So finden Sie alle Schnittpunkte zwischen zwei Kontur-Set auf effiziente Weise

Question

Ich frage mich über die beste Möglichkeit, alle Schnittpunkte (um Rundungsfehler) zwischen zwei Sätze von Konturlinien zu finden. Was ist die beste Methode? Hier ist das Beispiel:

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() 

Ich möchte einige ähnlich wie:

intersection_points = intersect(contour1,contour2) 
print intersection_points 
[(x1,y1),...,(xn,yn)] 

Answer 1

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() 

Ausbeuten

Answer 2

Ausgehend von der Antwort von @unutbu und einem von numpy-and-line-intersections gezupften Kreuzungsalgorithmus habe ich mir das ausgedacht. Es ist langsam wegen der brutalen Art des Wegs, die Kreuzungen und die Schleifen innerhalb von Schleifen innerhalb von Schleifen zu finden. Es könnte einen Weg geben, die Schleifen schneller zu machen, aber ich bin mir nicht sicher über den Schnittalgorithmus.

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() 

Edit:find_intersections oben nicht vertikale und horizontale Linien, noch überlappenden Liniensegmente nicht handhaben. Die folgende linelineintersect-Funktion behandelt diese Fälle, aber der gesamte Prozess ist immer noch langsam. Ich habe eine Zählung hinzugefügt, damit Sie eine Vorstellung davon haben, wie lange es noch dauern wird. Ich änderte auch contour1 = plt.contour(Z1,colors='k') zu contour1 = plt.contour(X,Y,Z1,colors='k'), also sind die Achsen und Ihre Kreuzungen in den tatsächlichen X- und Y-Koordinaten Ihres Rasters eher als eine Zählung Ihrer Rasterpunkte.

Ich denke, das Problem ist, dass Sie nur eine Menge Daten haben. In einem Kontursatz gibt es 24 Linien mit insgesamt 12818 Liniensegmenten, der andere Kontursatz hat 29 Linien mit 11411 Liniensegmenten. Das sind viele zu prüfende Liniensegmentkombinationen (696 Aufrufe an linelineintersect). Sie können ein Ergebnis schneller erhalten, indem Sie ein gröberes Gitter verwenden, um Ihre Funktionen zu berechnen (100 mal 100 war viel schneller als Ihre ursprünglichen 500 mal 500). Sie könnten vielleicht einen Zeilen-Sweep-Algorithmus ausführen, aber ich weiß nicht, wie Sie solche Dinge implementieren sollen. Das Linienkreuzungsproblem hat viele Anwendungen in Computerspielen und ist als Kollisionserkennung bekannt - es muss einige Grafikbibliotheken geben, die schnell alle Schnittpunkte bestimmen können.

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() 

Dieses Grundstück ist für 50x50-Raster mit der tatsächlichen X- und Y-Achsen: