y12) < np.ma.minimum(y21, y22)] = True
ignore[np.ma.minimum(y11, y12) > np.ma.maximum(y21, y22)] = True
#find intersection of segments, ignoring impossible line segment pairs when new info becomes available
denom_ = np.empty(ignore.shape, dtype=float)
denom = np.ma.array(denom_, mask=ignore)
denom_[:, :] = ((y22 - y21) * (x12 - x11)) - ((x22 - x21) * (y12 - y11))
ua_ = np.empty(ignore.shape, dtype=float)
ua = np.ma.array(ua_, mask=ignore)
ua_[:, :] = (((x22 - x21) * (y11 - y21)) - ((y22 - y21) * (x11 - x21)))
ua_[:, :] /= denom
ignore[ua < 0] = True
ignore[ua > 1] = True
ub_ = np.empty(ignore.shape, dtype=float)
ub = np.ma.array(ub_, mask=ignore)
ub_[:, :] = (((x12 - x11) * (y11 - y21)) - ((y12 - y11) * (x11 - x21)))
ub_[:, :] /= denom
ignore[ub < 0] = True
ignore[ub > 1] = True
nans_ = np.zeros(ignore.shape, dtype = bool)
nans = np.ma.array(nans_, mask = ignore)
nans_[:,:] = np.isnan(ua)
if not np.ma.any(nans):
#remove duplicate cases where intersection happens on an endpoint
# ignore[np.ma.where((ua[:, :-1] == 1) & (ua[:, 1:] == 0))] = True
# ignore[np.ma.where((ub[:-1, :] == 1) & (ub[1:, :] == 0))] = True
ignore[np.ma.where((ua[:, :-1] < 1.0 + atol) & (ua[:, :-1] > 1.0 - atol) & (ua[:, 1:] < atol) & (ua[:, 1:] > -atol))] = True
ignore[np.ma.where((ub[:-1, :] < 1 + atol) & (ub[:-1, :] > 1 - atol) & (ub[1:, :] < atol) & (ub[1:, :] > -atol))] = True
xi = x11 + ua * (x12 - x11)
yi = y11 + ua * (y12 - y11)
return np.ma.compressed(xi), np.ma.compressed(yi)
else:
n_nans = np.ma.sum(nans)
n_standard = np.ma.count(x11) - n_nans
#I initially tried using a set to get unique points but had problems with floating point equality
#create n by 2 array to hold all possible intersection points, check later for uniqueness
points = np.empty([n_standard + 2 * n_nans, 2], dtype = float) #each colinear segment pair has two intersections, hence the extra n_colinear points
#add standard intersection points
xi = x11 + ua * (x12 - x11)
yi = y11 + ua * (y12 - y11)
points[:n_standard,0] = np.ma.compressed(xi[~nans])
points[:n_standard,1] = np.ma.compressed(yi[~nans])
ignore[~nans]=True
#now add the appropriate intersection points for the colinear overlapping segments
#colinear overlapping segments intersect on the two internal points of the four points describing a straight line
#ua and ub have already serverd their purpose. Reuse them here to mean:
#ua is relative position of 1st b segment point along segment a
#ub is relative position of 2nd b segment point along segment a
use_x = x12 != x11 #avoid vertical lines diviiding by zero
use_y = ~use_x
ua[use_x] = (x21[use_x]-x11[use_x]) / (x12[use_x]-x11[use_x])
ua[use_y] = (y21[use_y]-y11[use_y]) / (y12[use_y]-y11[use_y])
ub[use_x] = (x22[use_x]-x11[use_x]) / (x12[use_x]-x11[use_x])
ub[use_y] = (y22[use_y]-y11[use_y]) / (y12[use_y]-y11[use_y])
np.ma.clip(ua, a_min=0,a_max = 1, out = ua)
np.ma.clip(ub, a_min=0,a_max = 1, out = ub)
xi = x11 + ua * (x12 - x11)
yi = y11 + ua * (y12 - y11)
points[n_standard:n_standard + n_nans,0] = np.ma.compressed(xi)
points[n_standard:n_standard + n_nans,1] = np.ma.compressed(yi)
xi = x11 + ub * (x12 - x11)
yi = y11 + ub * (y12 - y11)
points[n_standard+n_nans:,0] = np.ma.compressed(xi)
points[n_standard+n_nans:,1] = np.ma.compressed(yi)
points_unique = unique(points)
return points_unique[:,0], points_unique[:,1]
x = np.linspace(-1,1,500)
X,Y = np.meshgrid(x,x)
Z1 = np.abs(np.sin(2*X**2+Y))
Z2 = np.abs(np.cos(2*Y**2+X**2))
contour1 = plt.contour(X,Y,Z1,colors='k')
contour2 = plt.contour(X,Y,Z2,colors='r')
xi = np.array([])
yi = np.array([])
i=0
ncombos = (sum([len(x.get_paths()) for x in contour1.collections]) *
sum([len(x.get_paths()) for x in contour2.collections]))
for linecol1, linecol2 in itertools.product(contour1.collections, contour2.collections):
for path1, path2 in itertools.product(linecol1.get_paths(),linecol2.get_paths()):
i += 1
print('line combo %5d of %5d' % (i, ncombos))
xinter, yinter = linelineintersect(path1.vertices, path2.vertices)
xi = np.append(xi, xinter)
yi = np.append(yi, yinter)
plt.scatter(xi, yi, s=20)
plt.show()
此图适用于具有实际 X 和 Y 轴的 50x50 网格：
