mjhjmtu

I have a file filled with numbers, number per line. Each number consists of two or three digits.

I would like to filter out this file by any number has two or more sequential digits. These sequential digits can be consecutive (e.g. 127, 215, 781), or non-consecutive (e.g. 506). The order of the sequential digits is not important. It can be small to large (e.g. 127) or large to small (e.g. 215).

For example:

127
215
781
874
370
01
10
142
506
94

The expected output:

370
94

Because:

127 # Has two sequential and consecutive digits (1 and 2)
215 # Has two sequential and consecutive digits (1 and 2)
781 # Has two sequential and consecutive digits (7 and 8)
874 # Has two sequential and consecutive digits (7 and 8)
370 # Keep
01 # Has two sequential and consecutive digits (0 and 1)
10 # Has two sequential and consecutive digits (0 and 1)
142 # Has two sequential and non-consecutive digits (1 and 2)
506 # Has two sequential and non-consecutive digits (5 and 6)
94 # Keep

With GNU awk which supports empty string as field separator.

awk -F '' ' is_sequential=0;
 for (i=2; i<=NF; i++) is_sequential+=($0 ~ $i-1 ; 
 !is_sequential print ' infile

we are checking on each number $i for a number that it's equal with number-1 $i-1 or number+1 $i+1 against the whole line, meaning that if there was a number number-1 or number+1 or both seen in a line, so we found there are at least two numbers are next to each other (the first, the number $i itself and next one either $i-1 or $i+1 or both (sequential) and with Ternary condition it will increment the value of is_sequential variable otherwise always it will be 0.

In next block is_sequential print , we print that line where the value is unchanged (the value is still 0, no at least two numbers seen that were sequential).

You can try either

awk '
 split ("", N) # delete array N
 L = 1 # initialise boolean L to TRUE
 for (i=1; i<=length($1); i++) # for each digit
 P = substr($1, i, 1) 
 if (N[P-1] 
 
 L
' file

Output:

370
94

or

awk '
 split ("", N)
 L = 1
 for (i=1; i<=length; i++)
 N[substr($0,i,1)] = 1 # set all N elements for the digits in string

 for (i=0; i<9; i++)
 if (N[i] + N[i+1] == 2) # check for two adjacent elements to be TRUE
 L = 0 
 break
 
 
L
' file

Output:

370
94

Tested on Ubuntu 18.04

Here, as the list of combinations is relatively small, you might as well consider them all in a ERE alternation:

grep -vE '0.*1|1.*[02]|2.*[13]|3.*[24]|4.*[35]|5.*[46]|6.*[57]|7.*[68]|8.*[79]|9.*8'

The same with perl but using perl code in (??...) inside the regexp to match the next or previous digit:

perl -ne 'print unless /([0-8]).*(??$1+1)/ || /([1-9]).*(??$1-1)/'

With sed, you could append the list of consecutive pairs to the pattern space, and use back references to find the matches:

sed -ne '1x;s/$/0123456789876543210/;x;' -e 'G;/(.).*(.).*n.*12/!P'