Use same string part for regex inspection

Question

I'm pretty new to regex, but I did some study. I got into a problem that might turn out impossible to be solved by regex, so I need a piece of advice.

I have the following string:

some text key 12, 32, 311 ,465 and 345. some other text dog 612, 
12, 32, 9 and 10. some text key 1, 2.

I'm trying to figure if it possible (using regex only) to extract the numbers 12 32 311 465 345 1 2 only - as a set of individual matches.

When I approach this problem I tried to look for a pattern that matches only the relevant results. So I came up with :

get numbers that have prefix of "key" and NOT have prefix of "dog".

But I'm not sure if it is even possible. I mean that I know that for the number 1 I can use (?<=key )+[\d]+ and get it as a result, but for the rest of the numbers (i.e. 2..5), can I "use" the key prefix again?

Could you try to rewrite your question? It's not clear. Do you want to extract numbers or digits? What would you like to get from 11, 51, 1dog, dog1? — xenteros, Jul 23 '15 at 06:38
Exctracting numbers. the point is to extract only numbers that are followed by `key` string and NOT by `dog` string — Gil Peretz, Jul 23 '15 at 06:46
Are you expecting a single string match (i.e. `"12, 32, 311 ,465 and 345"`), or are you looking for a set of individual matches (i.e. `{12,32,311,465,345}`)? — Simon MᶜKenzie, Jul 23 '15 at 06:50
I'm expecting for set of individual matches. (edited my question). thank you — Gil Peretz, Jul 23 '15 at 06:54

Wiktor Stribiżew · Answer 1 · 2015-07-23T07:14:48.390

3

In Java, you can make use of a constrained width look-behind that accepts {n,m} limiting quantifier.

So, you can use

(?<=key(?:(?!dog)[^.]){0,100})[0-9]+

Or, if key and dog are whole words, use \b word boundary:

String pattern = "(?<=\\bkey\\b(?:(?!\\bdog\\b)[^.]){0,100})[0-9]+";

The only problem there may arise if the distance between the dog or key and the numbers is bigger than m. You may increase it to 1000 and I think that would work for most cases.

Sample IDEONE demo

String str = "some text key 12, 32, 311 ,465 and 345. some other text dog 612,\n12, 32, 9 and 10. some text key 1, 2.";
String str2 = "some text key 1, 2, 3 ,4 and 5. some other text dog 6, 7, 8, 9 and 10. some text, key 1, 2 dog 3, 4 key 5, 6";
Pattern ptrn = Pattern.compile("(?<=key(?:(?!dog)[^.]){0,100})[0-9]+");
Matcher m = ptrn.matcher(str);
while (m.find()) {
   System.out.println(m.group(0));
}
System.out.println("-----");
m = ptrn.matcher(str2);
while (m.find()) {
   System.out.println(m.group(0));
}

edited Jul 23 '15 at 07:14

answered Jul 23 '15 at 06:44

Wiktor Stribiżew

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Assumption being made here is that dog and key does not appear in the same "sentence" (delimited by `.`). – nhahtdh Jul 23 '15 at 06:49
Correct. That is what the sample input implies. – Wiktor Stribiżew Jul 23 '15 at 06:52
My issue with this solution is that the String: `key 1, 2 dog 3, 4 key 5, 6` will not work. I added it to the question example. – Gil Peretz Jul 23 '15 at 06:52
@GilPeretz: Write that test case in your question. – nhahtdh Jul 23 '15 at 06:56
@nhahtdh added it. Thank you. – Gil Peretz Jul 23 '15 at 06:58
Note that while it's possible to use look-behind this way in Java, it's extremely inefficient, since Java implements look-behind by *trying to match* at index (current - min_look_behind_length) to (current - max_look_behind_length). – nhahtdh Jul 23 '15 at 19:17
I have been noting the same everywhere. – Wiktor Stribiżew Jul 23 '15 at 19:18

score 2 · Answer 2 · edited May 23 '17 at 12:06

I wouldn't recommend using code that you can't understand and customize, but here is my one-pass solution, using the method described in this answer of mine. If you want to understand the construction method, please read the other answer.

