Related
I am trying to get the first letter of every word in a String:
String recInf = recursos.getString(nombre);
char[] tipoAbreviado = recInf.toCharArray();
tipoAbreviado[0] = Character.toUpperCase(tipoAbreviado[0]);
for (int i = 0; i < recInf.length() - 2; i++) {
// Es 'palabra'
if (tipoAbreviado[i] == ' ' || tipoAbreviado[i] == '.' || tipoAbreviado[i] == ',') {
// Reemplazamos
tipoAbreviado[i + 1] = Character.toUpperCase(tipoAbreviado[i + 1]);
}
nombre = tipoAbreviado.toString();
}
Finally the value of nombre is [C#3b1938ea, not the first letter of every word in recInf
You can use Arrays.toSting(your_array) to print your Array.
Take a look what toString() in Arrays do
public static String toString(long[] a) {
if (a == null)
return "null";
int iMax = a.length - 1;
if (iMax == -1)
return "[]";
StringBuilder b = new StringBuilder();
b.append('[');
for (int i = 0; ; i++) {
b.append(a[i]);
if (i == iMax)
return b.append(']').toString();
b.append(", ");
}
}
But when you use tipoAbreviado.toString(); it will call toString() method in Object class.
What toString() method in Object class do?
public String toString() {
return getClass().getName() + "#" + Integer.toHexString(hashCode());
}
That's why you are getting your current out put.
Instead of using toString on an Array which prints the memory address representation, you should create a String from char[] using new String(char[])
nombre = new String(tipoAbreviado);
You can use string's .split() with the correct regex and then just pick the first char:
String[] words = "Your String & !+ and some extraodrinary others".split("[^a-zA-Z]+");
for (String word: words ){
System.out.println(word.charAt(0));
}
You can try to fetch them with a Regular Expression.
When I use the RegEx (\w)\w+ on the string Hallo Welt, ich bin ein String, I get an array with H, W, i, b, e, S. Note: You have to run the expression as global expression (more than once).
Assuming the String object you want to process is recInf i would recommend to do it as follows:
String recInf = new String(text.replaceAll("[^\\p{L}\\p{Nd} ]+"," ")
.replaceAll("[\\s]+", " "));
String[] words = recInf.split(" ");
String[] firstLetters = new String[words.length];
for(int i=0; i<words.length;i++){
firstLetters[i]=words[i].getCharAt(0);
}
You could try this
nombre = (new StringBuilder(tipoAbreviado)).toString();
I'm trying to return strings in different lines given these conditions. Since I cannot use the += in Java with strings, how do I make one giant string that is spaced per line but "stacks?" In other words, how do I add a new string within a loop to an old string?
/**
Returns a String that concatenates all "offending"
words from text that contain letter; the words are
separated by '\n' characters; the returned string
does not contain duplicate words: each word occurs
only once; there are no punctuation or whitespace
characters in the returned string.
#param letter character to find in text
#return String containing all words with letter
*/
public String allWordsWith(char letter)
{
String result = "";
int i = 0;
while (i < text.length())
{
char newchar = text.charAt(i);
if (newchar == letter)
{
int index1 = text.lastIndexOf("",i);
int index2 = text.indexOf("",i);
String newstring = '\n' + text.substring(index2,index1);
}
i++;
}
return result;
}
Modify the result string, and fix your "word boundary" tests.
if (newchar == letter) {
int index1 = text.lastIndexOf(' ',i);
int index2 = text.indexOf(' ',i);
// TODO -- handle when index1 or index2 is < 0; that means it wasn't found,
// and you should use the string boundary (0 or length()) instead.
String word = text.substring( index2,index1);
result += "\n" + word;
}
If you were really concerned about performance you could use a StringBuilder and append(), but otherwise I strongly favour += for being concise & readable.
you are re-initializing your string in loop every time. Move the string declaration outsid eof loop:
Replace this
String newstring = '\n' + text.substring(index2,index1);
with
result = '\n' + text.substring(index2,index1);
First, use a StringBuilder.
Second, use System.getProperty("line.separator") to ensure proper line breaks are used.
Edited code:
public String allWordsWith(char letter)
{
StringBuilder sb = new StringBuilder();
int i = 0;
while (i < text.length())
{
char newchar = text.charAt(i);
if (newchar == letter)
{
int index1 = text.lastIndexOf("",i);
int index2 = text.indexOf("",i);
sb.Append(text.substring(index2,index1));
sb.Append(System.getProperty("line.separator"));
//I put the new line after the word so you don't get an empty
//line on top, but you can do what you need/want to do in this case.
