python – 熊猫：在最大距离内找到点

发布时间：2020-12-20 11:38:53 所属栏目：Python 来源：网络整理

导读：我试图在彼此的最大距离内找到成对的(x,y)点.我认为最简单的方法是生成一个DataFrame并逐个遍历每个点,计算在给定点(x_0,y_0)的距离r内是否存在坐标(x,y)的点.然后,将发现的对的总数除以2. %pylab inlineimport pandas as pddef find_nbrs(low,high,num,max_

我试图在彼此的最大距离内找到成对的(x,y)点.我认为最简单的方法是生成一个DataFrame并逐个遍历每个点,计算在给定点(x_0,y_0)的距离r内是否存在坐标(x,y)的点.然后,将发现的对的总数除以2.

%pylab inline
import pandas as pd

def find_nbrs(low,high,num,max_d):
    x = random.uniform(low,num)
    y = random.uniform(low,num)
    points = pd.DataFrame({'x':x,'y':y})

    tot_nbrs = 0

    for i in arange(len(points)):
        x_0 = points.x[i]
        y_0 = points.y[i]

        pt_nbrz = points[((x_0 - points.x)**2 + (y_0 - points.y)**2) < max_d**2]
        tot_nbrs += len(pt_nbrz)
        plot (pt_nbrz.x,pt_nbrz.y,'r-')

    plot (points.x,points.y,'b.')
    return tot_nbrs

print find_nbrs(0,1,50,0.1)

>首先,它并不总能找到合适的对(我看到在指定距离内没有标记的点).
>如果我写剧情(……,’或’),它会突出显示所有点.这意味着pt_nbrz = points [((x_0-points.x)** 2(y_0-points.y)** 2)< max_d ** 2]至少返回一个(x,y).为什么？如果比较为False,它不应该返回一个空数组吗？
>如何在熊猫中更优雅地完成上述所有操作？例如,无需遍历每个元素.

解决方法

您正在寻找的功能包含在 scipy’s spatial distance module中.

这是一个如何使用它的例子.真正的魔力在于方形(pdist(points)).

from scipy.spatial.distance import pdist,squareform
import numpy as np
import matplotlib.pyplot as plt

points = np.random.uniform(-.5,.5,(1000,2))

# Compute the distance between each different pair of points in X with pdist.
# Then,just for ease of working,convert to a typical symmetric distance matrix
# with squareform.
dists = squareform(pdist(points))

poi = points[4] # point of interest
dist_min = .1
close_points = dists[4] < dist_min

print("There are {} other points within a distance of {} from the point "
    "({:.3f},{:.3f})".format(close_points.sum() - 1,dist_min,*poi))

距离点0.1(0.194,0.160)距其他27个点

出于可视化目的：

f,ax = plt.subplots(subplot_kw=
    dict(aspect='equal',xlim=(-.5,.5),ylim=(-.5,.5)))
ax.plot(points[:,0],points[:,1],'b+ ')
ax.plot(poi[0],poi[1],ms=15,marker='s',mfc='none',mec='g')
ax.plot(points[close_points,points[close_points,marker='o',mec='r',ls='')  # draw all points within distance

t = np.linspace(0,2*np.pi,512)
circle = dist_min*np.vstack([np.cos(t),np.sin(t)]).T
ax.plot((circle+poi)[:,(circle+poi)[:,'k:') # Add a visual check for that distance
plt.show()

（编辑：李大同）

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