I have this homework problem where I need to use regex to remove every other character in a string.
In one part, I have to delete characters at index 1,3,5,... I have done this as follows:
String s = "1a2b3c4d5";
System.out.println(s.replaceAll("(.).", "$1"));
This prints 12345 which is what I want. Essentially I match two characters at a time, and replacing with the first character. I used group capturing to do this.
The problem is, I'm having trouble with the second part of the homework, where I need to delete characters at index 0,2,4,...
I have done the following:
String s = "1a2b3c4d5";
System.out.println(s.replaceAll(".(.)", "$1"));
This prints abcd5, but the correct answer must be abcd. My regex is only incorrect if the input string length is odd. If it's even, then my regex works fine.
I think I'm really close to the answer, but I'm not sure how to fix it.
You are indeed very close to the answer: just make matching the second char optional.
String s = "1a2b3c4d5";
System.out.println(s.replaceAll(".(.)?", "$1"));
// prints "abcd"
This works because:
Regex is greedy by default, it will take the second character if it's there
When the input is of odd length, the second char won't be there at the last replacement, but you'd still match one char (i.e. last char in input)
You can still use backreferences in substitution even if the group fails to match
It will substitute in the empty string, not "null"
This is different from Matcher.group(int), which returns null for failed groups
References
regular-expressions.info/Optional
A closer look at the first part
Let's take a closer look at the first part of the homework:
String s = "1a2b3c4d5";
System.out.println(s.replaceAll("(.).", "$1"));
// prints "12345"
Here you didn't have to use ? for the second char, but it "works" because even though you didn't match the last char, you didn't have to! The last char can remain unmatched, unreplaced, due to the problem specification.
Now suppose that we want to delete chars at index 1,3,5..., and put the chars at index 0,2,4... in brackets.
String s = "1a2b3c4d5";
System.out.println(s.replaceAll("(.).", "($1)"));
// prints "(1)(2)(3)(4)5"
A-ha!! Now you're experiencing the exact same problem with odd-length input! You couldn't match the last char with your regex, because your regex needs two chars, but there's only one char at the end for odd-length input!
The solution, again, is to make matching the second char optional:
String s = "1a2b3c4d5";
System.out.println(s.replaceAll("(.).?", "($1)"));
// prints "(1)(2)(3)(4)(5)"
my regex is only incorrect if the input string length is odd. if it's even, then my regex works fine.
Change your expresion to .(.)? - the question mark makes the second character optional, which means it doesn't matter if input is odd or even
Your regex needs 2 chars to match, so fails on the final char.
This regex:
".(.{0,1})"
Will make the second char optional, so it will match with your final '5' as well
Related
I have problem with character swap in string.
e.g. I have string "sdgk4e5s3gj6ds3h6fggh" and I need code that can swap numbers with character.
The result should look something like this: "sdgke4s5g3jd6sh3f6ggh"
I have got to the point where I make char array out of String, but I don't know what to do next. Any help?
If I understand correctly what you are asking, a simple regex could solve your problem:
String result = "sdgk4e5s3gj6ds3h6fggh".replaceAll("(\\d)(\\D)", "$2$1")
which basically inverts 2 characters every time it finds one digit followed by one non-digit.
I have a string of format: A-2-Q4567
More examples: AB-456-T12, A24-5-M12345, etc.
I want to extract the last numerical values out of these strings, which are: 4567, 12, 12345 respectively (which is the numerical value of the substring from the end till first non-numeric character is encountered)
I can split the string, get the last string from the splitted string array, and then do a parseInt after removing the non-numerical characters from it.
But is there a more elegant way of doing this?
You can use this regex: (\d+$). It returns the last sequence of digits in the string.
EDIT - some explanation:
The \d means any digit.
The + means one or more of the previous symbols. Since the previous symbol is a digit, then \d+ means "one or more digits".
The $ means the end of the string, so \d+$ is the last sequence of digits in the string.
you can do this :
String getLastNumeric(String input)
{
String str="";
char c;
for(int i=input.length()-1;i>=0 && Character.isDigit(c=input.charAt(i));i--)
str=c+str;
return str;
}
The regex solutions might be more elegant but performance-wise I think the above is the best because Regex match can be more expensive than a simple for loop with a simple condition to evaluate.
Ofcourse The Regex is more flexible, what if your requirements change and now a dash "-" must precede the numbers ? with Regex it should be just a matter of changing one regex expression.
I put the Regex version here but remember if you're sure your requirements won't change I think the above solution is better on the CPU :
Matcher matcher= Pattern.compile("(\\d+$)").matcher(input);
if(matcher.find())
return matcher.group();
return "";
for a test I created following regex by mistake:
|(\\w+)|
I was puzzled that this regex really works and I can't explain the result:
public static void main(String[] args) {
String toReplace="Hey I'm a lovely String an I'm giving my |value| worth!";
// String replacement1="2 cent"; // I planned to replace |value| with 2 cent
String replacement1="#"; // to produce a better Output
String regex="|(\\w+)|"; // I forgot to escape the |
replacement1="#";
result=toReplace.replaceAll(regex,replacement1);
System.out.println(result);
}
the result is:
#H#e#y# #I#'#m# #a# #l#o#v#e#l#y# #S#t#r#i#n#g# #a#n# #I#'#m# #g#i#v#i#n#g# #m#y# #|#v#a#l#u#e#|# #w#o#r#t#h#!#
My ideas so far are that java tries to replace "nothing" between the characters but why not the characters itself?
\\w+ should match the 'H'
I would expect that every char is replaced by 3 # signs or only by one but that the characters are not replaced puzzles me.
You're right, this regex matches the empty string between each character.
