所以我有一个名为mol2文件的文件。它基本上是一个文本文件，我试图弄清楚如何交换文件中的数据行。
数据的一个例子
1 N -0.5660 -1.0450 0.3550 N.am 1 TRP1 -0.7300 BACKBONE
2 H3 -0.3350 -1.9480 0.7600 H 1 TRP1 0.3700 BACKBONE
3 CA 0.3540 0.0650 0.6210 C.3 1 TRP1 0.3610 BACKBONE
4 HA 0.3150 0.7690 -0.2150 H 1 TRP1 0.0000 BACKBONE
5 C 1.7800 -0.5120 0.6800 C.2 1 TRP1 0.5690 BACKBONE
6 O 2.3250 -0.8480 1.7310 O.2 1 TRP1 -0.5700 BACKBONE
7 CB -0.0030 0.8580 1.8980 C.3 1 TRP1 0.1810
8 HB1 0.8740 1.4160 2.2510 H 1 TRP1 0.0000
9 HB2 -0.7440 1.6230 1.6310 H 1 TRP1 0.0000
10 CG -0.5640 0.0480 3.0300 C.2 1 TRP1 -0.1810
11 CD1 0.1540 -0.6770 3.9600 C.2 1 TRP1 -0.3020
12 HD1 1.2210 -0.8180 4.0790 H 1 TRP1 0.1500
13 CD2 -1.9510 -0.1190 3.3620 C.ar 1 TRP1 0.0000
14 NE1 -0.7260 -1.2960 4.8040 N.pl3 1 TRP1 0.0330
15 HE1 -0.4680 -1.9020 5.5700 H 1 TRP1 0.2700
16 CE2 -2.0170 -0.9760 4.4610 C.ar 1 TRP1 -0.1520
17 CE3 -3.1590 0.3750 2.8250 C.ar 1 TRP1 -0.1500
18 HE3 -3.1560 1.0570 1.9790 H 1 TRP1 0.1500
19 CZ2 -3.2280 -1.3780 5.0370 C.ar 1 TRP1 -0.1500
20 HZ2 -3.2550 -2.0510 5.8880 H 1 TRP1 0.1500
21 CZ3 -4.3780 -0.0150 3.3900 C.ar 1 TRP1 -0.1500
22 HZ3 -5.3060 0.3660 2.9720 H 1 TRP1 0.1500
23 CH2 -4.4090 -0.8820 4.4790 C.ar 1 TRP1 -0.1500
24 HH2 -5.3650 -1.1760 4.9040 H 1 TRP1 0.1500
25 CA -2.1700 0.2590 -0.9830 C.3 1 TRP1 0.0610 BACKBONE
26 CB -3.6350 0.1990 -1.3950 C.3 1 TRP1 0.3000
27 HA -2.0190 1.0810 -0.2770 H 1 TRP1 0.0000
28 HA1 -1.5370 0.4510 -1.8560 H 1 TRP1 0.0000 BACKBONE
29 N -4.0840 1.4530 -1.9680 N.am 1 TRP1 -0.7300
30 H2 -4.1770 1.5150 -2.9760 H 1 TRP1 0.3700
31 C -4.3910 2.5490 -1.1890 C.2 1 TRP1 0.5690
32 C -4.8370 3.7450 -1.9830 C.3 1 TRP1 0.0610
33 H1 -5.4650 4.3840 -1.3570 H 1 TRP1 0.0000
34 H2 -5.4180 3.4380 -2.8580 H 1 TRP1 0.0000
35 H3 -3.9580 4.3060 -2.3120 H 1 TRP1 0.0000
36 O -4.2780 2.5770 0.0330 O.2 1 TRP1 -0.5700
37 HB2 -3.7980 -0.5770 -2.1510 H 1 TRP1 0.0000
38 HB3 -4.2700 -0.0350 -0.5320 H 1 TRP1 0.0000
39 C -1.7340 -1.0430 -0.3530 C.2 1 TRP1 0.5690 BACKBONE
40 O -2.3720 -2.0860 -0.4820 O.2 1 TRP1 -0.5700 BACKBONE
41 CA 2.4750 0.9300 -2.3010 C.3 1 TRP1 0.0610 BACKBONE
42 CB 2.0300 -0.4580 -1.8610 C.3 1 TRP1 0.3000
43 HA 3.5610 1.0310 -2.1890 H 1 TRP1 0.0000
44 HA1 2.0080 1.7020 -1.6790 H 1 TRP1 0.0000 BACKBONE
45 N 2.4480 -0.7360 -0.4950 N.am 1 TRP1 -0.7300
46 H1 3.3610 -1.1580 -0.3590 H 1 TRP1 0.3700
47 HB2 0.9470 -0.5730 -1.9510 H 1 TRP1 0.0000
48 HB3 2.4940 -1.2240 -2.4920 H 1 TRP1 0.0000
49 C 2.1130 1.1940 -3.7420 C.2 1 TRP1 0.5690 BACKBONE
50 N 2.5940 2.3870 -4.2190 N.am 1 TRP1 -0.3940 BACKBONE
51 O 2.2950 2.6850 -5.5640 O.3 1 TRP1 -0.4360 BACKBONE
52 H 1.7580 1.9070 -5.8290 H 1 TRP1 0.4000 BACKBONE
53 H2 3.1370 3.0670 -3.6990 H 1 TRP1 0.3700 BACKBONE
54 O 1.4490 0.4330 -4.4420 O.2 1 TRP1 -0.5700 BACKBONE
我想要做的是找到原子的图案
O, H, H2, O
然后一旦发现是从第一个O线获取数据，所以在这种情况下，它将以51开始的行并用从47开始的行切换它。所以输出文件将具有
47 O 2.2950 2.6850 -5.5640 O.3 1 TRP1 -0.4360 BACKBONE
48 HB3 2.4940 -1.2240 -2.4920 H 1 TRP1 0.0000
49 C 2.1130 1.1940 -3.7420 C.2 1 TRP1 0.5690 BACKBONE
50 N 2.5940 2.3870 -4.2190 N.am 1 TRP1 -0.3940 BACKBONE
51 HB2 0.9470 -0.5730 -1.9510 H 1 TRP1 0.0000
所以现在行51开始用输出文件
中以47开头的行切换迄今为止的守则
def main():
f = open('test.mol2', "r")
newf = open('test2.mol2', "w")
lines = f.readlines()
for i in range(0, len(lines)):
line = lines[i]
if len(line) == 73 or len(line) == 64:
if line[1] == "4" and line[2] == "7":
line47 = line
if line[1] == "5" and line[2] == "0":
line50 = line
H = lines[i + 1]
H2 = lines[i + 2]
O = lines[i + 3]
if line[4] == "O" and H[4] == "H" and H2[4] == "H" and H2[5] == "2" and O[4] == "O":
oxygen1Line = line
oxygen2Line = O
else:
newf.write(line)
f.close()
main()
它读取文件并将行存储在内存中。线47和线50存储在变量中以成为新线。下一行需要遵循O，H，H2，O的模式，然后进行交换。我在代码中处理交换时遇到了一些困难。任何想法？