我有兩個檔案：bor.mol2 和 bor.com 在 bor.mol2 中，有原子的 X、Y、Z 座標@<TRIPOS>ATOM，並且@<TRIPOS>BOND該檔案中的關鍵字第 3、4 和 5 列是原子的座標。例如H1座標為：-0.1660 2.5890 -0.2030
另一個檔案是 bor.com，第 2、3 和 4 列是 X、Y、Z 座標（在 0 1和之間1 2 1.0）。我想用 bor.com 坐標替換 bor.mol2 坐標並獲得 boron.mol2 如下所示。
如何使用 awk 或 grep 來做到這一點？
硼摩爾2
@<TRIPOS>MOLECULE
MOL
19 18 1 0 0
SMALL
resp
@<TRIPOS>ATOM
1 H1 -0.1660 2.5890 -0.2030 H 1 MOL 0.425234
2 O1 -0.6950 1.8160 -0.3360 O 1 MOL -0.740851
3 B1 -0.0040 0.6800 -0.0410 B 1 MOL 0.916675
4 O2 1.2760 0.7930 0.3900 O 1 MOL -0.584834
5 C1 2.0810 -0.3200 0.7070 C 1 MOL 0.351772
6 C2 2.8020 -0.8460 -0.5220 C 1 MOL -0.254733
7 H2 3.3950 -0.0620 -0.9800 H 1 MOL 0.065923
8 H3 2.0930 -1.2150 -1.2550 H 1 MOL 0.065923
9 H4 3.4660 -1.6630 -0.2510 H 1 MOL 0.065923
10 H5 1.4780 -1.0990 1.1550 H 1 MOL -0.005252
11 H6 2.7960 0.0200 1.4450 H 1 MOL -0.005252
12 O3 -0.5850 -0.5340 -0.1770 O 1 MOL -0.584834
13 C3 -1.9220 -0.7000 -0.5940 C 1 MOL 0.351772
14 C4 -2.8780 -0.6420 0.5840 C 1 MOL -0.254733
15 H7 -3.8950 -0.8350 0.2550 H 1 MOL 0.065923
16 H8 -2.8530 0.3360 1.0520 H 1 MOL 0.065923
17 H9 -2.6110 -1.3870 1.3260 H 1 MOL 0.065923
18 H10 -2.1780 0.0530 -1.3290 H 1 MOL -0.005252
19 H11 -1.9740 -1.6700 -1.0740 H 1 MOL -0.005252
@<TRIPOS>BOND
1 1 2 1
2 2 3 1
3 3 4 1
4 3 12 1
5 4 5 1
6 5 6 1
7 5 10 1
8 5 11 1
9 6 7 1
10 6 8 1
11 6 9 1
12 12 13 1
13 13 14 1
14 13 18 1
15 13 19 1
16 14 15 1
17 14 16 1
18 14 17 1
@<TRIPOS>SUBSTRUCTURE
1 MOL 1 TEMP 0 **** **** 0 ROOT
bor.com：
%nprocshared=4
%mem=1GB
# am1 geom=connectivity sp
MOL
0 1
H -0.16720146 2.58919775 -0.19942423
O -0.69500000 1.81600000 -0.33600000
B -0.00400000 0.68000000 -0.04100000
O -0.38867986 -0.48241992 -0.62214658
C 0.24973028 -1.71425091 -0.37253088
C -0.34829932 -2.40893346 0.83855738
H -1.41561875 -2.54983268 0.70799890
H -0.18501334 -1.82371627 1.73688335
H 0.11053291 -3.38414892 0.98087325
H 1.31216868 -1.56088464 -0.23520857
H 0.10760188 -2.31766293 -1.25975262
O 1.04303104 0.71384972 0.81482310
C 1.49768870 1.90335555 1.42093890
C 2.50478033 2.62078365 0.54000342
H 2.89233808 3.49938587 1.04758216
H 2.04460091 2.94105867 -0.38832398
H 3.33724896 1.96604913 0.30506364
H 0.65864568 2.55057309 1.64429105
H 1.95688815 1.61234302 2.35819638
1 2 1.0
2 3 1.0
3 4 1.0 12 1.0
4 5 1.0
5 6 1.0 10 1.0 11 1.0
6 7 1.0 8 1.0 9 1.0
7
8
9
10
11
12 13 1.0
13 14 1.0 18 1.0 19 1.0
14 15 1.0 16 1.0 17 1.0
15
16
17
18
19
結果必須是這樣的 boron.mol2：
@<TRIPOS>MOLECULE
MOL
19 18 1 0 0
SMALL
resp
