Python 将x、y点集与另一个缩放、旋转、平移且缺少元素的集相匹配
（我为什么要这样做？请参见下面的说明） 考虑两组点，Python 将x、y点集与另一个缩放、旋转、平移且缺少元素的集相匹配,python,performance,geometry,affinetransform,astronomy,Python,Performance,Geometry,Affinetransform,Astronomy,（我为什么要这样做？请参见下面的说明） 考虑两组点，A和B，如下所示 它可能看起来不像，但setA在setB中是“隐藏”的。由于B中的点相对于A在（x，y）中缩放、旋转和平移，因此不容易看到。更糟糕的是，A中存在的一些点在B中缺失，B中包含许多不在A中的点 我需要找到必须应用于B集合的适当缩放、旋转和平移，以便将其与集合A匹配。在上述情况下，正确的值为： scale = 0.14, rot_angle = 0.0, x_transl = 35.0, y_transl = 2.0 哪个产生了（
ABABBA（x，y）ABBABAscale = 0.14, rot_angle = 0.0, x_transl = 35.0, y_transl = 2.0
B10001）我将通过标记所有点并从每组中找到3个点的所有可能组合来解决此问题
# normalize B data to same format as A
set_Bx, set_By = (set_B)
set_B = []
for i in range(len(set_Bx)):
set_B.append([set_Bx[i],set_By[i]])
'''
set_B = [[2689.28, 2061.19], [3507.04, 3700.27], [2895.67, 2131.2],
[1051.3, 1837.3], [1929.49, 2017.52], [1035.97, 80.96], [752.44,
3524.61], [130.62, 3821.22], [620.06, 3711.53], [2769.06, 1812.12],
[1580.77, 1868.33], [281.76, 3865.77], [224.54, 3273.77], [3848.3,
2100.71]]
'''
list(itertools.combinations(range(len(set_A)), 3))
list(itertools.combinations(range(len(set_B)), 3))
2） 对于每个3点编组，计算相应三角形的边；对A组和B组重复此过程
dist = sqrt( (x2 - x1)**2 + (y2 - y1)**2 )
3） 然后减小每个三角形的边比，使每个三角形的最小边等于1；其他方面的比例也相应降低
在两个相似的三角形中：
这两个三角形的周长与三角形的周长之比相同
边。相应的侧面、中间带和高度都将处于
这个比例是相同的
4） 最后，对于a组中的每个三角形，使用元素减法与B组中的每个三角形进行比较。然后对结果元素求和，从A和B中找出总和最小的三角形
list(numpy.array(list1)-numpy.array(list2))
5） 给定匹配三角形；就CPU/RAM而言，找到适当的缩放、平移和旋转应该相对简单。
ETA1：草稿脚本
ETA2:注释中讨论的修补错误：使用sum（abs（））而不是abs（sum（））。现在它工作了，速度也很快
'''
known correct solution
A = [[1894.41, 1957.065],[1967.31, 1960.865],[2015.81, 1981.665]]
B = [[1051.3, 1837.3],[1580.77, 1868.33],[1929.49, 2017.52]]
'''
import numpy as np
import itertools
import math
import operator
set_A = [[2015.81, 1981.665], [1967.31, 1960.865], [1962.91, 1951.365],
[1964.91, 1994.565], [1894.41, 1957.065]]
set_B = [[2689.28, 3507.04, 2895.67, 1051.3, 1929.49, 1035.97, 752.44, 130.62,
620.06, 2769.06, 1580.77, 281.76, 224.54, 3848.3],
[2061.19, 3700.27, 2131.2, 1837.3, 2017.52, 80.96, 3524.61, 3821.22,
3711.53, 1812.12, 1868.33, 3865.77, 3273.77, 2100.71]]
# normalize set B data to set A format
set_Bx, set_By = (set_B)
set_B = []
for i in range(len(set_Bx)):
set_B.append([set_Bx[i],set_By[i]])
'''
set_B = [[2689.28, 2061.19], [3507.04, 3700.27], [2895.67, 2131.2],
[1051.3, 1837.3], [1929.49, 2017.52], [1035.97, 80.96], [752.44, 3524.61],
[130.62, 3821.22], [620.06, 3711.53], [2769.06, 1812.12], [1580.77, 1868.33],
[281.76, 3865.77], [224.54, 3273.77], [3848.3, 2100.71]]
'''
print set_A
print set_B
print len(set_A)
print len(set_B)
set_A_tri = list(itertools.combinations(range(len(set_A)), 3))
set_B_tri = list(itertools.combinations(range(len(set_B)), 3))
print set_A_tri
print set_B_tri
print len(set_A_tri)
print len(set_B_tri)
'''
set_A = [[2015.81, 1981.665], [1967.31, 1960.865], [1962.91, 1951.365],
[1964.91, 1994.565], [1894.41, 1957.065]]
set_A_tri = [(0, 1, 2), (0, 1, 3), (0, 1, 4), (0, 2, 3), (0, 2, 4), (0, 3, 4),
(1, 2, 3), (1, 2, 4), (1, 3, 4), (2, 3, 4)]
