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I need help finding a regular expression to match all digit sequences which has no repeated digit. Example:

198345 -> matched.

198315 -> not matched (1 repeated twice )

How can I achieve this in grep\egrep?

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up vote 3 down vote accepted

If you need to do it with regex, see Stephane's wonderful answer to a similar question.

If regex is not a must, I offer you a Perl alternative:

perl -nle '
    $digits{$_}++ for /([0-9])/g;
    @repeated = grep { $digits{$_} > 1 } keys %digits;
    print @repeated? "":"$_"
    %digits = ();
' your_file

This assumes your_file has one number per line and will print out only those numbers whose digits are unique.

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Interesting question. Thank you. The answer I found a little bit creepy:

egrep -v '([0-9])[0-9]*\1'

pattern find the occurrence of digit [0-9], save it to \1 and check for non repetition. So it finds any digit followed by the same digit. Use -v for inverse

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Below the same numbers must appear consecutively: they match 1123456 (consecutive 1s) but not 1213456 (non-consecutive 1s):

A lame solution is this:

cat testfile |  grep [0-9] | grep -v "0\{2,\}" | grep -v "1\{2,\}" | grep -v "2\{2,\}"  | grep -v "3\{2,\}"  | grep -v "4\{2,\}"  | grep -v "5\{2,\}"  | grep -v "6\{2,\}"  | grep -v "7\{2,\}"  | grep -v "8\{2,\}"  | grep -v "9\{2,\}"  

The first grep matches numbers and the rest 10 grepensure that every number appears only once.

A more compact but still lame way is this:

cat test |  grep [0-9] | grep -v "1\{2,\}\|2\{2,\}\|3\{2,\}\|4\{2,\}\|5\{2,\}\|6\{2,\}\|7\{2,\}\|8\{2,\}\|9\{2,\}\|0\{2,\}"

testfile must have one word per line.

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The greps are not actually doing what you think: they discard numbers with no repeating consecutive digits. That's not quite what the OP wants. In any case, your assumption (1 number per line) renders grep [0-9] redundant. – Joseph R. Jun 3 '14 at 20:11
@Joseph Yeap, I should have said "one word per line" instead of "one number per line". Thanks! – pidosaurus Jun 3 '14 at 20:15
Ah, they match consecutive digits. :( Yes it is not correct. – pidosaurus Jun 3 '14 at 20:42

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