[英]Python: 4 for cycles, comparing list items
我有一个非常可怕的算法来比较原子之间的距离,但它不起作用我想让它起作用。 这是代码:
for k in ResListA:
for n in ResListB:
for m in ResListA[counter3].atoms:
for z in ResListB[counter4].atoms:
coordDist = distance.distance(ResListA[counter3].atoms[counter4],ResListB[counter2].atoms[counter1])
counter1 = counter1 + 1
counter1 = 0
counter4 = counter4 + 1
counter4 = 0
counter2 = counter2 + 1
counter2 = 0
counter3 = counter3 + 1
基本上我想要之间的最小距离
ResListA [0] .atoms [0,..,n]的
ResListB [0,..,K] .atoms [0,..,M]
要计算。 但是,它计算
ResListA [0] .atoms [0]
至
ResListB [0,..,K] .atoms [0,..,M]
例如:
ResListA [N,P,C,N,C] ResListB [C,C] [P,P] ......
它应该是
dist(N,C)dist(N,C)dist(P,C)dist(P,C)
不
dist(N,C)dist(N,C)dist(N,P)dist(N,P)
先感谢您。
我认为你的代码可以写得更像这样。
for k in ResListA:
for n in ResListB:
for m in k.atoms:
for z in n.atoms:
coordDist = distance.distance(m.atoms, z.atoms)
不知道distance.distance
做了什么。 你不应该用coordDist
做一些涉及min()
吗?
虽然gnibbler可能是正确的,但这是你应该做的,这是你当前的代码简化为:
for k in ResListA:
for n in ResListB:
for counter4, m in enumerate(k.atoms):
for counter1, z in enumerate(ResListB[counter4].atoms):
coordDist = distance.distance(m, n.atoms[counter1])
你的问题是你需要:
for z in ResListB[counter2].atoms:
代替
for z in ResListB[counter4].atoms:
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