@<TRIPOS>ATOM
1 H1 -0.16720146 2.58919775 -0.19942423 H 1 MOL 0.425234
2 O1 -0.69500000 1.81600000 -0.33600000 O 1 MOL -0.740851
3 B1 -0.00400000 0.68000000 -0.04100000 B 1 MOL 0.916675
4 O2 -0.38867986 -0.48241992 -0.62214658 O 1 MOL -0.584834
5 C1 0.24973028 -1.71425091 -0.37253088 C 1 MOL 0.351772
6 C2 -0.34829932 -2.40893346 0.83855738 C 1 MOL -0.254733
7 H2 -1.41561875 -2.54983268 0.70799890 H 1 MOL 0.065923
8 H3 -0.18501334 -1.82371627 1.73688335 H 1 MOL 0.065923
9 H4 0.11053291 -3.38414892 0.98087325 H 1 MOL 0.065923
10 H5 1.31216868 -1.56088464 -0.23520857 H 1 MOL -0.005252
11 H6 0.10760188 -2.31766293 -1.25975262 H 1 MOL -0.005252
12 O3 1.04303104 0.71384972 0.81482310 O 1 MOL -0.584834
13 C3 1.49768870 1.90335555 1.42093890 C 1 MOL 0.351772
14 C4 2.50478033 2.62078365 0.54000342 C 1 MOL -0.254733
15 H7 2.89233808 3.49938587 1.04758216 H 1 MOL 0.065923
16 H8 2.04460091 2.94105867 -0.38832398 H 1 MOL 0.065923
17 H9 3.33724896 1.96604913 0.30506364 H 1 MOL 0.065923
18 H10 0.65864568 2.55057309 1.64429105 H 1 MOL -0.005252
19 H11 1.95688815 1.61234302 2.35819638 H 1 MOL -0.005252
@<TRIPOS>BOND
1 1 2 1
2 2 3 1
3 3 4 1
4 3 12 1
5 4 5 1
6 5 6 1
7 5 10 1
8 5 11 1
9 6 7 1
10 6 8 1
11 6 9 1
12 12 13 1
13 13 14 1
14 13 18 1
15 13 19 1
16 14 15 1
17 14 16 1
18 14 17 1
@<TRIPOS>SUBSTRUCTURE
1 MOL 1 TEMP 0 **** **** 0 ROOT
答案1
我不太了解 awk，所以我使用 sed。
sed -rn '/^ [A-Z]/{H;x;s/^\n//;x};/^ *[0-9]+ +[A-Z]+[0-9]+/{G;s/^( *[^ ]+ +[^ ]+) +[^ ]+ +[^ ]+ +[^ ]+([^\n]+)\n *[^ ]+( *[^ ]+ +[^ ]+ +[^\n]+).*/\1\3\2/;x;s/^[^\n]+\n//;x};/MOLECULE/,$p' bor.com bor.mol2 > boron.mol2
輸出：硼摩爾2
@<TRIPOS>MOLECULE
MOL
19 18 1 0 0
SMALL
resp
@<TRIPOS>ATOM
1 H1 -0.16720146 2.58919775 -0.19942423 H 1 MOL 0.425234
2 O1 -0.69500000 1.81600000 -0.33600000 O 1 MOL -0.740851
3 B1 -0.00400000 0.68000000 -0.04100000 B 1 MOL 0.916675
4 O2 -0.38867986 -0.48241992 -0.62214658 O 1 MOL -0.584834
5 C1 0.24973028 -1.71425091 -0.37253088 C 1 MOL 0.351772
6 C2 -0.34829932 -2.40893346 0.83855738 C 1 MOL -0.254733
7 H2 -1.41561875 -2.54983268 0.70799890 H 1 MOL 0.065923
8 H3 -0.18501334 -1.82371627 1.73688335 H 1 MOL 0.065923
9 H4 0.11053291 -3.38414892 0.98087325 H 1 MOL 0.065923
10 H5 1.31216868 -1.56088464 -0.23520857 H 1 MOL -0.005252
11 H6 0.10760188 -2.31766293 -1.25975262 H 1 MOL -0.005252
12 O3 1.04303104 0.71384972 0.81482310 O 1 MOL -0.584834
13 C3 1.49768870 1.90335555 1.42093890 C 1 MOL 0.351772
14 C4 2.50478033 2.62078365 0.54000342 C 1 MOL -0.254733
