Regex match string that end with character but exclusive of that character

Question

I need to match the quality of a film and order them. The value can be found in the title

Ex. "The Wolf of Wall Street (2014) (HD, 1080p).mov"

I have a regex that matches the quality:

"\\d\\d\\d[\\d]*p"

For some reason when i compare (compareTo) the values they are ordered:

Ex.

480p lists higher than 720p. I believe that this happens because of the p(?). I wonder how I can change my regex to check that there is a p (so the year in the title isn't matched, only the quality) but excludes the p in the matched string.

What is your java code? It could be done without using regex — Yassin Hajaj, Mar 19 '16 at 18:41
Also note that being listed "higher" means that the value compares as *smaller*. Thus "1080" < "480" < "720". — RealSkeptic, Mar 19 '16 at 18:41
You might also take a look at http://stackoverflow.com/questions/1262239/natural-sort-order-string-comparison-in-java-is-one-built-in concering natural sorting (what you might search for, if you don't want to compare just the numbers as numbers) — Sebastian Proske, Mar 19 '16 at 18:45

score 2 · Answer 1 · answered Mar 19 '16 at 18:43

2

Use can use regex "(\\d{3,4})p" to extract number from string with Matcher.group(1) and sort them as numbers, not strings. Strings are compared lexicographically.

answered Mar 19 '16 at 18:43

Vladimir Petrakovich

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Please explain more! What does matcher.group(1) do? – Mar 19 '16 at 19:01
@kletan it extracts the first capturing group (the string inside parentheses) – Vladimir Petrakovich Mar 19 '16 at 19:31
@kletan Matcher created with `Pattern.compile("(\\d{3,4})p)").matcher("The Wolf of Wall Street (2014) (HD, 1080p).mov")` must return `1080` from `matcher.group(1)` after `matcher.find()`. – Vladimir Petrakovich Mar 20 '16 at 04:43

Alex Salauyou · Answer 2 · 2016-03-19T19:38:43.100

1

You need a special String comparator, which firstly compare by length, then by value:

Comparator<String> LENGTH_FIRST = (s1, s2) -> s1.length() == s2.length() 
                                      ? s1.compareTo(s2) 
                                      : (s1.length() - s2.length());

When sorting using such comparator, longer strings (meaning higher quality) become located after shorter:

List<String> qualities = Arrays.asList("1080p", "480p", "720p");
Collections.sort(qualities, LENGTH_FIRST); // will re-order as [480p, 720p, 1080p]

edited Mar 19 '16 at 19:38

answered Mar 19 '16 at 18:42

Alex Salauyou

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This actually works, but I can't really understand why? – Mar 19 '16 at 19:02
@kletan very simple. For positive integers, any number with more digits is bigger than any number having less digits (`1000` > `999`). Among numbers of same count of digits, bigger is number that is bigger lexicographically. – Alex Salauyou Mar 19 '16 at 19:27

score 0 · Answer 3 · answered Mar 19 '16 at 18:43

0

I haven't used Java that much, but maybe you could use a group inside the regex? I.e.:

"(\\d\\d\\d[\\d]*)p"

And then make a backreference to the group (content inside the parentheses). I know you can do that with perl regex

Hope it helped!

answered Mar 19 '16 at 18:43

JuanKman94

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Regex match string that end with character but exclusive of that character

3 Answers3