在大型 numpy 三维数组上计算二维成对距离答案

【问题标题】：Calculate two dimensional pairwise distance on a large numpy three dimensional array在大型 numpy 三维数组上计算二维成对距离
【发布时间】：2016-02-15 13:13:32
【问题描述】：

我有一个包含 300 万个点的 numpy 数组，格式为 [pt_id, x, y, z]。目标是返回所有具有欧几里得距离两个数字min_d 和max_d 的点对。

欧几里得距离在x 和y 之间，而不是在z 上。但是，我想保留具有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=',')

如何返回表单数组：[pt_id_from, pt_id_to, distance]？

【问题讨论】：

您能否也为示例案例添加预期输出？
@Divakar 我修复了输出格式，所以 `['pt1', 'pt2', distance_as_number]

标签： python-2.7 numpy scipy pdist

【解决方案1】：

您只需通过循环所有可能的组合从数据中创建一个新数组。 itertools 模块非常适合这一点。

n = coords_arr.shape[0] # number of points
D = scipy.spatial.distance.squareform(dists) # distance matrix

data = []
for i, j in itertools.combinations(range(n), 2):
    pt_a = coords_arr[i, 0]
    pt_b = coords_arr[j, 0]
    d_ab = D[i,j]
    data.append([pt_a, pt_b, d_ab])

result_arr = np.array(data)

如果内存有问题，您可能希望将距离查找从使用巨大矩阵 D 更改为使用 i 和 j 索引直接在 dists 中查找值。

【讨论】：

【解决方案2】：

嗯，['pt1', 'pt2', distance_as_number] 是不可能的。使用混合数据类型可以获得的最接近的是结构化数组，但是您不能执行result[:2,0] 之类的操作。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]]

【讨论】：

感谢您的精彩回答。我认为经过进一步检查，我希望答案为 (x_2, y_2, z_2), (x_4, y_4, z_4), (dist_2, 4)
@dassouki done :) 虽然我不确定你是否还想要结构化数组，所以我提供了两种输出格式。

【解决方案3】：

您可以使用np.where 获取范围内的距离坐标，然后以您的格式生成一个新列表，过滤相同的对。像这样：

>>> import scipy.spatial.distance
>>> import numpy as np
>>> 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')
>>> dists = scipy.spatial.distance.squareform(dists)
>>> x, y = np.where((dists >= 8000) & (dists <= 30000))
>>> [(coords_arr[x[i]][0], coords_arr[y[i]][0], dists[y[i]][x[i]]) for i in xrange(len(x)) if x[i] < y[i]]
[('pt1', 'pt2', 28959.576688895162), ('pt1', 'pt3', 29042.897927032005)]

【讨论】：