Regex in Java - Removing unnecessary spaces

Question

Every time I need something with regular expressions I have such a hard time...

Now, I need turn some fuzzy text... In an old, very old database, certain system didn't allow to users format their texts... so, users got creative, entering expressions like:

S O M E   T E X T   I   W O U L D   L I K E   T O   H I G H L I G H T

My question is, how can I turn that text in:

SOME TEXT I WOULD LIKE TO HIGHLIGHT

with regular expressions in Java.

Sorry about the silly question, but I've spent a lot more time trying figure this out than I was supposed to.

Each character was followed by one white-space... even the spaces. But you have to consider that this was made by hand... so, sometimes, instead of three white-spaces between words, it's possible you find two, four or even more. — Alex Gouvêa Vasconcelos, Mar 06 '15 at 17:44

score 3 · Answer 1 · answered Mar 06 '15 at 14:22

3

This regex will gives you a single space at the middle that is, a single space between words.

String r = "S O M E   T E X T   I   W O U L D   L I K E   T O   H I G H L I G H T";
System.out.println(r.replaceAll("(\\s){2,}|\\s", "$1"));

Output:

SOME TEXT I WOULD LIKE TO HIGHLIGHT

The idea behind this is, the above regex would capture a a single space from two or more consecutive spaces and all other spaces or further matched. Replacing the matched spaces with the character inside group index 1 will give you the desired output.

Regex Demo

answered Mar 06 '15 at 14:22

Avinash Raj

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It really works, but I couldn't understand it... As I can read, the expression says: [2 or more white-spaces OR 1 white-space] replaces for [white-space], since the group 1 is always a white-space... Did I understand wrong? – Alex Gouvêa Vasconcelos Mar 06 '15 at 18:21
@AlexGouvêaVasconcelos Idea is to let regex test if there are more than one spaces first, and if that is true replace it with first one. In case there is only one space match in group `1` will be empty so you will replace this single space with nothing (you will remove it). – Pshemo Mar 06 '15 at 18:26
I see... I wasn't realizing that the quantifying that follows the group creates a condition for it's value, and because of this the '$1' that represents the group could be empty... – Alex Gouvêa Vasconcelos Mar 06 '15 at 18:40
It is similar to `(foo)+` and `(foo+)`. For data like `ZfoofooZ` both regexes will match `foofoo` part, but in first case since group doesn't include repeating (`+`) its value will be each time replaced with new `foo` it will find. So I lied a little by saying that "and if that is true replace it with first one" because it will replace with last one (but since both values will be same it doesn't change anything). – Pshemo Mar 06 '15 at 18:47

CoronA · Accepted Answer · 2015-03-06T15:00:11.653

2

With one Pattern, no Lookaheads, no word border anchors

text.replaceAll("\\s(\\s?)\\s*", "$1")

Explanation:

replace any whitespace sequence with a minimum length of 1 (\s)
if the next char is a whitespace ((\s?) is matched) => replace with whitespace
else ((\s?) is not matched) replace with empty String
capture all whitespaces after (\s*)

edited Mar 06 '15 at 15:00

answered Mar 06 '15 at 13:54

CoronA

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In fact, this is exactly what I was looking for. Thank you StefanA. Now, I have to figure out when do I need apply the replace, since, sometimes, the text is written using that technique, sometimes it's not (did I mention the user has been creative :-) )... I mean, sometimes you find [A...T.E.X.T] sometimes [A.TEXT], and, if I always make the replacement, in the second case I'll have [ATEXT]. (dots instead of white-spaces, so they can be seen in this comment). – Alex Gouvêa Vasconcelos Mar 06 '15 at 17:54

score 1 · Answer 3 · edited Mar 06 '15 at 14:37

1

If the words are seperated by multiple spaces, you can use negatve look ahead as

\s(?!\s)

Regex demo

Test

"S O M E   T E X T   I   W O U L D   L I K E   T O   H I G H L I G H T"
.replaceAll("\\s(?!\\s)", "")
.replaceAll("\\s+", " ");
=> SOME TEXT I WOULD LIKE TO HIGHLIGHT

edited Mar 06 '15 at 14:37

Pshemo

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answered Mar 06 '15 at 13:53

nu11p01n73R

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This could leave uneven spacing in the resulting sentence if the users also tried to align stuff with multiple spaces. – Ravi K Thapliyal Mar 06 '15 at 14:06
@RaviThapliyal Yes, but in that case we will have to use another `replaceAll` to align everything. I skiped that as OP has not clearly mentioned anything related to that. – nu11p01n73R Mar 06 '15 at 14:08
@nu11p01n73R this would give you two spaces at the middle. – Avinash Raj Mar 06 '15 at 14:20
It works... but there are other answers more optimized. Thank you anyway. – Alex Gouvêa Vasconcelos Mar 06 '15 at 18:13

Pshemo · Answer 4 · 2015-03-06T18:37:02.253

So you can use replaceAll("(.)\\s", "$1")

Example:

String s = "S O M E   T E X T   I   W O U L D   L I K E   T O   H I G H L I G H T";
s = s.replaceAll("(.)\\s", "$1");
System.out.println(s);

Output: SOME TEXT I WOULD LIKE TO HIGHLIGHT

Explanation:

Think of your text as two characters chunks (I will mark them with ^^ and ##).

S O M E   T E X T
^^##^^##^^##^^##

If you look closely you will notice that you want to remove second character from each pair (which is space), and leave first character:

S O M E   T E X T
^ # ^ # ^ # ^ # T - T will not be affected (will stay) 
                    because it doesn't have space after it.

You can achieve it with (.)\s regex where

. represents any character (including space)
\s represents any whitespace

This way first character will be placed in group (indexed as 1) which allows us to use match from this part in replacement part via $x where x represents group index.

Ver.2 (in case spaces to remove are not only on odd indexed positions)

Other way to solve this problem is to remove only these spaces which

are placed right after non-space character (?<=\\S)\\s
```
S O M E       T E X T
 ^ ^ ^ ^       ^ ^ ^
```

are placed before other spaces \\s(?=\\s)

S O M E       T E X T
 ^ ^ ^ ^#####  ^ ^ ^

This way as you can see one space is left (the one right before word) so your solution can look like

s = s.replaceAll("(?<=\\S)\\s|\\s+(?=\\s)", "");

@StefanA Which is what OP gave us in his question. I suspect that string from OP example was created precisely by adding space after every character in original string, except last one. — Pshemo, Mar 06 '15 at 14:57
Actually, it's a very clever way to "undo" what the user may have done. But @StepfanA is right... it's possible that this is not the only case we can find in those texts. Thank you anyway, Pshemo. — Alex Gouvêa Vasconcelos, Mar 06 '15 at 18:18

score 0 · Answer 5 · answered Mar 06 '15 at 13:56

If all the words are guaranteed to have two or more spaces in between then:

First, remove all the spaces between the characters as
```
input.replaceAll("(?<=\\S)\\s(?=\\S)", "");
```
Then, replace all the multiple spaces between the words with just one
```
input.replaceAll("\\s{2,}", " ");
```

So, the complete code would look like

String input = "S O M E   T E X T   I   W O U L D   L I K E   T O   H I G H L I G H T";
input = input.replaceAll("(?<=\\S)\\s(?=\\S)", "").replaceAll("\\s{2,}", " ");

System.out.println(input); // SOME TEXT I WOULD LIKE TO HIGHLIGHT

It works... but there are other answers more optimized. Thank you anyway. — Alex Gouvêa Vasconcelos, Mar 06 '15 at 18:13

Regex in Java - Removing unnecessary spaces

5 Answers5