}
i++;
}
return result;
}
Use StringBuilder as following:
public String allWordsWith(char letter){
//String result = "";
StringBuilder result = new StringBuilder();
int i = 0;
while (i < text.length()){
char newchar = text.charAt(i);
if (newchar == letter){
int index1 = text.lastIndexOf("",i);
int index2 = text.indexOf("",i);
result.append('\n' + text.substring(index2,index1));
}
i++;
}
return result.toString();
}
String text = "I have android code with many different java, bmp and xml files everywhere in my project that I used during the drafting phase of my project.";
String letter = "a";
Set<String> duplicateWordsFilter = new HashSet<String>(Arrays.asList(text.split(" ")));
StringBuilder sb = new StringBuilder(text.length());
for (String word : duplicateWordsFilter) {
if (word.contains(letter)) {
sb.append(word);
sb.append("\n");
}
}
return sb.toString();
result is:
android
have
java,
drafting
and
many
that
phase
For accessing individual characters of a String in Java, we have String.charAt(2). Is there any inbuilt function to remove an individual character of a String in java?
Something like this:
if(String.charAt(1) == String.charAt(2){
//I want to remove the individual character at index 2.
}
You can also use the StringBuilder class which is mutable.
StringBuilder sb = new StringBuilder(inputString);
It has the method deleteCharAt(), along with many other mutator methods.
Just delete the characters that you need to delete and then get the result as follows:
String resultString = sb.toString();
This avoids creation of unnecessary string objects.
You can use Java String method called replace, which will replace all characters matching the first parameter with the second parameter:
String a = "Cool";
a = a.replace("o","");
One possibility:
String result = str.substring(0, index) + str.substring(index+1);
Note that the result is a new String (as well as two intermediate String objects), because Strings in Java are immutable.
No, because Strings in Java are immutable. You'll have to create a new string removing the character you don't want.
For replacing a single char c at index position idx in string str, do something like this, and remember that a new string will be created:
String newstr = str.substring(0, idx) + str.substring(idx + 1);
String str = "M1y java8 Progr5am";
deleteCharAt()
StringBuilder build = new StringBuilder(str);
System.out.println("Pre Builder : " + build);
build.deleteCharAt(1); // Shift the positions front.
build.deleteCharAt(8-1);
build.deleteCharAt(15-2);
System.out.println("Post Builder : " + build);
replace()
StringBuffer buffer = new StringBuffer(str);
buffer.replace(1, 2, ""); // Shift the positions front.
buffer.replace(7, 8, "");
buffer.replace(13, 14, "");
System.out.println("Buffer : "+buffer);
char[]
char[] c = str.toCharArray();
String new_Str = "";
for (int i = 0; i < c.length; i++) {
if (!(i == 1 || i == 8 || i == 15))
new_Str += c[i];
}
System.out.println("Char Array : "+new_Str);
To modify Strings, read about StringBuilder because it is mutable except for immutable String. Different operations can be found here https://docs.oracle.com/javase/tutorial/java/data/buffers.html. The code snippet below creates a StringBuilder and then append the given String and then delete the first character from the String and then convert it back from StringBuilder to a String.
StringBuilder sb = new StringBuilder();
sb.append(str);
sb.deleteCharAt(0);
str = sb.toString();
Consider the following code:
public String removeChar(String str, Integer n) {
String front = str.substring(0, n);
String back = str.substring(n+1, str.length());
return front + back;
}
You may also use the (huge) regexp machine.
inputString = inputString.replaceFirst("(?s)(.{2}).(.*)", "$1$2");
"(?s)" - tells regexp to handle newlines like normal characters (just in case).
"(.{2})" - group $1 collecting exactly 2 characters
"." - any character at index 2 (to be squeezed out).
"(.*)" - group $2 which collects the rest of the inputString.
"$1$2" - putting group $1 and group $2 together.
If you want to remove a char from a String str at a specific int index:
public static String removeCharAt(String str, int index) {
// The part of the String before the index:
String str1 = str.substring(0,index);
// The part of the String after the index:
String str2 = str.substring(index+1,str.length());
// These two parts together gives the String without the specified index
return str1+str2;
}
By the using replace method we can change single character of string.
string= string.replace("*", "");
Use replaceFirst function of String class. There are so many variants of replace function that you can use.
If you need some logical control over character removal, use this
String string = "sdsdsd";
char[] arr = string.toCharArray();
// Run loop or whatever you need
String ss = new String(arr);
If you don't need any such control, you can use what Oscar orBhesh mentioned. They are spot on.