'''
def distance(x1,y1,x2,y2):
return math.sqrt((x2 - x1)**2 + (y2 - y1)**2 )
def tri_sides(set_x, set_x_tri):
triangles = []
for i in range(len(set_x_tri)):
point1 = set_x_tri[i][0]
point2 = set_x_tri[i][1]
point3 = set_x_tri[i][2]
point1x, point1y = set_x[point1][0], set_x[point1][1]
point2x, point2y = set_x[point2][0], set_x[point2][1]
point3x, point3y = set_x[point3][0], set_x[point3][1]
len1 = distance(point1x,point1y,point2x,point2y)
len2 = distance(point1x,point1y,point3x,point3y)
len3 = distance(point2x,point2y,point3x,point3y)
min_side = min(len1,len2,len3)
len1/=min_side
len2/=min_side
len3/=min_side
t=[len1,len2,len3]
t.sort()
triangles.append(t)
return triangles
A_triangles = tri_sides(set_A, set_A_tri)
B_triangles = tri_sides(set_B, set_B_tri)
print A_triangles
'''
[[1.0, 5.0405616860744304, 5.822935502560814],
[1.0, 1.5542012854321234, 1.5619803879976761],
[1.0, 1.3832883678507584, 2.347214708755337],
[1.0, 1.2141910838179129, 1.4096730529373076],
[1.0, 1.1275138587537166, 2.0318412465223665],
[1.0, 1.5207417600732074, 2.3589630093994876],
[1.0, 3.2270326342163584, 4.13069930678442],
[1.0, 6.565440477766354, 6.972550347780966],
[1.0, 2.1606693015281997, 2.3635387983160885],
[1.0, 1.589425903498476, 1.846471085870448]]
'''
print B_triangles
def list_subtract(list1,list2):
return np.absolute(np.array(list1)-np.array(list2))
sums = []
threshold = 1
for i in range(len(A_triangles)):
for j in range(len(B_triangles)):
k = sum(list_subtract(A_triangles[i], B_triangles[j]))
if k < threshold:
sums.append([i,j,k])
# sort by smallest sum
sums = sorted(sums, key=operator.itemgetter(2))
print sums
print 'winner %s' % sums[0]
print sums[0][0]
print sums[0][1]
match_A = set_A_tri[sums[0][0]]
match_B = set_B_tri[sums[0][1]]
print 'triangle A %s matches triangle B %s' % (match_A, match_B)
match_A_pts = []
match_B_pts = []
for i in range(3):
match_A_pts.append(set_A[match_A[i]])
match_B_pts.append(set_B[match_B[i]])
print 'triangle A has points %s' % match_A_pts
print 'triangle B has points %s' % match_B_pts
'''
winner [2, 204, 0.031415474959738399]
2
204
triangle A (0, 1, 4) matches triangle B (3, 4, 10)
triangle A has points [[2015.81, 1981.665], [1967.31, 1960.865], [1894.41, 1957.065]]
triangle B has points [[1051.3, 1837.3], [1929.49, 2017.52], [1580.77, 1868.33]]
'''
“”
已知正确解
A=[[1894.411957.065]、[1967.311960.865]、[2015.811981.665]]
B=[[1051.31837.3]、[1580.771868.33]、[1929.492017.52]]
'''
将numpy作为np导入
进口itertools
输入数学
进口经营者
set_A=[2015.811981.665]、[1967.311960.865]、[1962.911951.365]，
[1964.91, 1994.565], [1894.41, 1957.065]]
集合B=[[2689.283507.042895.671051.31929.491035.97752.44130.62，
620.06, 2769.06, 1580.77, 281.76, 224.54, 3848.3],
[2061.19, 3700.27, 2131.2, 1837.3, 2017.52, 80.96, 3524.61, 3821.22,
3711.53, 1812.12, 1868.33, 3865.77, 3273.77, 2100.71]]
#规范化SETB数据以设置格式
set_Bx，set_By=（set_B）
set_B=[]
对于范围内的i（len（set_Bx））：
set_B.append（[set_Bx[i]，set_By[i]]））
'''
集合B=[2689.282061.19]、[3507.043700.27]、[2895.672131.2]，