Since the first alternative (the empty string left of |) matches, the rest of the pattern isn't even tried, so the \w+ isn't even reached by the matching engine. You could have written any (valid) pattern to the right of that first |, it wouldn't ever be reached.
The engine works the following way: It has a current position cursor in the subject string. It tries to match starting at that current position. Since your regex is a match, it will perform the replacement at this point, and then move the current position cursor after the found match.
But since the match is zero-width, it simply advances to the next character, because not doing so would result in an infinite loop.
I have string with spaces and some non-informative characters and substrings required to be excluded and just to keep some important sections. I used the split as below:
String myString[]={"01: Hi you look tired today? Can I help you?"};
myString=myString[0].split("[\\s+]");// Split based on any white spaces
for(int ii=0;ii<myString.length;ii++)
System.out.println(myString[ii]);
The result is :
01:
Hi
you
look
tired
today?
Can
I
help
you?
The spaces appeared after the split as sub strings when the regex is “[\s+]” but disappeared when the regex is "\s+". I am confused and not able to find answer in the related stack overflow pages. The link regex-Pattern made me more confused.
Please help, I am new with java.
19/1/2015:Edit
After your valuable advice, I reached to point in my program where a conditional statements is required to be decomposed and processed. The case I have is:
String s1="01:IF rd.h && dq.L && o.LL && v.L THEN la.VHB , av.VHR with 0.4610;";
String [] s2=s1.split(("[\\s\\&\\,]+"));
for(int ii=0;ii<s2.length;ii++)System.out.println(s2[ii]);
The result is fine till now as:
01:IF
rd.h
dq.L
o.LL
v.L
THEN
la.VHB
av.VHR
with
0.4610;
My next step is to add string "with" to the regex and get rid of this word while doing the split.
I tried it this way:
String s1="01:IF rd.h && dq.L && o.LL && v.L THEN la.VHB , av.VHR with 0.4610;";
String [] s2=s1.split(("[\\s\\&\\, with]+"));
for(int ii=0;ii<s2.length;ii++)System.out.println(s2[ii]);
The result not perfect, because I got unwonted extra split at every "h" letter as:
01:IF
rd.
dq.L
o.LL
v.L
THEN
la.VHB
av.VHR
0.4610;
Any advice on how to specify string with mixed white spaces and separation marks?
Many thanks.
inside square brackets, [\s+] will represent the whitespace character class with the plus sign added. it is only one character so a sequence of spaces will split many empty strings as Todd noted, and will also use + as separator.
you should use \s+ (without brackets) as the separator. that means one or more whitespace characters.
myString=myString[0].split("\\s+");
Your biggest problem is not understanding enough about regular expressions to write them properly. One key point you don't comprehend is that [...] is a character class, which is a list of characters any one of which can match. For example:
[abc] matches either a, b or c (it does not match "abc")
[\\s+] matches any whitespace or "+" character
[with] matches a single character that is either w, i, t or h
[.$&^?] matches those literal characters - most characters lose their special regex meaning when in a character class
To split on any number of whitespace, comma and ampersand and consume "with" (if it appears), do this:
String [] s2 = s1.split("[\\s,&]+(with[\\s,&]+)?");
You can try it easily here Online Regex and get useful comments.
I want to split my string on every occurrence of an alpha-beta character.
for example:
"s1l1e13" to an array of: ["s1","l1","e13"]
when trying to use this simple split by regex i get some weird results:
testStr = "s1l1e13"
Arrays.toString(testStr.split("(?=[a-z])"))
gives me the array of:
["","s1","l1","e13"]
how can i create the split without the empty array element?
I tried a couple more things:
testStr = "s1"
Arrays.toString(testStr.split("(?=[a-z])"))
does return the currect array: ["s1"]
but when trying to use substring
testStr = "s1l1e13"
Arrays.toString(testStr.substring(1).split("(?=[a-z])")
i get in return ["1","l1","e13"]
what am i missing?
Your Lookahead marks each position before any character of a to z; marking the following positions:
s1 l1 e13
^ ^ ^
So by spliting using just the Lookahead, it returns ["", "s1", "l1", "e13"]
You can use a Negative Lookbehind here. This looks behind to see if there is not the beginning of the string.
String s = "s1l1e13";
String[] parts = s.split("(?<!\\A)(?=[a-z])");
System.out.println(Arrays.toString(parts)); //=> [s1, l1, e13]
Your problem is that (?=[a-z]) means "place before [a-z]" and in your text
s1l1e13
you have 3 such places. I will mark them with |
|s1|l1|e13
so split (unfortunately correctly) produces "" "s1" "l1" "e13" and doesn't automatically remove for you first empty elements.
To solve this problem you have at least two options:
make sure that there is something before your place you need to split on (it is not at start of your string). You can use for instance (?<=\\d)(?=[a-z]) if you want to split after digit but before character
(PREFFERED SOLUTION) start using Java 8 which automatically removes empty strings at start of result array if regex used on split is zero-length (look-arounds are zero length).
The first match finds "" to be okay because its looking ahead for any alpha character, which is called zero-width lookahead, so it doesn't need to actually match anything. So "s" at the beginning is alphanumeric, and it matches that at a probable spot.
If you want the regex to match something always, use ".+(?=[a-z])"
The problem is that the initial "s" counts as an alphabetic character. So, the regex is trying to split at s.
The issue is that there is nothing before the s, so the regex machine instead decides to show that there is nothing by adding the null element. It'll do the same thing at the end if you ended with "s" (or any other letter).
If this is the only string you're splitting, or if every array you had starts with a letter but does not end with one, just truncate the array to omit the first element. Otherwise, you'll probably need to loop through each array as you make it so that you can drop empty elements.
So it seems your matches has the pattern x###, where x is a letter, and # is a number.
I'd make the following Regex:
([a-z][0-9]+)