I'm trying to split a string in Java, but keep the newline characters as elements in the array.
For example, with input: "Hello \n\n\nworld!"
I want the output to be: ["Hello", "\n", "\n", "\n", "world", "!"]
The regex I have in place right now is this:
String[] parsed = input.split(" +|(?=\\p{Punct})|(?<=\\p{Punct})");
This gets me the punctuation separation I want, but its output looks like this:["Hello", "\n\n\nworld", "!"]
Is there a way to unclump the newlines in Java?
You could first replace all \n with \n (newline and a space) and then do a simple split on the space character.
String input = "Hello \n\n\nworld!";
String replacement = input.replace("\n", "\n ");
String[] result = replacement.split(" ");
input: "Hello \n\n\nworld!"
replacement: "Hello \n \n \n world!"
result: ["Hello", "\n", "\n", "\n", "world!"]
Note: my example does not handle the final exclamation mark - but it seems you already know how to handle that.
The trick is to add whitespace after each "\n" and then apply your regex.
String line = "Hello \n\n\nworld!";
line = line.replaceAll("\n", "\n "); // here we replace all "\n" to "\n "
String[] items = line.split(" +|(?=\\p{Punct})|(?<=\\p{Punct})");
or shorter version:
String line = "Hello \n\n\nworld!";
String[] items = line.replaceAll("\n", "\n ").split(" +|(?=\\p{Punct})|(?<=\\p{Punct})");
So, in this context the result is: ["Hello", "\n", "\n", "\n", "world", "!"]
Using the find method makes things easier:
String str = "Hello \n\n\nworld!";
List<String> myList = new ArrayList<String>();
Pattern pat = Pattern.compile("\\w+|\\H");
Matcher m = pat.matcher(str);
while (m.find()) {
myList.add(m.group(0));
}
If you use Java 7, change \\H to [\\S\\n].
Note that using this approach, you obtain a pattern easier to write and to edit since you don't need to use lookarounds.
I want to surround all tokens in a text with tags in the following manner:
Input: " abc fg asd "
Output:" <token>abc</token> <token>fg</token> <token>asd</token> "
This is the code I tried so far:
String regex = "(\\s)([a-zA-Z]+)(\\s)";
String text = " abc fg asd ";
text = text.replaceAll(regex, "$1<token>$2</token>$3");
System.out.println(text);
Output:" <token>abc</token> fg <token>asd</token> "
Note: for simplicity we can assume that the input starts and ends with whitespaces
Use lookaround:
String regex = "(?<=\\s)([a-zA-Z]+)(?=\\s)";
...
text = text.replaceAll(regex, "<token>$1</token>");
If your tokens are only defined with a character class you don't need to describe what characters are around. So this should suffice since the regex engine walks from left to right and since the quantifier is greedy:
String regex = "[a-zA-Z]+";
text = text.replaceAll(regex, "<token>$0</token>");
// meaning not a space, 1+ times
String result = input.replaceAll("([^\\s]+)", "<token>$1</token>");
this matches everything that isn't a space. Prolly the best fit for what you need. Also it's greedy meaning it will never leave out a character that it shouldn't ( it will never find the string "as" in the string "asd" when there is another character with which it matches)
I want to check wheter a string is containing two words/string directly followed in a specific order.
The punctuation should also be included in the word/string. (i.e. "word" and "word." should be handeled as different words).
As an example:
String word1 = "is";
String word1 = "a";
String text = "This is a sample";
Pattern p = Pattern.compile(someregex+"+word1+"someregex"+word2+"someregex");
System.out.println(p.matcher(text).matches());
This should print out true.
With the following variables, it should also print true.
String word1 = "sample.";
String word1 = "0END";
String text = "This is a sample. 0END0";
But the latter should return false when setting word1 = "sample" (without punctuation).
Does anyone have an idea how the regex string should look like (i.e. what i should write instead of "someregex" ?)
Thank you!
Looks like you're just splitting on whitespace, try:
Pattern p = Pattern.compile("(\\s|^)" + Pattern.quote(word1) + "\\s+" + Pattern.quote(word2) + "(\\s|$)");
Explaination
(\\s|^) matches any whitespace before the first word, or the start of the string
\\s+ matches the whitespace between the words
(\\s|$) matches any whitespace after the second word, or the end of the string
Pattern.quote(...) ensures that any regex special characters in your input strings are properly escapes.
You also need to call find(), not match(). match() will only return true if the whole string matches the pattern.
Complete example
String word1 = "is";
String word2 = "a";
String text = "This is a sample";
String regex =
"(\\s|^)" +
Pattern.quote(word1) +
"\\s+" +
Pattern.quote(word2) +
"(\\s|$)";
Pattern p = Pattern.compile(regex);
System.out.println(p.matcher(text).find());
You can concatenate the two words with a whitespace and use that as the regexp.
the only thing, you have to do, is to replace "." with "." so the point does not match as any character.
