Question

我有一个点列表（数量级为数万），我需要使用python识别两件事：

1-这些点中的连续点组（abs（x2-x1）＆lt; = 1和abs（y2-y1）＆lt; = 1）

2-每组的弧度/半径

以下是一组示例：

[[331,400]，[331,1200]，[332,400]，[332,486]，[332,522]，[332， 655]，[332,1200]，[332,3800]，[332,3877]，[332,3944]，[332,3963]， [332,3992]，[332,4050]，[333,400]，[333,486]，[333,522]，[333， 560]，[333,588]，[333,655]，[333,700]，[333,1200]，[333,3800]， [333,3877]，[333,3944]，[333,3963]，[333,3992]，[333,4050]，[334， 400]，[334,486]，[334,522]，[334,558]，[334,586]，[334,654]， [334,697]，[334,1200]，[334,3800]，[334,3877]，[334,3944]，[334， 3963]，[334,3992]，[334,4050]，[335,400]，[335,486]，[335,521]， [335,556]，[335,585]，[335,653]，[335,695]，[335,1200]，[335， 3800]，[335,3877]，[335,3944]，[335,3963]，[335,3992]，[335， 4050]，[336,400]，[336,486]，[336,520]，[336,555]，[336,584]， [336,651]，[336,693]，[336,1200]，[336,3800]，[336,3877]，[336， 3944]，[336,3963]，[336,3992]，[336,4050]，[337,400]，[337,486]， [337,554]，[337,583]，[337,649]，[337,692]，[337,1200]，[337， 3800]，[337,3877]，[337,3944]，[337,3963]，[337,3992]，[337， 4050]，[338,377]，[338,400]，[338,486]，[338,553]，[338,582]， [338,647]，[338,691]，[338,1200]，[338,3800]，[338,3877]，[338， 3944]，[338,3963]，[338,3992]，[338,4050]，[339,377]，[339,400]， [339,486]，[339,553]，[339,581]，[339,585]，[339,644]，[339， 654]，[339,690]，[339,706]，[339,1200]，[339,3800]，[339,3877]， [339,3944]，[339,3963]，[339,3992]，[339,4050]，[340,376]，[340， 400]，[340,486]，[340,552]，[340,580]，[340,585]，[340,641]， [340,655]，[340,689]，[340,713]，[340,1200]，[340,3800]，[340， 3877]，[340,3944]，[340,3963]，[340,3992]，[340,4050]，[341,376]， [341,400]，[341,486]，[341,552]，[341,579]，[341,585]，[341， 639]，[341,655]，[341,688]，[341,715]，[341,1200]，[341,3800]， [341,3877]，[341,3944]，[341,3963]，[341,3992]，[341,4050]，[342， 375]，[342,400]，[342,486]，[342,552]，[342,578]，[342,585]， [342,637]，[342,655]，[342,688]，[342,717]，[342,1200]，[342， 3800]，[342,3858]，[342,3925]，[342,3954]，[342,4011]，[342， 4050]，[342,4107]，[343,374]，[343,400]，[343,486]，[343,521]， [343,552]，[343,577]，[343,585]，[343,635]，[343,642]，[343， [687]，[343,718]，[343,1200]，[343,3800]，[343,3858]，[343,3925]， [343,3954]，[343,4011]，[343,4050]，[343,4107]，[344,373]，[344， 400]，[344,486]，[344,521]，[344,552]，[344,576]，[344,585]， [344,633]，[344,642]，[344,687]，[344,719]，[344,1200]，[344， 3800]，[344,3858]，[344,3925]，[344,3954]，[344,4011]，[344， 4050]，[344,4107]，[345,372]，[345,400]，[345,486]，[345,521]， [345,552]，[345,575]，[345,585]，[345,630]，[345,642]，[345， [687]，[345,720]，[345,1200]，[345,3800]，[345,3858]，[345,3925]， [345,3954]，[345,4011]，[345,4050]，[345,4107]，[346,370]，[346， 400]，[346,486]，[346,521]，[346,552]，[346,574]，[346,585]， [346,628]，[346,642]，[346,686]，[346,721]，[346,1200]，[346， 3800]，[346,3858]，[346,3925]，[346,3954]，[346,4011]，[346， 4050]，[346,4107]，[347,368]，[347,400]，[347,486]，[347,521]， [347,552]，[347,572]，[347,585]，[347,626]，[347,642]，[347， 686]，[347,721]，[347,1200]，[347,3800]，[347,3858]，[347,3925]， [347,3954]，[347,4011]，[347,4050]，[347,4107]，[348,366]，[348， 400]，[348,487]，[348,521]，[348,552]，[348,570]，[348,585]， [348,624]，[348,642]，[348,686]，[348,721]，[348,1200]，[348， 3800]，[348,3858]，[348,3925]，[348,3954]，[348,4011]，[348， 4050]，[348,4107]，[349,364]，[349,400]，[349,487]，[349,521]， [349,553]，[349,568]，[349,585]，[349,622]，[349,642]，[349， 686]，[349,722]，[349,1200]，[349,3800]，[349,3858]，[349,3925]， [349,3954]，[349,4011]，[349,4050]，[349,4107]，[350,362]，[350， 400]，[350,487]，[350,521]，[350,553]，[350,585]，[350,619]， [350,642]，[350,686]，[350,722]，[350,1200]，[350,3800]，[350， 3858]，[350,3925]，[350,3954]，[350,4011]，[350,4050]，[350， 4107]，[351,357]，[351,400]，[351,487]，[351,521]，[351,554]， [351,585]，[351,619]，[351,642]，[351,686]，[351,722]，[351， 1200]，[351,3800]，[351,3819]，[351,3858]，[351,3877]，[351， 3915]，[351,3934]，[351,3963]，[351,3992]，[351,4050]，[351， 4069]，[351,4107]，[352,355]，[352,373]，[352,400]，[352,487]， [352,520]，[352,555]，[352,585]，[352,621]，[352,642]，[352， 686]，[352,722]，[352,1200]，[352,3800]，[352,3819]，[352,3858]， [352,3877]，[352,3915]，[352,3934]，[352,3963]，[352,3992]，[352， 4050]，[352,4069]，[352,4107]，[353,353]，[353,375]，[353,400]， [353,487]，[353,520]，[353,556]，[353,585]，[353,623]，[353， 642]，[353,686]，[353,722]，[353,1200]，[353,3800]，[353,3819]， [353,3858]，[353,3877]，[353,3915]，[353,3934]，[353,3963]，[353， 3992]，[353,4050]，[353,4069]，[353,4107]，[354,351]，[354,376]， [354,400]，[354,487]，[354,520]，[354,558]，[354,584]，[354， 625]，[354,642]，[354,686]，[354,721]，[354,1200]，[354,3800]， [354,3819]，[354,3858]，[354,3877]]

