python中xyz坐标的欧几里德距离

Question

我有一个像这样的数据文件。在

HETATM    1  H10 XSHQ    0      10.139   2.231   0.091  1.00  0.00           H
HETATM    2   N1 XSHQ    0       9.641   1.386  -0.104  1.00  0.00           N
HETATM    3   H9 XSHQ    0       9.773   1.133  -1.063  1.00  0.00           H
HETATM    4   C1 XSHQ    0       8.245   1.531   0.230  1.00  0.00           H

其中XYZ坐标在第6、7、8列和最后一列中，与点相关联的字母在最后一列中。我想确定在最后一列有字母H的点之间的距离。我该怎么做？我知道这是我执行该操作所需的代码，但我不清楚如何使用第6、7和8列中的值，而且只适用于最后一列是H的情况：

^{pr2}$

Answer 1

当然，@smonener的答案已经是正确的，使用OrderedDict这样的数据结构是个好主意，但是如果只想使用标准方法，可以尝试：

from scipy.spatial import distance

# Load data from file
with open('datafile.txt') as datafile: 
    contents = [line.split() for line in datafile]

# Extract the xyz coordiantes, if there is an H in the last column
coords = []
for i, item in enumerate(contents):
    if item[-1] == 'H':
        coords.append([[float(x) for x in item[5:8]], i+1])

# Show results
for i in range(len(coords)):
    for j in range(i+1, len(coords)):
        dist = distance.euclidean(coords[i][0], coords[j][0])
        print('({}, {}): {:.5f}'.format(coords[i][1], coords[j][1], dist))

Answer 2

我使用regexp来提取日期，然后根据规则过滤它们。在

演示代码如下：

Answer 3

使用生成器表达式的简单解决方案

From PEP 289 Generator Expressions
Rationale
Experience with list comprehensions has shown their widespread utility throughout Python. However, many of the use cases do not need to have a full list created in memory. Instead, they only need to iterate over the elements one at a time.

因为

1. 你不需要保存中间结果
2. 可能你有一个大的数据集

以及itertools标准库模块中的^{}，因为您需要计算数据集中每对有趣的点的距离。在

$ cat euclid.py
from scipy.spatial.distance import euclidean
from itertools import combinations

lines = ['HETATM 1 H10 XSHQ 0  10.139 2.231  0.091 1.00 0.00 H',
         'HETATM 2  N1 XSHQ 0   9.641 1.386 -0.104 1.00 0.00 N',
         'HETATM 3  H9 XSHQ 0   9.773 1.133 -1.063 1.00 0.00 H',
         'HETATM 4  C1 XSHQ 0   8.245 1.531  0.230 1.00 0.00 H']

H_lines = (line for line in lines if line[-1]=='H')
H_lists = (line.split() for line in H_lines)
H_data = ((int(tok[1]), [float(x) for x in tok[5:8]]) for tok in H_lists)
H_dist = {(i[0], j[0]):euclidean(i[1], j[1])
             for i, j in combinations(H_data, 2)}

for m, n in H_dist:
    print('Distance between points %d and %d is %.6f'%(
             m, n, H_dist[m, n]))
$ python3 euclid.py
Distance between points 1 and 3 is 1.634404
Distance between points 1 and 4 is 2.023995
Distance between points 3 and 4 is 2.040842
$