[1051.3, 1837.3], [1929.49, 2017.52], [1035.97, 80.96], [752.44, 3524.61],
[130.62, 3821.22], [620.06, 3711.53], [2769.06, 1812.12], [1580.77, 1868.33],
[281.76, 3865.77], [224.54, 3273.77], [3848.3, 2100.71]]
'''
打印集
打印集
打印透镜（设置A）
打印透镜（套）
set_A_tri=列表（itertools.组合（范围（len（set_A）），3））
set_B_tri=列表（itertools.组合（范围（len（set_B）），3））
打印集
打印集
打印透镜（设置为三个）
打印透镜（设置为三个）
'''
set_A=[2015.811981.665]、[1967.311960.865]、[1962.911951.365]，
[1964.91, 1994.565], [1894.41, 1957.065]]
集合A_tri=[（0,1,2），（0,1,3），（0,1,4），（0,2,3），（0,2,4），（0,3,4），
(1, 2, 3), (1, 2, 4), (1, 3, 4), (2, 3, 4)]
'''
def距离（x1、y1、x2、y2）：
返回数学sqrt（（x2-x1）**2+（y2-y1）**2）
def tri_侧面（设置_x，设置_x_tri）：
三角形=[]
对于范围内的i（len（set_x_tri））：
第1点=集合[i][0]
第2点=集合[i][1]
第3点=集合[i][2]
点1X，点1Y=set_x[point1][0]，set_x[point1][1]
点2x，点2y=set_x[point2][0]，set_x[point2][1]
点3x，点3y=set_x[point3][0]，set_x[point3][1]
len1=距离（点1X、点1Y、点2X、点2Y）
len2=距离（点1X、点1Y、点3X、点3Y）
len3=距离（点2x、点2y、点3x、点3y）
最小侧=最小（透镜1、透镜2、透镜3）
len1/=最小侧
len2/=最小侧
len3/=最小侧
t=[len1，len2，len3]
t、 排序（）
三角形。附加（t）
返回三角形
A_三角形=三边（集合A，集合A_三边）
B_三角形=三边（集合B，集合B_tri）
打印A_三角形
'''
[[1.0, 5.0405616860744304, 5.822935502560814],
[1.0, 1.5542012854321234, 1.5619803879976761],
[1.0, 1.3832883678507584, 2.347214708755337],
[1.0, 1.2141910838179129, 1.4096730529373076],
[1.0, 1.1275138587537166, 2.0318412465223665],
[1.0, 1.5207417600732074, 2.3589630093994876],
[1.0, 3.2270326342163584, 4.13069930678442],
[1.0, 6.565440477766354, 6.972550347780966],
[1.0, 2.1606693015281997, 2.3635387983160885],
[1.0, 1.589425903498476, 1.846471085870448]]
'''
打印B_三角形
def list_减法（列表1、列表2）：
返回np.absolute（np.array（list1）-np.array（list2））
总和=[]
阈值=1
对于范围内的i（len（A_三角形））：
对于范围内的j（len（B_三角形））：
k=和（列表减去（A_三角形[i]，B_三角形[j]））
如果k<阈值：
sums.append（[i，j，k]）
#按最小和排序
sums=排序（sums，key=operator.itemgetter（2））
打印金额
打印“赢家%s”%s[0]
打印总和[0][0]
打印金额[0][1]
match_A=set_A_tri[总和[0][0]]
match_B=set_B_tri[总和[0][1]]
打印“三角形A%s匹配三角形B%s%”（匹配A，匹配B）
匹配A_pts=[]
匹配_B_pts=[]
对于范围（3）中的i：
match_A_pts.append（set_A[match_A[i]]）
match_B_pts.append（set_B[match_B[i]]）
打印“三角形A有点%s”%match\u A\u pts
打印“三角形B有点%s”%match\u B\u pts
'''
获胜者[2204，
dist = sqrt( (x2 - x1)**2 + (y2 - y1)**2 )
list(numpy.array(list1)-numpy.array(list2))
'''
known correct solution
A = [[1894.41, 1957.065],[1967.31, 1960.865],[2015.81, 1981.665]]
B = [[1051.3, 1837.3],[1580.77, 1868.33],[1929.49, 2017.52]]
'''
import numpy as np
import itertools
import math
import operator
set_A = [[2015.81, 1981.665], [1967.31, 1960.865], [1962.91, 1951.365],
[1964.91, 1994.565], [1894.41, 1957.065]]
set_B = [[2689.28, 3507.04, 2895.67, 1051.3, 1929.49, 1035.97, 752.44, 130.62,
620.06, 2769.06, 1580.77, 281.76, 224.54, 3848.3],
[2061.19, 3700.27, 2131.2, 1837.3, 2017.52, 80.96, 3524.61, 3821.22,
3711.53, 1812.12, 1868.33, 3865.77, 3273.77, 2100.71]]
# normalize set B data to set A format
set_Bx, set_By = (set_B)
set_B = []
for i in range(len(set_Bx)):
set_B.append([set_Bx[i],set_By[i]])
'''
set_B = [[2689.28, 2061.19], [3507.04, 3700.27], [2895.67, 2131.2],
