根据查询ID的列和列表从固定格式的无空格文件中提取行

Question

I have a quite large fixed format file without spaces (file1):

file1:

0808563800555550000367120000500000
0005555566369330000078020000500000
01066666780000000008933600009000005635
0904251263088000000786590056500000
0000469011009904440425120444444440

I want to extract lines with fields 4-8,11-15 and 20-24 when fields 4-8 (only) are in a list of IDs in file2

file2:

55555
42512

The desired outputs are:

55555 36933 07802
42512 08800 78659

I have tried the following combination of cut | grepcut | grep commands:

cut -c 4-8,11-15,20-24 file1 --output-delimiter=' ' | grep -w -F -f file2

It works fine and the speed is very good, but the problem is that I am getting columns where the lookup ID (fields 4-8) is not in the first column of the cutted data, and that is because grep checks the three columns after cut, not only the first one.

Here are the outputs of the command above:

85638 55555 36712
55555 36933 07802
66666 00000 89336
42512 08800 78659
04690 00990 42512

I know one may write the output to a file and then use, for example awk, but I thought there could be a much simpler approach to avoid longer processing time (for example, makes grep picks only the match in a specific cutted column).

Any help will be very appreciated and many thanks!

Answer 1

Would you please try the following:

cut -c 4-8,11-15,20-24 file1 --output-delimiter=' ' | grep -wf <(sed 's/^/^/' file2)

Each line in file2 is prepended by a caret ^ character to anchor to the start of the line of the output by cut .
It may be a bit slower than before due to the lack of -F option.

Answer 2

With GNU awk for FIELDWIDTHS :

$ awk -v FIELDWIDTHS='3 5 2 5 4 5 *' 'NR==FNR{a[$0]; next} $2 in a{ print $2, $4, $6 }' file2 file1
55555 36933 07802
42512 08800 78659