在大型 numpy 三维数组上计算二维成对距离

Question

我有一个包含 300 万个点的 numpy 数组，格式为[pt_id, x, y, z]. 目标是返回具有欧几里德距离两个数字min_d和 的所有点对max_d。

欧几里得距离在x和之间，y而不是在 上z。但是，我想保留带有pt_id_from, pt_id_to,distance属性的数组。

我正在使用 scipy 的 dist 来计算距离：

import scipy.spatial.distance
coords_arr = np.array([['pt1', 2452130.000, 7278106.000, 25.000],
                       ['pt2', 2479539.000, 7287455.000, 4.900],
                       ['pt3', 2479626.000, 7287458.000, 10.000],
                       ['pt4', 2484097.000, 7292784.000, 8.800],
                       ['pt5', 2484106.000, 7293079.000, 7.300],
                       ['pt6', 2484095.000, 7292891.000, 11.100]])

dists = scipy.spatial.distance.pdist(coords_arr[:,1:3], 'euclidean')
np.savetxt('test.out', scipy.spatial.distance.squareform(dists), delimiter=',')

我应该怎么做才能返回一个表单数组：[pt_id_from, pt_id_to, distance]？

Answer 1

嗯，['pt1', 'pt2', distance_as_number]这并不完全可能。您可以使用混合数据类型获得的最接近的是结构化数组，但是您不能执行诸如result[:2,0]. 您必须分别索引字段名称和数组索引，例如：result[['a','b']][0]。

这是我的解决方案：

import numpy as np
import scipy.spatial.distance

coords_arr = np.array([['pt1', 2452130.000, 7278106.000, 25.000],
                       ['pt2', 2479539.000, 7287455.000, 4.900],
                       ['pt3', 2479626.000, 7287458.000, 10.000],
                       ['pt4', 2484097.000, 7292784.000, 8.800],
                       ['pt5', 2484106.000, 7293079.000, 7.300],
                       ['pt6', 2484095.000, 7292891.000, 11.100]])

dists = scipy.spatial.distance.pdist(coords_arr[:,1:3], 'euclidean')

# Create a shortcut for `coords_arr.shape[0]` which is basically
# the total amount of points, hence `n`
n = coords_arr.shape[0]

# `a` and `b` contain the indices of the points which were used to compute the
# distances in dists. In this example:
# a = [0, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 3, 3, 4]
# b = [1, 2, 3, 4, 5, 2, 3, 4, 5, 3, 4, 5, 4, 5, 5]
a = np.arange(n).repeat(np.arange(n-1, -1, -1))
b = np.hstack([range(x, n) for x in xrange(1, n)])

min_d = 1000
max_d = 10000

# Find out which distances are in range.
in_range = np.less_equal(min_d, dists) & np.less_equal(dists, max_d)

# Define the datatype of the structured array which will be the result.
dtype = [('a', '<f8', (3,)), ('b', '<f8', (3,)), ('dist', '<f8')]

# Create an empty array. We fill it later because it makes the code cleaner.
# Its size is given by the sum over `in_range` which is possible
# since True and False are equivalent to 1 and 0.
result = np.empty(np.sum(in_range), dtype=dtype)

# Fill the resulting array.
result['a'] = coords_arr[a[in_range], 1:4]
result['b'] = coords_arr[b[in_range], 1:4]
result['dist'] = dists[in_range]

print(result)

# In caste you don't want a structured array at all, this is what you can do:
result = np.hstack([coords_arr[a[in_range],1:],
                    coords_arr[b[in_range],1:],
                    dists[in_range, None]]).astype('<f8')
print(result)

结构化数组：

[([2479539.0, 7287455.0, 4.9], [2484097.0, 7292784.0, 8.8], 7012.389393067102)
 ([2479539.0, 7287455.0, 4.9], [2484106.0, 7293079.0, 7.3], 7244.7819152821985)
 ([2479539.0, 7287455.0, 4.9], [2484095.0, 7292891.0, 11.1], 7092.75912462844)
 ([2479626.0, 7287458.0, 10.0], [2484097.0, 7292784.0, 8.8], 6953.856268287403)
 ([2479626.0, 7287458.0, 10.0], [2484106.0, 7293079.0, 7.3], 7187.909362255481)
 ([2479626.0, 7287458.0, 10.0], [2484095.0, 7292891.0, 11.1], 7034.873843929257)]

ndarray：

[[2479539.0, 7287455.0, 4.9, 2484097.0, 7292784.0, 8.8, 7012.3893],
 [2479539.0, 7287455.0, 4.9, 2484106.0, 7293079.0, 7.3, 7244.7819],
 [2479539.0, 7287455.0, 4.9, 2484095.0, 7292891.0, 11.1, 7092.7591],
 [2479626.0, 7287458.0, 10.0, 2484097.0, 7292784.0, 8.8, 6953.8562],
 [2479626.0, 7287458.0, 10.0, 2484106.0, 7293079.0, 7.3, 7187.9093],
 [2479626.0, 7287458.0, 10.0, 2484095.0, 7292891.0, 11.1, 7034.8738]]