更新:
所以,我想出了一個這樣做的方法。我實際上將每個相鄰原子轉換成它們的笛卡爾座標。然後,我減去相鄰原子所連接的中心原子的座標。最後,我通過計算每兩個矢量的點積除以它們的幅度乘積來得到角度,然後取這些值的反餘弦。我使用的代碼如下。這可能不是最優雅的做事方式,但它確實完成了工作。如果有人有改進,請隨時發表評論。
import random
import itertools as iter
import pymatgen as mg
import numpy as np
import math
def all_angles(POSCAR,amin,amax):
s = mg.Structure.from_file(POSCAR)
atoms = s.atomic_numbers
van = [x for x in atoms if x == 23]
length = len(van)
nb = ['NONE']*length
x = 0
n = 0
while x < length:
nb[x] = s.get_neighbors(s[x],2.4)
x += 1
w = 100
h = len(van)
oxygen = [[0 for x in range(w)] for y in range(h)]
x = 0
y = 0
while x < len(nb):
y = 0
n = 0
while y < len(nb[x]):
oxygen[x][n] = (nb[x][y][0]).coords
n += 1
y += 1
x += 1
x = 0
while x < len(oxygen):
oxygen[x] = [n for n in oxygen[x] if not isinstance(n,int)]
x += 1
van = [x for x in range(0,len(van))]
x = 0
while x < len(van):
van[x] = s[van[x]].coords
x += 1
van = [np.array(x) for x in van]
x = 0
while x < len(van):
oxygen[x] = [np.subtract(oxygen[x][y],van[x]) for y in range(0,len(oxygen[x]))]
x += 1
combo = [[0 for x in range(0,1000)] for y in range(0,len(van))]
r = 0
while r < len(van):
x = 0
for subset in iter.combinations(oxygen[r],2):
combo[r][x] = subset
x += 1
r += 1
x = 0
while x < len(combo):
combo[x] = [c for c in combo[x] if c != 0]
x += 1
angles = [[0 for x in range(0,1000)] for y in range(0,len(van))]
x = 0
while x < len(combo):
group = combo[x]
y = 0
while y < len(group):
angles[x][y] = math.degrees(math.acos(np.dot(group[y][0],group[y][1])/(np.linalg.norm(group[y][0])*np.linalg.norm(group[y][1]))))
y += 1
angles[x] = [round(n,3) for n in angles[x] if n != 0 and n > amin and n < amax]
x += 1
angles = np.concatenate(angles,axis=0)
return angles