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在python中将距离平方转换为长格式

distance scipy numpy pandas python

Mr_and_Mrs_D · 技术社区 · 6 年前

代码:

import numpy as np
import pandas as pd
from scipy.spatial.distance import pdist, squareform

ids = ['1', '2', '3']
points=[(0,0), (1,1), (3,3)]
distances = pdist(np.array(points), metric='euclidean')
print(distances)
distance_matrix = squareform(distances)
print(distance_matrix)

印刷品:

[1.41421356 4.24264069 2.82842712]
[[0.         1.41421356 4.24264069]
 [1.41421356 0.         2.82842712]
 [4.24264069 2.82842712 0.        ]]

如预期

我想把它变成一个长格式,用csv来写,就像

id1,id2,distance
1,1,0
1,2,1.41421356
1,3,4.24264069
2,1,1.41421356
2,2,0
2,3,2.82842712

等等-我该怎么做才能最大限度地提高效率?使用熊猫是一种选择

2 回复 | 直到 6 年前

jezrael 6 年前

使用 DataFrame 承包商 stack :

df = pd.DataFrame(distance_matrix, index=ids, columns=ids).stack().reset_index()
df.columns=['id1','id2','distance']
print (df)
  id1 id2  distance
0   1   1  0.000000
1   1   2  1.414214
2   1   3  4.242641
3   2   1  1.414214
4   2   2  0.000000
5   2   3  2.828427
6   3   1  4.242641
7   3   2  2.828427
8   3   3  0.000000

或 数据帧 承包商 numpy.repeat , numpy.tile 和 ravel :

df = pd.DataFrame({'id1':np.repeat(ids, len(ids)), 
                   'id2':np.tile(ids, len(ids)),
                   'dist':distance_matrix.ravel()})
print (df)
  id1 id2      dist
0   1   1  0.000000
1   1   2  1.414214
2   1   3  4.242641
3   2   1  1.414214
4   2   2  0.000000
5   2   3  2.828427
6   3   1  4.242641
7   3   2  2.828427
8   3   3  0.000000

Divakar 6 年前

我建议使用 indices_merged_arr_generic_using_cp -

助手函数-

import numpy as np
import functools

# https://stackoverflow.com/a/46135435/ by @unutbu
def indices_merged_arr_generic_using_cp(arr):
    """
    Based on cartesian_product
    http://stackoverflow.com/a/11146645/190597 (senderle)
    """
    shape = arr.shape
    arrays = [np.arange(s, dtype='int') for s in shape]
    broadcastable = np.ix_(*arrays)
    broadcasted = np.broadcast_arrays(*broadcastable)
    rows, cols = functools.reduce(np.multiply, broadcasted[0].shape), len(broadcasted)+1
    out = np.empty(rows * cols, dtype=arr.dtype)
    start, end = 0, rows
    for a in broadcasted:
        out[start:end] = a.reshape(-1)
        start, end = end, end + rows
    out[start:] = arr.flatten()
    return out.reshape(cols, rows).T

使用-

In [169]: out = indices_merged_arr_generic_using_cp(distance_matrix)

In [170]: np.savetxt('out.txt', out, fmt="%i,%i,%f")

In [171]: !cat out.txt
0,0,0.000000
0,1,1.414214
0,2,4.242641
1,0,1.414214
1,1,0.000000
1,2,2.828427
2,0,4.242641
2,1,2.828427
2,2,0.000000

得到 distance_matrix 我们也可以使用 Scipy's cdist : cdist(points, points) . 还有 eucl_dist 包(免责声明:我是它的作者),它包含各种计算欧几里得距离的方法,这些方法比 SciPy's cdist 尤其是对于大型阵列。