String regexp = " " + word1 + " " + word2 + " ";
regexp = regexp.replaceAll("\\.", "\\\\.");
Looking for quick, simple way in Java to change this string
" hello there "
to something that looks like this
"hello there"
where I replace all those multiple spaces with a single space, except I also want the one or more spaces at the beginning of string to be gone.
Something like this gets me partly there
String mytext = " hello there ";
mytext = mytext.replaceAll("( )+", " ");
but not quite.
Try this:
String after = before.trim().replaceAll(" +", " ");
See also
String.trim()
Returns a copy of the string, with leading and trailing whitespace omitted.
regular-expressions.info/Repetition
No trim() regex
It's also possible to do this with just one replaceAll, but this is much less readable than the trim() solution. Nonetheless, it's provided here just to show what regex can do:
String[] tests = {
" x ", // [x]
" 1 2 3 ", // [1 2 3]
"", // []
" ", // []
};
for (String test : tests) {
System.out.format("[%s]%n",
test.replaceAll("^ +| +$|( )+", "$1")
);
}
There are 3 alternates:
^_+ : any sequence of spaces at the beginning of the string
Match and replace with $1, which captures the empty string
_+$ : any sequence of spaces at the end of the string
Match and replace with $1, which captures the empty string
(_)+ : any sequence of spaces that matches none of the above, meaning it's in the middle
Match and replace with $1, which captures a single space
See also
regular-expressions.info/Anchors
You just need a:
replaceAll("\\s{2,}", " ").trim();
where you match one or more spaces and replace them with a single space and then trim whitespaces at the beginning and end (you could actually invert by first trimming and then matching to make the regex quicker as someone pointed out).
To test this out quickly try:
System.out.println(new String(" hello there ").trim().replaceAll("\\s{2,}", " "));
and it will return:
"hello there"
Use the Apache commons StringUtils.normalizeSpace(String str) method. See docs here
This worked perfectly for me : sValue = sValue.trim().replaceAll("\\s+", " ");
trim() method removes the leading and trailing spaces and using replaceAll("regex", "string to replace") method with regex "\s+" matches more than one space and will replace it with a single space
myText = myText.trim().replaceAll("\\s+"," ");
The following code will compact any whitespace between words and remove any at the string's beginning and end
String input = "\n\n\n a string with many spaces, \n"+
" a \t tab and a newline\n\n";
String output = input.trim().replaceAll("\\s+", " ");
System.out.println(output);
This will output a string with many spaces, a tab and a newline
Note that any non-printable characters including spaces, tabs and newlines will be compacted or removed
For more information see the respective documentation:
String#trim() method
String#replaceAll(String regex, String replacement) method
For information about Java's regular expression implementation see the documentation of the Pattern class
"[ ]{2,}"
This will match more than one space.
String mytext = " hello there ";
//without trim -> " hello there"
//with trim -> "hello there"
mytext = mytext.trim().replaceAll("[ ]{2,}", " ");
System.out.println(mytext);
OUTPUT:
hello there
To eliminate spaces at the beginning and at the end of the String, use String#trim() method. And then use your mytext.replaceAll("( )+", " ").
You can first use String.trim(), and then apply the regex replace command on the result.
Try this one.
Sample Code
String str = " hello there ";
System.out.println(str.replaceAll("( +)"," ").trim());
OUTPUT
hello there
First it will replace all the spaces with single space. Than we have to supposed to do trim String because Starting of the String and End of the String it will replace the all space with single space if String has spaces at Starting of the String and End of the String So we need to trim them. Than you get your desired String.
String blogName = "how to do in java . com";
String nameWithProperSpacing = blogName.replaceAll("\\\s+", " ");
trim()
Removes only the leading & trailing spaces.
From Java Doc,
"Returns a string whose value is this string, with any leading and trailing whitespace removed."
System.out.println(" D ev Dum my ".trim());
"D ev Dum my"
replace(), replaceAll()
Replaces all the empty strings in the word,
System.out.println(" D ev Dum my ".replace(" ",""));
System.out.println(" D ev Dum my ".replaceAll(" ",""));
System.out.println(" D ev Dum my ".replaceAll("\\s+",""));
Output:
"DevDummy"
"DevDummy"
"DevDummy"
Note: "\s+" is the regular expression similar to the empty space character.
Reference : https://www.codedjava.com/2018/06/replace-all-spaces-in-string-trim.html
In Kotlin it would look like this
val input = "\n\n\n a string with many spaces, \n"
val cleanedInput = input.trim().replace(Regex("(\\s)+"), " ")
A lot of correct answers been provided so far and I see lot of upvotes. However, the mentioned ways will work but not really optimized or not really readable.
I recently came across the solution which every developer will like.
String nameWithProperSpacing = StringUtils.normalizeSpace( stringWithLotOfSpaces );
You are done.
This is readable solution.
You could use lookarounds also.
test.replaceAll("^ +| +$|(?<= ) ", "");
OR
test.replaceAll("^ +| +$| (?= )", "")
<space>(?= ) matches a space character which is followed by another space character. So in consecutive spaces, it would match all the spaces except the last because it isn't followed by a space character. This leaving you a single space for consecutive spaces after the removal operation.