15 H7 2.89233808 3.49938587 1.04758216 H 1 MOL 0.065923
16 H8 2.04460091 2.94105867 -0.38832398 H 1 MOL 0.065923
17 H9 3.33724896 1.96604913 0.30506364 H 1 MOL 0.065923
18 H10 0.65864568 2.55057309 1.64429105 H 1 MOL -0.005252
19 H11 1.95688815 1.61234302 2.35819638 H 1 MOL -0.005252
@<TRIPOS>BOND
1 1 2 1
2 2 3 1
3 3 4 1
4 3 12 1
5 4 5 1
6 5 6 1
7 5 10 1
8 5 11 1
9 6 7 1
10 6 8 1
11 6 9 1
12 12 13 1
13 13 14 1
14 13 18 1
15 13 19 1
16 14 15 1
17 14 16 1
18 14 17 1
@<TRIPOS>SUBSTRUCTURE
1 MOL 1 TEMP 0 **** **** 0 ROOT
答案2
這是一種方法：
$ awk '{
if(/@<TRIPOS>ATOM/){a=1; print; next}
if(/@<TRIPOS>BOND/){a=0}
if(NR==FNR){
val[FNR][2]=$2;
val[FNR][3]=$3;
val[FNR][4]=$4;
}
else{
if(a){
OFS="\t"
$3=val[a][2];
$4=val[a][3];
$5=val[a][4]
a++;
}
print $0
}
}' <(grep '^ [A-Z]' bor.com) bor.mol2
以上回：
@<TRIPOS>MOLECULE
MOL
19 18 1 0 0
SMALL
resp
@<TRIPOS>ATOM
1 H1 -0.16720146 2.58919775 -0.19942423 H 1 MOL 0.425234
2 O1 -0.69500000 1.81600000 -0.33600000 O 1 MOL -0.740851
3 B1 -0.00400000 0.68000000 -0.04100000 B 1 MOL 0.916675
4 O2 -0.38867986 -0.48241992 -0.62214658 O 1 MOL -0.584834
5 C1 0.24973028 -1.71425091 -0.37253088 C 1 MOL 0.351772
6 C2 -0.34829932 -2.40893346 0.83855738 C 1 MOL -0.254733
7 H2 -1.41561875 -2.54983268 0.70799890 H 1 MOL 0.065923
8 H3 -0.18501334 -1.82371627 1.73688335 H 1 MOL 0.065923
9 H4 0.11053291 -3.38414892 0.98087325 H 1 MOL 0.065923
10 H5 1.31216868 -1.56088464 -0.23520857 H 1 MOL -0.005252
11 H6 0.10760188 -2.31766293 -1.25975262 H 1 MOL -0.005252
12 O3 1.04303104 0.71384972 0.81482310 O 1 MOL -0.584834
13 C3 1.49768870 1.90335555 1.42093890 C 1 MOL 0.351772
14 C4 2.50478033 2.62078365 0.54000342 C 1 MOL -0.254733
15 H7 2.89233808 3.49938587 1.04758216 H 1 MOL 0.065923
16 H8 2.04460091 2.94105867 -0.38832398 H 1 MOL 0.065923
17 H9 3.33724896 1.96604913 0.30506364 H 1 MOL 0.065923
18 H10 0.65864568 2.55057309 1.64429105 H 1 MOL -0.005252
19 H11 1.95688815 1.61234302 2.35819638 H 1 MOL -0.005252
@<TRIPOS>BOND
1 1 2 1
2 2 3 1
3 3 4 1
4 3 12 1
5 4 5 1
6 5 6 1
7 5 10 1
8 5 11 1
9 6 7 1
10 6 8 1
11 6 9 1
12 12 13 1
13 13 14 1
14 13 18 1
15 13 19 1
16 14 15 1
17 14 16 1
18 14 17 1
@<TRIPOS>SUBSTRUCTURE
1 MOL 1 TEMP 0 **** **** 0 ROOT
該命令將只列印以空格開頭然後是大寫字母的grep '^ [A-Z]' bor.com那些行。bor.com這些是我們想要的唯一線路bor.com。使用 bash 的建構將輸出grep作為檔案句柄傳遞給 awk <()。然後，val如果腳本正在讀取的檔案是其輸入檔案中的第一個，則腳本會將值保存在陣列中（NR==FNR)如果是第二個，並且如果我們位於@<TRIPOS>ATOM和@<TRIPOS>BOND字串之間，則它將替換第三個、第四個和第五個欄位與數組中的值val。