remove first column from hexdump output - bash

Question

I have a hexdump output that looks like this
0101f10 64534 64943 00568 00262 01077 00721 00297 00140
0101f20 00748 00288 02211 01124 02533 01271 02451 00997
0101f30 03056 01248 02894 01026 02397 00696 00646 65114
0101f40 00943 64707 01113 64179 01135 64179 00805 64109
0101f50 00514 64045 64654 63037 63026 62014 62173 61625
I want to remove the first column, but I don't know what delimiter has been used by the hexdump command. I tried with awk and cut, but cant figure it out. Any help is appreciated.
Output I want is
64534 64943 00568 00262 01077 00721 00297 00140
00748 00288 02211 01124 02533 01271 02451 00997
03056 01248 02894 01026 02397 00696 00646 65114
00943 64707 01113 64179 01135 64179 00805 64109
00514 64045 64654 63037 63026 62014 62173 61625

Answer 1

With sed
sed 's/[^[:blank:]]*[[:blank:]]*//' infile
With gnu sed
sed 's/\S*\s*//' infile

Answer 2

input... | sed -E $'s/ +/\t/g' | cut -f2-
(Assuming Bash for $'\t', but GNU sed supports \t directly anyway.)

Answer 3

hexdump /path/to/file | awk '{sub(/[^ ]+ /, ""); print $0}'
This will do the job.

Answer 4

If the the delimiter is really a bunch of space, use tr to squeeze-repeats (-s) of psace to a tab and use cut for getting rid of the first column:
$ cat file | tr -s ' ' '\t' | cut -f 2-
64534 64943 00568 00262 01077 00721 00297 00140
00748 00288 02211 01124 02533 01271 02451 00997
03056 01248 02894 01026 02397 00696 00646 65114
00943 64707 01113 64179 01135 64179 00805 64109
00514 64045 64654 63037 63026 62014 62173 61625

Answer 5

All solution above works fine, Just adding an awk solution.
So, you only need to omit first column, but get the rest of it, you can try this :
awk '{$1=""; print $0}' /path/to/hexfile
It works perfectly, except that it leaves a space at beginning of each line. If that bothers you, there is a workaround using the substr() function in awk itself.
awk '{$1=""; print substr($0,2)}' /path/to/hexfile
To see more possible ways to do it, follow this link