Easiest way to remove a char from string
String str="welcome";
str=str.replaceFirst(String.valueOf(str.charAt(2)),"");//'l' will replace with ""
System.out.println(str);//output: wecome
public class RemoveCharFromString {
public static void main(String[] args) {
String output = remove("Hello", 'l');
System.out.println(output);
}
private static String remove(String input, char c) {
if (input == null || input.length() <= 1)
return input;
char[] inputArray = input.toCharArray();
char[] outputArray = new char[inputArray.length];
int outputArrayIndex = 0;
for (int i = 0; i < inputArray.length; i++) {
char p = inputArray[i];
if (p != c) {
outputArray[outputArrayIndex] = p;
outputArrayIndex++;
}
}
return new String(outputArray, 0, outputArrayIndex);
}
}
In most use-cases using StringBuilder or substring is a good approach (as already answered). However, for performance critical code, this might be a good alternative.
/**
* Delete a single character from index position 'start' from the 'target' String.
*
* ````
* deleteAt("ABC", 0) -> "BC"
* deleteAt("ABC", 1) -> "B"
* deleteAt("ABC", 2) -> "C"
* ````
*/
public static String deleteAt(final String target, final int start) {
return deleteAt(target, start, start + 1);
}
/**
* Delete the characters from index position 'start' to 'end' from the 'target' String.
*
* ````
* deleteAt("ABC", 0, 1) -> "BC"
* deleteAt("ABC", 0, 2) -> "C"
* deleteAt("ABC", 1, 3) -> "A"
* ````
*/
public static String deleteAt(final String target, final int start, int end) {
final int targetLen = target.length();
if (start < 0) {
throw new IllegalArgumentException("start=" + start);
}
if (end > targetLen || end < start) {
throw new IllegalArgumentException("end=" + end);
}
if (start == 0) {
return end == targetLen ? "" : target.substring(end);
} else if (end == targetLen) {
return target.substring(0, start);
}
final char[] buffer = new char[targetLen - end + start];
target.getChars(0, start, buffer, 0);
target.getChars(end, targetLen, buffer, start);
return new String(buffer);
}
*You can delete string value use the StringBuilder and deletecharAt.
String s1 = "aabc";
StringBuilder sb = new StringBuilder(s1);
for(int i=0;i<sb.length();i++)
{
char temp = sb.charAt(0);
if(sb.indexOf(temp+"")!=1)
{
sb.deleteCharAt(sb.indexOf(temp+""));
}
}
To Remove a Single character from The Given String please find my method hope it will be usefull. i have used str.replaceAll to remove the string but their are many ways to remove a character from a given string but i prefer replaceall method.
Code For Remove Char:
import java.util.ArrayList;
import java.util.Collection;
import java.util.Collections;
public class Removecharacter
{
public static void main(String[] args)
{
String result = removeChar("Java", 'a');
String result1 = removeChar("Edition", 'i');
System.out.println(result + " " + result1);
}
public static String removeChar(String str, char c) {
if (str == null)
{
return null;
}
else
{
return str.replaceAll(Character.toString(c), "");
}
}
}
Console image :
please find The Attached image of console,
Thanks For Asking. :)
public static String removechar(String fromString, Character character) {
int indexOf = fromString.indexOf(character);
if(indexOf==-1)
return fromString;
String front = fromString.substring(0, indexOf);
String back = fromString.substring(indexOf+1, fromString.length());
return front+back;
}
BufferedReader input=new BufferedReader(new InputStreamReader(System.in));
String line1=input.readLine();
String line2=input.readLine();
char[] a=line2.toCharArray();
char[] b=line1.toCharArray();
loop: for(int t=0;t<a.length;t++) {
char a1=a[t];
for(int t1=0;t1<b.length;t1++) {
char b1=b[t1];
if(a1==b1) {
StringBuilder sb = new StringBuilder(line1);
sb.deleteCharAt(t1);
line1=sb.toString();
b=line1.toCharArray();
list.add(a1);
continue loop;
}
}
When I have these kinds of questions I always ask: "what would the Java Gurus do?" :)
And I'd answer that, in this case, by looking at the implementation of String.trim().
Here's an extrapolation of that implementation that allows for more trim characters to be used.
However, note that original trim actually removes all chars that are <= ' ', so you may have to combine this with the original to get the desired result.
String trim(String string, String toTrim) {
// input checks removed
if (toTrim.length() == 0)
return string;
final char[] trimChars = toTrim.toCharArray();
Arrays.sort(trimChars);
int start = 0;
int end = string.length();
while (start < end &&
Arrays.binarySearch(trimChars, string.charAt(start)) >= 0)
start++;
while (start < end &&
Arrays.binarySearch(trimChars, string.charAt(end - 1)) >= 0)
end--;
return string.substring(start, end);
}
public String missingChar(String str, int n) {
String front = str.substring(0, n);
// Start this substring at n+1 to omit the char.
// Can also be shortened to just str.substring(n+1)
// which goes through the end of the string.
String back = str.substring(n+1, str.length());
return front + back;
}
I just implemented this utility class that removes a char or a group of chars from a String. I think it's fast because doesn't use Regexp. I hope that it helps someone!
package your.package.name;
/**
* Utility class that removes chars from a String.
*
*/
public class RemoveChars {
public static String remove(String string, String remove) {
return new String(remove(string.toCharArray(), remove.toCharArray()));
}
public static char[] remove(final char[] chars, char[] remove) {
int count = 0;
char[] buffer = new char[chars.length];
for (int i = 0; i < chars.length; i++) {
boolean include = true;
for (int j = 0; j < remove.length; j++) {
if ((chars[i] == remove[j])) {
include = false;
break;
}
}
if (include) {
buffer[count++] = chars[i];
}
}
char[] output = new char[count];
System.arraycopy(buffer, 0, output, 0, count);
return output;
}
/**
* For tests!
*/
public static void main(String[] args) {
String string = "THE QUICK BROWN FOX JUMPS OVER THE LAZY DOG";
String remove = "AEIOU";
System.out.println();
System.out.println("Remove AEIOU: " + string);
System.out.println("Result: " + RemoveChars.remove(string, remove));
}
}
This is the output:
Remove AEIOU: THE QUICK BROWN FOX JUMPS OVER THE LAZY DOG
Result: TH QCK BRWN FX JMPS VR TH LZY DG
For example if you want to calculate how many a's are there in the String, you can do it like this:
if (string.contains("a"))
{
numberOf_a++;
string = string.replaceFirst("a", "");
}
How would I replace a string 10100 with 10010 using the algorithm "replace the last substring 10 with 01."
I tried
s=s.replace(s.substring(a,a+2), "01");
but this returns 01010, replacing both the first and the second substring of "10".
"a" represents s.lastindexOf("10");
Here's a simple and extensible function you can use. First its use/output and then its code.
String original = "10100";
String toFind = "10";
String toReplace = "01";
int ocurrence = 2;
String replaced = replaceNthOcurrence(original, toFind, toReplace, ocurrence);
System.out.println(replaced); // Output: "10010"
original = "This and This and This";
toFind = "This";
toReplace = "That";
ocurrence = 3;
replaced = replaceNthOcurrence(original, toFind, toReplace, ocurrence);
System.out.println(replaced); // Output: "This and This and That"
Function code:
public static String replaceNthOcurrence(String str, String toFind, String toReplace, int ocurrence) {
Pattern p = Pattern.compile(Pattern.quote(toFind));
Matcher m = p.matcher(str);
StringBuffer sb = new StringBuffer(str);
int i = 0;
while (m.find()) {
if (++i == ocurrence) { sb.replace(m.start(), m.end(), toReplace); break; }
}
return sb.toString();
}
If you want to access the last two indices of a string, then you can use: -
str.substring(str.length() - 2);
This gives you string from index str.length() - 2 to the last character, which is exactly the last two character.
Now, you can replace the last two indices with whatever string you want.
UPDATE: -
Of you want to access the last occurrence of a character or substring, you can use String#lastIndexOf method: -
str.lastIndexOf("10");
Ok, you can try this code: -
String str = "10100";
int fromIndex = str.lastIndexOf("10");
str = str.substring(0, fromIndex) + "01" + str.substring(fromIndex + 2);
System.out.println(str);
10100 with 10010
String result = "10100".substring(0, 2) + "10010".substring(2, 4) + "10100".substring(4, 5);
You can get the last index of a character or substring using string's lastIndexOf method. See the documentation link below for how to use it.
http://docs.oracle.com/javase/1.4.2/docs/api/java/lang/String.html#lastIndexOf(java.lang.String)
Once you know the index of your substring, you can get the substring of all characters before that index, and the substring of all characters after the last character in your search string, and concatenate.