Example:
String[] tests = {
" x ", // [x]
" 1 2 3 ", // [1 2 3]
"", // []
" ", // []
};
for (String test : tests) {
System.out.format("[%s]%n",
test.replaceAll("^ +| +$| (?= )", "")
);
}
See String.replaceAll.
Use the regex "\s" and replace with " ".
Then use String.trim.
String str = " hello world"
reduce spaces first
str = str.trim().replaceAll(" +", " ");
capitalize the first letter and lowercase everything else
str = str.substring(0,1).toUpperCase() +str.substring(1,str.length()).toLowerCase();
you should do it like this
String mytext = " hello there ";
mytext = mytext.replaceAll("( +)", " ");
put + inside round brackets.
String str = " this is string ";
str = str.replaceAll("\\s+", " ").trim();
This worked for me
scan= filter(scan, " [\\s]+", " ");
scan= sac.trim();
where filter is following function and scan is the input string:
public String filter(String scan, String regex, String replace) {
StringBuffer sb = new StringBuffer();
Pattern pt = Pattern.compile(regex);
Matcher m = pt.matcher(scan);
while (m.find()) {
m.appendReplacement(sb, replace);
}
m.appendTail(sb);
return sb.toString();
}
The simplest method for removing white space anywhere in the string.
public String removeWhiteSpaces(String returnString){
returnString = returnString.trim().replaceAll("^ +| +$|( )+", " ");
return returnString;
}
check this...
public static void main(String[] args) {
String s = "A B C D E F G\tH I\rJ\nK\tL";
System.out.println("Current : "+s);
System.out.println("Single Space : "+singleSpace(s));
System.out.println("Space count : "+spaceCount(s));
System.out.format("Replace all = %s", s.replaceAll("\\s+", ""));
// Example where it uses the most.
String s = "My name is yashwanth . M";
String s2 = "My nameis yashwanth.M";
System.out.println("Normal : "+s.equals(s2));
System.out.println("Replace : "+s.replaceAll("\\s+", "").equals(s2.replaceAll("\\s+", "")));
}
If String contains only single-space then replace() will not-replace,
If spaces are more than one, Then replace() action performs and removes spacess.
public static String singleSpace(String str){
return str.replaceAll(" +| +|\t|\r|\n","");
}
To count the number of spaces in a String.
public static String spaceCount(String str){
int i = 0;
while(str.indexOf(" ") > -1){
//str = str.replaceFirst(" ", ""+(i++));
str = str.replaceFirst(Pattern.quote(" "), ""+(i++));
}
return str;
}
Pattern.quote("?") returns literal pattern String.
My method before I found the second answer using regex as a better solution. Maybe someone needs this code.
private String replaceMultipleSpacesFromString(String s){
if(s.length() == 0 ) return "";
int timesSpace = 0;
String res = "";
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if(c == ' '){
timesSpace++;
if(timesSpace < 2)
res += c;
}else{
res += c;
timesSpace = 0;
}
}
return res.trim();
}
Stream version, filters spaces and tabs.
Stream.of(str.split("[ \\t]")).filter(s -> s.length() > 0).collect(Collectors.joining(" "))
I know replaceAll method is much easier but I wanted to post this as well.
public static String removeExtraSpace(String input) {
input= input.trim();
ArrayList <String> x= new ArrayList<>(Arrays.asList(input.split("")));
for(int i=0; i<x.size()-1;i++) {
if(x.get(i).equals(" ") && x.get(i+1).equals(" ")) {
x.remove(i);
i--;
}
}
String word="";
for(String each: x)
word+=each;
return word;
}
String myText = " Hello World ";
myText = myText.trim().replace(/ +(?= )/g,'');
// Output: "Hello World"
string.replaceAll("\s+", " ");
If you already use Guava (v. 19+) in your project you may want to use this:
CharMatcher.whitespace().trimAndCollapseFrom(input, ' ');
or, if you need to remove exactly SPACE symbol ( or U+0020, see more whitespaces) use:
CharMatcher.anyOf(" ").trimAndCollapseFrom(input, ' ');
public class RemoveExtraSpacesEfficient {
public static void main(String[] args) {
String s = "my name is mr space ";
char[] charArray = s.toCharArray();
char prev = s.charAt(0);
for (int i = 0; i < charArray.length; i++) {
char cur = charArray[i];
if (cur == ' ' && prev == ' ') {
} else {
System.out.print(cur);
}
prev = cur;
}
}
}
The above solution is the algorithm with the complexity of O(n) without using any java function.
Please use below code
package com.myjava.string;
import java.util.StringTokenizer;
public class MyStrRemoveMultSpaces {
public static void main(String a[]){
String str = "String With Multiple Spaces";
StringTokenizer st = new StringTokenizer(str, " ");
StringBuffer sb = new StringBuffer();
while(st.hasMoreElements()){
sb.append(st.nextElement()).append(" ");
}
System.out.println(sb.toString().trim());
}
}