how to grep a N (7) column row before a grepped number from a long text file - awk

Question

I have a text file with a spacial format.
After the top "N" rows, the file will have a 7 column row ans then there will be "X" rows (X is the value from column number 6 in this 7 column row). Then there will be another row with 7 column and it will have further "Y" sub-rows (Y is the value from column number 6 in this 7 column row). and it occurance of rows will go upto some fixed numbers, say 40.
En example is here
(I am skipping top few rows).
2.857142857143E-01 2.857142857143E-01-2.857142857143E-01 1 1533 9 1.0
1 -3.52823873905418
2 -3.52823873905417
3 -1.77620635653680
4 -1.77620635653680
5 -1.77620570068355
6 -1.77620570068354
7 -1.77620570066112
8 -1.77620570066112
9 -1.60388273192418
1.428571428571E-01 1.428571428571E-01-1.428571428571E-01 2 1506 14 8.0
1 -3.52823678441811
2 -3.52823678441810
3 -1.77620282216865
4 -1.77620282216865
5 -1.77619365786042
6 -1.77619365786042
7 -1.77619324280126
8 -1.77619324280125
9 -1.60387130881086
10 -1.60387130881086
11 -1.60387074066972
12 -1.60387074066972
13 -1.51340357895078
14 -1.51340357895078
1.000000000000E+00 4.285714285714E-01-1.428571428571E-01 20 1524 51 24.0
1 -3.52823580096110
2 -3.52823580096109
3 -1.77624472106293
4 -1.77624472106293
5 -1.77623455229910
6 -1.77623455229909
7 -1.77620473017160
8 -1.77620473017159
9 -1.60387169115834
10 -1.60387169115834
11 -1.60386634866654
12 -1.60386634866654
13 -1.51340851656332
14 -1.51340851656332
15 -1.51340086887553
16 -1.51340086887553
17 -1.51321967923767
18 -1.51321967923766
19 -1.40212716813452
20 -1.40212716813451
21 -1.40187887062753
22 -1.40187887062753
23 -0.749391485667459
24 -0.749391485667455
25 -0.740712218931955
26 -0.740712218931954
27 -0.714030906779278
28 -0.714030906779278
29 -0.689087278411268
30 -0.689087278411265
31 -0.687054399753234
32 -0.687054399753233
33 -0.677686868127079
34 -0.677686868127075
35 -0.405343895324740
36 -0.405343895324739
37 -0.404786479693490
38 -0.404786479693488
39 -0.269454266134757
40 -0.269454266134755
41 -0.267490250650300
42 -0.267490250650296
43 -0.262198373307171
44 -0.262198373307170
45 -0.260912148881762
46 -0.260912148881761
47 -9.015623907768122E-002
48 -9.015623907767983E-002
49 0.150591609452852
50 0.150591609452856
51 0.201194203960446
I want to grep a particular number from my text file and to do so, I use
awk -v c=2 -v t=$GREP 'NR==1{d=$c-t;d=d<0?-d:d;v=$c;next}{m=$c-t;m=m<0?-m:m}m<d{d=m;v=$c}END{print v}' case.dat
Here $GREP is 0.2011942 which prints the last row (it will change according to different file)
51 0.201194203960446
I want to print the header row also with this number, i.e., my script should print,
51 0.201194203960446
1.000000000000E+00 4.285714285714E-01-1.428571428571E-01 20 1524 51 24.0.
How can I print this header row of the grepped numbers?
I have idea, but I could not implement it in script format.
Simply, grep the number using my script and print the first row before this number that have 7 columns.

Answer 1

This may be what you're looking for
awk -v t="$GREP" '
BEGIN { sub("\\.", "\\.", t) }
NF > 2 { header=$0; next }
NF == 2 && $2 ~ t { printf("%s %s\n%s\n", $1, $2, header) }
' file
You can replace the NF > 2 with NF == 7 if you want the strictly seven-column header to be printed (that header contains 6 columns in your sample data, not 7).
Update after the comment "Can you please modify my script so that it should grep upto 13 decimal number":
awk -v t="$GREP" '
BEGIN { if (match(t, "\\.")) {
t = substr(t, 1, RSTART + 13)
sub("\\.", "\\.", t)
}
}
NF > 2 { header=$0; next }
NF == 2 && $2 ~ t { printf("%s %s\n%s\n", $1, $2, header) }
' file