Answer 1

这将为您提供群集和list of angles：

from sklearn.cluster import DBSCAN
from scipy.spatial import distance
from scipy.optimize import curve_fit
import numpy as np, math
data = [[331, 400], [331, 1200], [332, 400], [332, 486], [332, 522]] #....

def angle(pt1, pt2):
    x1, y1 = pt1
    x2, y2 = pt2
    inner_product = x1*x2 + y1*y2
    len1 = math.hypot(x1, y1)
    len2 = math.hypot(x2, y2)
    return math.acos(inner_product/(len1*len2))

db=DBSCAN(eps=1,min_samples=2,metric='precomputed').fit(
  distance.squareform(distance.pdist(data)))
core_samples = db.core_sample_indices_
labels = db.labels_
n_clusters_ = len(set(labels)) - (1 if -1 in labels else 0)
unique_labels = set(labels)

for k in unique_labels:
    class_members = [index[0] for index in np.argwhere(labels == k)]
    cluster_core_samples = [index for index in core_samples
                            if labels[index] == k]
    curve = np.array([data[index] for index in class_members])
    print k, curve, [angle(p1,p2) for p1,p2 in zip(curve,curve[1:])]

Answer 2

获得群集后，你必须得到角度。

现在获得这个角度有点棘手，因为正如我在评论中所说，它不太可能保持不变。

所以我打算做的是将每个簇转换为多项式并得到它的导数通过这种方式，您将能够首先获得给定点的角度polyfit。

polyfit将获取一个点列表并将其转换为给定顺序的多项式。基本上，订单越高，您获得的准确性就越高，但它有问题。您可能需要进行实验。

所以

z = np.polyfit(ClusterXCoordinates, ClusterYCoordinates, 5)// last part are the degree, this is where you experiment

没有一些ploy1d魔法

p = np.poly1d

我们采取衍生工具现在获取曲线中两点之间的角度

point = p(x_coordinate)//will give you the y of this cord after polyfit so it's continues and derivable

所以请derivative

// remember p computes the value of the function
    def df(x, h=0.1e-5):
//All above code to create p
       return ( p(x+h/2) - p(x-h/2) )/h
//Get slope
slope = df(pointToCheckFromCluster)
//calculate angle in rad by defintion

math.atan(slope);

http://docs.scipy.org/doc/numpy/reference/generated/numpy.poly1d.html#numpy.poly1d

测量连续点的曲率

2 个答案: