Re: RegEx Question

match a range of numbers with each digit just matching once?

Please clarify the question by providing some sample input and some desired output.

I've tried something like ^0-9{1,1} but its not working.

Please use 'code' tags around your code because it renders poorly. See Writeup Formatting Tips.

Here is what your regular expression means, according to YAPE::Regex::Explain:

use warnings;
use strict;
use YAPE::Regex::Explain;
print YAPE::Regex::Explain->new('[^0-9]{1,1}')->explain();

__END__
The regular expression:

(?-imsx:[^0-9]{1,1})

matches as follows:

NODE                     EXPLANATION
----------------------------------------------------------------------
(?-imsx:                 group, but do not capture (case-sensitive)
                         (with ^ and $ matching normally) (with . not
                         matching \n) (matching whitespace and #
                         normally):
----------------------------------------------------------------------
  [^0-9]{1,1}              any character except: '0' to '9' (between
                           1 and 1 times (matching the most amount
                           possible))
----------------------------------------------------------------------
)                        end of grouping
----------------------------------------------------------------------
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Comment on Re: RegEx Question Download Code

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Re^2: RegEx Question by yoda54 (Monk) on Sep 09, 2009 at 20:06 UTC
Thanks for the reply. I like to match permutations of 0-8 with each digits occurring once using all 8 digits. I've tried something like the follow but I can't get it straight... :( `[0-8]{1, 1}` [download]	[reply] [d/l]
Re^3: RegEx Question by CountZero (Bishop) on Sep 09, 2009 at 21:28 UTC
`[0-8]{1, 1}` matches one digit out of the set {0 1 2 3 4 5 6 7 8}, so it doesn't do what you need at all. Actually, I am not sure that you can do this in one regex. I will show a simpler example only using permutations of the digits 0 - 3 (to make it easier to follow): First you have to check that your string contains 4 digits in the range 0 - 3. That one is easy: `/^[0-3]{4}$/` If your string passes this test, then you check whether each digit occurs only once. `/^(\d)(?!\d\1)(\d)(?!\d\2)(\d)(?!\d\3)\d$/;` [download] This works by capturing each digit and using negative look-aheads to check that this digit does not re-occur again in the string. The following program proves that it works: `use warnings; use strict; for my $one (0 .. 3) { for my $two (0 .. 3) { for my $three (0 .. 3) { for my $four (0 .. 4) { my $test = "$one$two$three$four"; print "$test\n" if $test=~m/^[0-3]{4}$/ and $test=~m/^ +(\d)(?!\d\1)(\d)(?!\d\2)(\d)(?!\d\3)\d$/; } } } }` [download] Output: `0123 0132 0213 0231 0312 0321 1023 1032 1203 1230 1302 1320 2013 2031 2103 2130 2301 2310 3012 3021 3102 3120 3201 3210` [download] CountZero A program should be light and agile, its subroutines connected like a string of pearls. The spirit and intent of the program should be retained throughout. There should be neither too little or too much, neither needless loops nor useless variables, neither lack of structure nor overwhelming rigidity." - The Tao of Programming, 4.1 - Geoffrey James	[reply] [d/l] [select]
Re^4: RegEx Question by ikegami (Patriarch) on Sep 10, 2009 at 00:46 UTC
Actually, I am not sure that you can do this in one regex. Assuming he meant all 9 digits, `/ ^ (?=[0-8]{9}\z) (?=.0) (?=.1) (?=.2) (?=.3) (?=.4) (?=.5) (?=.6) (?=.7) (?=.*8) /sx;` [download]	[reply] [d/l]
Re^5: RegEx Question by CountZero (Bishop) on Sep 10, 2009 at 06:03 UTC
Re^6: RegEx Question by ikegami (Patriarch) on Sep 10, 2009 at 06:32 UTC
Re^4: RegEx Question by yoda54 (Monk) on Sep 09, 2009 at 21:34 UTC
Thanks for the reply... I was really hoping for a easy way out...!! :)	[reply]
Re^5: RegEx Question by GrandFather (Saint) on Sep 09, 2009 at 22:12 UTC
Re^3: RegEx Question by graff (Chancellor) on Sep 09, 2009 at 23:41 UTC
I like to match permutations of 0-8 with each digits occurring once using all 8 digits. Um... the way I learned it, "0-8" represents 9 digits. (1-8 would be 8 digits, as would 0-7.) Is there some digit between 0 and 8 that you intend to leave out, and if so, which one? Anyway, if you had said "permutations of 0-8 with each digit occurring once, using all 9 digits", then I would understand that you are looking only for strings of nine digit characters, such that all nine characters are distinct, and none of them is the digit "9": `#!/usr/bin/perl use strict; while (<DATA>) { chomp; if ( length() == 9 and not ( /[^0-8]/ or /(.).*\1/ )) { print "$_\n"; } else { warn "rejected input: $_\n"; } } __DATA__ 01234 123456780 1234567890 223456781 345678012 234567890 a12345678 012345678 456782011 456782013` [download] Update: I suspect that the regex used as the last stage in my conditional is a fairly expensive operation; for strings that actually meet the criteria (are not rejected), it has to do 8+7+6+...+1 (total of 36) character comparisons to finish. (There should be some sort of "O(...n...)" expression for that, but it escapes me.) So, it would most likely be better to use a split/hash solution, as suggested by others, especially if you'll be handling large quantities of input with a relatively high "success" rate. Something like this: `while (<DATA>) { chomp; if ( length() == 9 and not /[^0-8]/ ) { my %c = map { $_ => undef } split //; if ( keys %c == 9 ) { print "$_\n"; next; } } warn "rejected input: $_\n"; }` [download]	[reply] [d/l] [select]