I am trying to use perl regexp to normalize search strings in a search log against a library database. I need to remove all digit occurrences, would be:
s/\d*//g
except when I have a birth date like 1964- or a lifetime like 1903-1970 or 1903-70. How do I do that?
You could use lookaround assertions.
For example, the following pattern
/\b(?<!-)\d+(?!-)\b/
would match a number like 42 or 1970 but not match:
For example, given an input:
42 foo 123 1964- 1903-1970 456 bar 1970
using the above regex to remove the intended strings:
$ echo 42 foo 123 1964- 1903-1970 456 bar 1970 | perl -pe 's/\b(?<!-)\d+(?!-)\b//g'
foo 1964- 1903-1970 bar
A complicated regex could solve this, for sure. However, I believe the easiest solution is to take advantage of one of regular expressions most powerful tools, namely greedy matching, and break this into two steps.
s{([-\d]+)}{my $num = $1; $num =~ /^(?:\d+-\d*|-+)$/ ? $num : ''}eg;
The LHS pulls any number and/or dashes. Then the RHS leaves them if they match the specific exception that you requested.
I like the two step solution because it's quicker to see what's happening, and also the regex is less fragile so it's easier to adjust it at a later time with less risk of introducing a bug. All you'd have to do is add any additional exceptions you'd want to the RHS.
It is possible to duplicate the above using just the LHS by adding a lot of boundary conditions that mirror the effect of greedy matching. The below demonstrates that:
s{
(?<![-\d]) # Start Boundary Condition to Enforce Greedy Matching
(?!
(?: # Old RHS: List of expressions we don't want to match
\d+-\d*
|
-+
)
(?![-\d]) # End Boundary Condition to Enforce Greedy Matching
)
([-\d]+) # Old LHS: What we want to match
(?![-\d]) # End Boundary Condition to Enforce Greedy Matching
}{}xg;
Did you mean to replace all the digits except the digits are in format of 1000- or 1000-90?
Try this one:
(?<!\d)(?<!-)\d+(?!-\d*)(?!\d)