This is a little drawn out, and I didn't actually run it (so I might have a syntax error), but it gives you the point of what I'm trying to convey at least. You could do this all in one line if you want, but it wouldn't illustrate the point as well.
string s = "10100";
string searchString = "10";
string replacementString = "01";
string charsBeforeSearchString = s.substring(0, s.lastIndexOf(searchString) - 1);
string charsAfterSearchString = s.substring(s.lastIndexIf(searchString) + 2);
s = charsBeforeSearchString + replacementString + charsAfterSearchString;
The easiest way:
String input = "10100";
String result = Pattern.compile("(10)(?!.*10.*)").matcher(input).replaceAll("01");
System.out.println(result);
I just want to add a space between each character of a string. Can anyone help me figuring out how to do this?
E.g. given "JAYARAM", I need "J A Y A R A M" as the result.
Unless you want to loop through the string and do it "manually" you could solve it like this:
yourString.replace("", " ").trim()
This replaces all "empty substrings" with a space, and then trims off the leading / trailing spaces.
ideone.com demonstration
An alternative solution using regular expressions:
yourString.replaceAll(".(?=.)", "$0 ")
Basically it says "Replace all characters (except the last one) with with the character itself followed by a space".
ideone.com demonstration
Documentation of...
String.replaceAll (including the $0 syntax)
The positive look ahead (i.e., the (?=.) syntax)
StringBuilder result = new StringBuilder();
for (int i = 0; i < input.length(); i++) {
if (i > 0) {
result.append(" ");
}
result.append(input.charAt(i));
}
System.out.println(result.toString());
Iterate over the characters of the String and while storing in a new array/string you can append one space before appending each character.
Something like this :
StringBuilder result = new StringBuilder();
for(int i = 0 ; i < str.length(); i++)
{
result = result.append(str.charAt(i));
if(i == str.length()-1)
break;
result = result.append(' ');
}
return (result.toString());
Blow up your String into array of chars, loop over the char array and create a new string by succeeding a char by a space.
Create a StringBuilder with the string and use one of its insert overloaded method:
StringBuilder sb = new StringBuilder("JAYARAM");
for (int i=1; i<sb.length(); i+=2)
sb.insert(i, ' ');
System.out.println(sb.toString());
The above prints:
J A Y A R A M
This would work for inserting any character any particular position in your String.
public static String insertCharacterForEveryNDistance(int distance, String original, char c){
StringBuilder sb = new StringBuilder();
char[] charArrayOfOriginal = original.toCharArray();
for(int ch = 0 ; ch < charArrayOfOriginal.length ; ch++){
if(ch % distance == 0)
sb.append(c).append(charArrayOfOriginal[ch]);
else
sb.append(charArrayOfOriginal[ch]);
}
return sb.toString();
}
Then call it like this
String result = InsertSpaces.insertCharacterForEveryNDistance(1, "5434567845678965", ' ');
System.out.println(result);
I am creating a java method for this purpose with dynamic character
public String insertSpace(String myString,int indexno,char myChar){
myString=myString.substring(0, indexno)+ myChar+myString.substring(indexno);
System.out.println(myString);
return myString;
}
This is the same problem as joining together an array with commas. This version correctly produces spaces only between characters, and avoids an unnecessary branch within the loop:
String input = "Hello";
StringBuilder result = new StringBuilder();
if (input.length() > 0) {
result.append(input.charAt(0));
for (int i = 1; i < input.length(); i++) {
result.append(" ");
result.append(input.charAt(i));
}
}
public static void main(String[] args) {
String name = "Harendra";
System.out.println(String.valueOf(name).replaceAll(".(?!$)", "$0 "));
System.out.println(String.valueOf(name).replaceAll(".", "$0 "));
}
This gives output as following use any of the above:
H a r e n d r a
H a r e n d r a
One can use streams with java 8:
String input = "JAYARAM";
input.toString().chars()
.mapToObj(c -> (char) c + " ")
.collect(Collectors.joining())
.trim();
// result: J A Y A R A M
A simple way can be to split the string on each character and join the parts using space as the delimiter.
Demo:
public class Main {
public static void main(String[] args) {
String s = "JAYARAM";
s = String.join(" ", s.split(""));
System.out.println(s);
}
}
Output:
J A Y A R A M
ONLINE DEMO
Create a char array from your string
Loop through the array, adding a space +" " after each item in the array(except the last one, maybe)
BOOM...done!!
If you use a stringbuilder, it would be efficient to initalize the length when you create the object. Length is going to be 2*lengthofString-1.
Or creating a char array and converting it back to the string would yield the same result.
Aand when you write some code please be sure that you write a few test cases as well, it will make your solution complete.
I believe what he was looking for was mime code carrier return type code such as %0D%0A (for a Return or line break)
and
\u00A0 (for spacing)
or alternatively
$#032