regex to modify the string

Question

I have one string like below

A:82% X 18% Y B:100% X C:82% X 18% Y AB:82% X 18% Y

Want to convert above string into below format (Required Format)

#A:82% X,18% Y #B:100% X #C:82% X,18% Y #AB:82% X,18% Y

I'm able to achieve this

#A:82% X 18% Y #B:100% X #C:82% X 18% Y #AB:82% X 18% Y

using this code

String inp = "A:82% X 18% Y B:100% X C:82% X 18% Y AB:82% X 18% Y";
String regex = "(\\b[A-Za-z]{1,}\\:\\b)";   
System.out.println(inp.replaceAll(regex, "#$1"));

But am not able to find a way where I could place commas the way I want mentioned in required format.

Is there any way I could select only words except "#word:", "Number%" and "spaces", so that in resultant I could select only Xs and Ys (X and Y is just an example, other word are also possible). Because once am able to select those words then I can replace them with "$1,"

Thanks

is the number of (Number% Word) always fixed to 2 ?

Nahuel Fouilleul
– Nahuel Fouilleul

2018-02-26 12:59:25 +00:00
Commented Feb 26, 2018 at 12:59 — Nahuel Fouilleul
– Nahuel Fouilleul, Commented Feb 26, 2018 at 12:59
@NahuelFouilleul, No..!

Squeez
– Squeez

2018-02-26 14:21:46 +00:00
Commented Feb 26, 2018 at 14:21 — Squeez
– Squeez, Commented Feb 26, 2018 at 14:21
ok, see last update

Nahuel Fouilleul
– Nahuel Fouilleul

2018-02-26 14:31:32 +00:00
Commented Feb 26, 2018 at 14:31 — Nahuel Fouilleul
– Nahuel Fouilleul, Commented Feb 26, 2018 at 14:31

Nahuel Fouilleul · Accepted Answer · 2018-02-26 14:47:33Z

1

EDIT: updated after comment regex

(\d+%\s[A-Z]+)\s(?=\d+%\s[A-Z]+)

will match pattern (Number% Word) followed by (Number% Word), without consuming the second as it is a lookahead.

\1,

Java code:

String inp = "A:82% X 18% Y B:100% X C:82% X 18% Y AB:60% X 20% Y 20% ZZ";
String regex = "(\\b[A-Za-z]{1,}\\:\\b)";
String inp2 = inp.replaceAll(regex, "#$1");
String regex2 = "(\\d+%\\s[A-Z]+)\\s(?=\\d+%\\s[A-Z]+)";
System.out.println(inp2.replaceAll(regex2, "$1,"));

Try it online

edited Feb 26, 2018 at 14:47

answered Feb 26, 2018 at 13:02

Nahuel Fouilleul

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5 Comments

Shen Yudong Over a year ago

this answer seems great

Squeez Over a year ago

@Nahuel, But if I use this string "A:82% X 18% Y B:100% X C:82% X 18% Y AB:60% X 20% Y 20% ZZ", I'm not able to get comma before "20% ZZ".

Nahuel Fouilleul Over a year ago

@Squeez, i let the old answer at the end, but i will remove to make more clear

Squeez Over a year ago

@NahuelFouilleul, Its working now...! Can you please explain little more your second regex...?

Nahuel Fouilleul Over a year ago

lookahead (?= .. ) is a zero width assertion which ensures that pattern matches at current position without moving forward position in input.

Lance Toth · Accepted Answer · 2018-02-26 13:02:52Z

0

This should give you the part between X and Y (if there is one) [^X]*% Y

answered Feb 26, 2018 at 13:02

Lance Toth

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Comments

Thomas · Accepted Answer · 2018-02-26 13:07:09Z

0

Assuming you always have input in the form #word:number% label number% label ... you could try the following regex to match the spaces between labels and numbers: (?!<:)\s+(?=\d+%)

Short breakdown:

(?!<:) is a negative lookbehing meaning the match must not be preceded by a colon
\s+ is a sequence of one or more spaces which you want to match
(?=\d+%) is a positive lookahead meaning the match must be followed by number%

In your test input that would match the space after each X.

answered Feb 26, 2018 at 13:07

Thomas

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Comments

kfairns · Accepted Answer · 2018-02-26 13:34:09Z

0

([a-zA-Z]+:)([0-9]{1,3}%) ([a-zA-Z]+)(?: ([0-9]{1,3}%) ([a-zA-Z]+))?

With substitution:

#$1 $2 $3, $4 $5

You would need to have an if statement to make sure capture group 4 has captured something. If it hasn't, then use substitution:

#$1 $2 $3

See the Demo

This regex should be language agnostic (no escape characters used)

answered Feb 26, 2018 at 13:34

kfairns

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