(?:key(?>\s+and\s+|[\s,]+)|(?!^)\G(?>\s+and\s+|[\s,]+))(\d+)

Compared to the method described in the other post, I dropped the look-ahead, since in this case, we don't need to check against a suffix.

The separator here is (?>\s+and\s+|[\s,]+). It currently allows "and" with spaces on both sides, or any mix of spaces and commas. I use (?>pattern) to inhibit backtracking, so the order of the alternation is significant. Change it back to (?:pattern) if you want to modify it and you are unsure of what you are doing.

Sample code:

String input = "some text key 12, 32, 311 ,465 and 345. some other text dog 612,\n12, 32, 9 and 10. some text key 1, 2. key 1, 2 dog 3, 4 key 5, 6. key is dog 23, 45. key 4";
Pattern p = Pattern.compile("(?:key(?>\\s+and\\s+|[\\s,]+)|(?!^)\\G(?>\\s+and\\s+|[\\s,]+))(\\d+)");
Matcher m = p.matcher(input);
List<String> numbers = new ArrayList<>();

while (m.find()) {
    numbers.add(m.group(1));
}

System.out.println(numbers);

Demo on ideone

This is really the correct regex +1. Was going to post similar but dropped due to loose nature of input data structure. — anubhava, Jul 23 '15 at 08:13

Mazdak · Answer 3 · 2015-07-23T07:22:10.593

1

You can use a positive look behind which ensure that your sequence doesn't precede by any word except key :

(?<=key)\s(?:\d+[\s,]+)+(?:and )?\d+

Note that here you don't need to use a negative look behind for dog because this regex will just match if your sequence precede by key.

See demo https://regex101.com/r/gZ4hS4/3

edited Jul 23 '15 at 07:22

answered Jul 23 '15 at 06:38

Mazdak

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In the last it should be \d+, otherwise it would not match two digit number and higher in last. – Rahul Jul 23 '15 at 06:47
@Rahul Actually after `and` we write one digit! but its better to use `\d+`! – Mazdak Jul 23 '15 at 06:48
it will not match completly if you use only \d in `some text key 12, 32, 311 ,465 and 345. some other text dog 612, 12, 32, 9 and 10. some text` – Rahul Jul 23 '15 at 06:50
Now your regex is good but earlier you used `(?<=key)[^\d]*[\d, ]+(?:and )?\d`. – Rahul Jul 23 '15 at 06:52
@Kasramvd i'm looking for individual set of results rather than 1 single result. is it possible? – Gil Peretz Jul 23 '15 at 07:00
@GilPeretz Do you mean the word `key` can be various? – Mazdak Jul 23 '15 at 07:01
@Kasramvd I mean that the `key` word can appear multiple times and the `dog` word can appear multiple time as well (without any particular order) – Gil Peretz Jul 23 '15 at 07:06
@GilPeretz Yep, fixed! – Mazdak Jul 23 '15 at 07:15
@nhahtdh Yep, I think there is no need to `[^\d]*` here! because the numbers should come right after the `key`! – Mazdak Jul 23 '15 at 07:22

vks · Accepted Answer · 2015-07-23T11:49:55.197

1

You can do it in 2 steps.

(?<=key\\s)\\d+(?:\\s*(?:,|and)\\s*\\d+)*

Grab all the numbers.See demo.

https://regex101.com/r/uK9cD8/6

Then split or extract \\d+ from it.See demo.

https://regex101.com/r/uK9cD8/7

edited Jul 23 '15 at 11:49

answered Jul 23 '15 at 06:38

vks

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I'm afraid two steps is not an option for me. is it possible in single step? – Gil Peretz Jul 23 '15 at 07:23
1

@GilPeretz 2 steps is a sure shot way of achieving what you want.You anyways cannot do it in a single regex with 100 % accuracy. – vks Jul 23 '15 at 07:25
Initially, you were looking for a single regex to do it. I see the limitations of a constrained width lookbehind and nested lookarounds scared you off. Anyway, it's great you have solutions to choose from! – Wiktor Stribiżew Jul 23 '15 at 14:51

Use same string part for regex inspection

4 Answers4