[1051.3, 1837.3], [1929.49, 2017.52], [1035.97, 80.96], [752.44, 3524.61],
[130.62, 3821.22], [620.06, 3711.53], [2769.06, 1812.12], [1580.77, 1868.33],
[281.76, 3865.77], [224.54, 3273.77], [3848.3, 2100.71]]
'''
print set_A
print set_B
print len(set_A)
print len(set_B)
set_A_tri = list(itertools.combinations(range(len(set_A)), 3))
set_B_tri = list(itertools.combinations(range(len(set_B)), 3))
print set_A_tri
print set_B_tri
print len(set_A_tri)
print len(set_B_tri)
'''
set_A = [[2015.81, 1981.665], [1967.31, 1960.865], [1962.91, 1951.365],
[1964.91, 1994.565], [1894.41, 1957.065]]
set_A_tri = [(0, 1, 2), (0, 1, 3), (0, 1, 4), (0, 2, 3), (0, 2, 4), (0, 3, 4),
(1, 2, 3), (1, 2, 4), (1, 3, 4), (2, 3, 4)]
'''
def distance(x1,y1,x2,y2):
return math.sqrt((x2 - x1)**2 + (y2 - y1)**2 )
def tri_sides(set_x, set_x_tri):
triangles = []
for i in range(len(set_x_tri)):
point1 = set_x_tri[i][0]
point2 = set_x_tri[i][1]
point3 = set_x_tri[i][2]
point1x, point1y = set_x[point1][0], set_x[point1][1]
point2x, point2y = set_x[point2][0], set_x[point2][1]
point3x, point3y = set_x[point3][0], set_x[point3][1]
len1 = distance(point1x,point1y,point2x,point2y)
len2 = distance(point1x,point1y,point3x,point3y)
len3 = distance(point2x,point2y,point3x,point3y)
min_side = min(len1,len2,len3)
len1/=min_side
len2/=min_side
len3/=min_side
t=[len1,len2,len3]
t.sort()
triangles.append(t)
return triangles
A_triangles = tri_sides(set_A, set_A_tri)
B_triangles = tri_sides(set_B, set_B_tri)
print A_triangles
'''
[[1.0, 5.0405616860744304, 5.822935502560814],
[1.0, 1.5542012854321234, 1.5619803879976761],
[1.0, 1.3832883678507584, 2.347214708755337],
[1.0, 1.2141910838179129, 1.4096730529373076],
[1.0, 1.1275138587537166, 2.0318412465223665],
[1.0, 1.5207417600732074, 2.3589630093994876],
[1.0, 3.2270326342163584, 4.13069930678442],
[1.0, 6.565440477766354, 6.972550347780966],
[1.0, 2.1606693015281997, 2.3635387983160885],
[1.0, 1.589425903498476, 1.846471085870448]]
'''
print B_triangles
def list_subtract(list1,list2):
return np.absolute(np.array(list1)-np.array(list2))
sums = []
threshold = 1
for i in range(len(A_triangles)):
for j in range(len(B_triangles)):
k = sum(list_subtract(A_triangles[i], B_triangles[j]))
if k < threshold:
sums.append([i,j,k])
# sort by smallest sum
sums = sorted(sums, key=operator.itemgetter(2))
print sums
print 'winner %s' % sums[0]
print sums[0][0]
print sums[0][1]
match_A = set_A_tri[sums[0][0]]
match_B = set_B_tri[sums[0][1]]
print 'triangle A %s matches triangle B %s' % (match_A, match_B)
match_A_pts = []
match_B_pts = []
for i in range(3):
match_A_pts.append(set_A[match_A[i]])
match_B_pts.append(set_B[match_B[i]])
print 'triangle A has points %s' % match_A_pts
print 'triangle B has points %s' % match_B_pts
'''
winner [2, 204, 0.031415474959738399]
2
204
triangle A (0, 1, 4) matches triangle B (3, 4, 10)
triangle A has points [[2015.81, 1981.665], [1967.31, 1960.865], [1894.41, 1957.065]]
triangle B has points [[1051.3, 1837.3], [1929.49, 2017.52], [1580.77, 1868.33]]
'''