我正在解析PDB文件,我有一个链名列表以及格式(链,[坐标])的XYZ坐标。我有很多坐标,但只有3个不同的链。我想将同一链中的所有坐标压缩到一个列表中,这样我得到链= [坐标],[坐标],[坐标]等等。我查看了biopython文档,但我很难理解如何获得我想要的坐标,所以我决定手动提取坐标。这是我到目前为止的代码:
pdb_file = open('1adq.pdb')
import numpy as np
chainids = []
chainpos= []
for line in pdb_file:
if line.startswith("ATOM"):
# get x, y, z coordinates for Cas
chainid =str((line[20:22].strip()))
atomid = str((line[16:20].strip()))
pdbresn= int(line[23:26].strip())
x = float(line[30:38].strip())
y = float(line[38:46].strip())
z = float(line[46:54].strip())
if line[12:16].strip() == "CA":
chainpos.append((chainid,[x, y, z]))
chainids.append(chainid)
allchainids = np.unique(chainids)
print(chainpos)
和一些输出:
[('A', [1.719, -25.217, 8.694]), ('A', [2.934, -21.997, 7.084]), ('A', [5.35, -19.779, 8.986])
我理想的输出是:
A = ([1.719, -25.217, 8.694]), ([2.934, -21.997, 7.084]),(5.35, -19.779,8.986])...
谢谢!
Here is a section of PDB file:
ATOM 1 N PRO A 238 1.285 -26.367 7.882 0.00 25.30 N
ATOM 2 CA PRO A 238 1.719 -25.217 8.694 0.00 25.30 C
ATOM 3 C PRO A 238 2.599 -24.279 7.885 0.00 25.30 C
ATOM 4 O PRO A 238 3.573 -24.716 7.275 0.00 25.30 O
ATOM 5 CB PRO A 238 2.469 -25.791 9.881 0.00 25.30 C
A是第4列中的链名称。我不知道链名称是先验的,但由于我逐行解析,因此我使用前面提到的格式的坐标来保持链名称。现在我想在它们之前用“A”拉出所有坐标并将它们粘贴在一个名为“A”的列表中。我不能只是硬编码“A”,因为它并不总是“A”。我也有“L”和“H”,但我想,一旦我克服了理解,我就可以得到它们。
答案 0 :(得分:1)
你想要这样的东西:
import numpy as np
chain_dict = {}
for line in open('input'):
if line.startswith("ATOM"):
line = line.split()
# get x, y, z coordinates for Cas
chainid = line[4]
atomid = line[2]
pdbresn= line[5]
xyz = [line[6],line[7],line[8]]
if chainid not in chain_dict:
chain_dict[chainid]=[xyz]
else:
chain_dict[chainid].append(xyz)
,对于您的示例数据,它给出:
>>> chain_dict
{'A': [['1.285', '-26.367', '7.882'], ['1.719', '-25.217', '8.694'], ['2.599', '-24.279', '7.885'], ['3.573', '-24.716', '7.275'], ['2.469', '-25.791', '9.881']]
因为它是字典,显然你可以这样做:
>>> chain_dict['A']
[['1.285', '-26.367', '7.882'], ['1.719', '-25.217', '8.694'], ['2.599', '-24.279', '7.885'], ['3.573', '-24.716', '7.275'], ['2.469', '-25.791', '9.881']]
获得您感兴趣的链条的xyz坐标。
答案 1 :(得分:1)
只列出元组
>>> chainpos.append((chainid,x, y, z))
>>> chainpos
[('A', 1.719, -25.217, 8.694), ('A', 2.934, -21.997, 7.084)]
>>> import itertools
>>> for id, coor in itertools.groupby(chainpos,lambda x:x[0]):
... print(id, [c[1:] for c in coor])
答案 2 :(得分:0)
您可以使用列表理解:
>>> print chainpos
[('A', [1.719, -25.217, 8.694]), ('A', [2.934, -21.997, 7.084]), ('A', [5.35, -19.779, 8.986])]
>>> print "A =", [ t[1] for t in chainpos]