Pandas：在最大距离内寻找点

Question

我试图在最大距离内找到（x，y）点对。我认为最简单的方法就是生成一个数据帧，逐个遍历每个点，计算在给定点（x_0，y_0）的距离r内是否有坐标（x，y）的点。然后，将发现的对的总数除以2。在

%pylab inline
import pandas as pd

def find_nbrs(low, high, num, max_d):
    x = random.uniform(low, high, num)
    y = random.uniform(low, high, 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)

1. 首先，它并不总是能找到正确的对（我看到在指定距离内的点没有标记）。

2. 如果我写plot(..., 'or')，它会突出显示所有的点。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？如果比较结果为False，它不应该返回一个空数组吗？

3. 我如何在熊猫身上更优雅地完成以上所有的事情？例如，不必遍历每个元素。

Answer 1

您要查找的功能包含在scipy's spatial distance module中。在

下面是一个如何使用它的示例。真正的魔力在于squareform(pdist(points))。在

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))

There are 27 other points within a distance of 0.1 from the point (0.194, 0.160)

出于可视化目的：

^{pr2}$