Develop Reference

split a string when there is a change in character without a regular expression

There is a way to split a string into repeating characters using a regex function but I want to do it without using it.
for example, given a string like: "EE B" my output will be an array of strings e.g
{"EE", " ", "B"}
my approach is:
given a string I will first find the number of unique characters in a string so I know the size of the array. Then I will change the string to an array of characters. Then I will check if the next character is the same or not. if it is the same then append them together if not begin a new string.
my code so far..
String myinput = "EE B";
char[] cinput = new char[myinput.length()];
cinput = myinput.toCharArray(); //turn string to array of characters
int uniquecha = myinput.length();
for (int i = 0; i < cinput.length; i++) {
if (i != myinput.indexOf(cinput[i])) {
uniquecha--;
} //this should give me the number of unique characters
String[] returninput = new String[uniquecha];
Arrays.fill(returninput, "");
for (int i = 0; i < uniquecha; i++) {
returninput[i] = "" + myinput.charAt(i);
for (int j = 0; j < myinput.length - 1; j++) {
if (myinput.charAt(j) == myinput.charAt(j + 1)) {
returninput[j] += myinput.charAt(j + 1);
} else {
break;
}
}
} return returninput;
but there is something wrong with the second part as I cant figure out why it is not beginning a new string when the character changes.

You question says that you don't want to use regex, but I see no reason for that requirement, other than this is maybe homework. If you are open to using regex here, then there is a one line solution which splits your input string on the following pattern:
(?<=\S)(?=\s)|(?<=\s)(?=\S)
This pattern uses lookarounds to split whenever what precedes is a non whitespace character and what proceeds is a whitespace character, or vice-versa.
String input = "EE B";
String[] parts = input.split("(?<=\\S)(?=\\s)|(?<=\\s)(?=\\S)");
System.out.println(Arrays.toString(parts));
[EE, , B]
^^ a single space character in the middle
Demo

If I understood correctly, you want to split the characters in a string so that similar-consecutive characters stay together. If that's the case, here is how I would do it:
public static ArrayList<String> splitString(String str)
{
ArrayList<String> output = new ArrayList<>();
String combo = "";
//iterates through all the characters in the input
for(char c: str.toCharArray()) {
//check if the current char is equal to the last added char
if(combo.length() > 0 && c != combo.charAt(combo.length() - 1)) {
output.add(combo);
combo = "";
}
combo += c;
}
output.add(combo); //adds the last character
return output;
}
Note that instead of using an array (has a fixed size) to store the output, I used an ArrayList, which has a variable size. Also, instead of checking the next character for equality with the current one, I preferred to use the last character for that. The variable combo is used to temporarily store the characters before they go to output.
Now, here is one way to print the result following your guidelines:
public static void main(String[] args)
{
String input = "EEEE BCD DdA";
ArrayList<String> output = splitString(input);
System.out.print("[");
for(int i = 0; i < output.size(); i++) {
System.out.print("\"" + output.get(i) + "\"");
if(i != output.size()-1)
System.out.print(", ");
}
System.out.println("]");
}
The output when running the above code will be:
["EEEE", " ", "B", "C", "D", " ", "D", "d", "A"]

Find space separated names using Apache OpenNLP

I am using NER of Apache Open NLP. I have successfully trained my custom data. And while using the name finder, I am splitting the given string based on white space and passing the string array as given below.
NameFinderME nameFinder = new NameFinderME(model);
String []sentence = input.split(" "); //eg:- input = Give me list of test case in project X
Span nameSpans[] = nameFinder.find(sentence);
Here, when I use split, test and case are given as separate values and is never detected by the namefinder. How would I possibly overcome the above issue. Is there a way by which I can pass the complete string (without splitting it into array) such that, test case will be considered as a whole by itself ?

You can do it using regular expressions. Try replacing the second line with this:
String []sentence = input.split("\\s(?<!(\\stest\\s(?=case\\s)))");
Maybe there is a better way to write the expression, but this works for me and the output is:
Give
me
list
of
test case
in
project
X
EDIT: If you are interested in the details check here where I split: https://regex101.com/r/6HLBnL/1
EDIT 2: If you have many words that don't get separated, I wrote a method that generates the regex for you. This is how the regex in this case should look like (if you don't want to separate 'test case' and 'in project'):
\s(?<!(\stest\s(?=case\s))|(\sin\s(?=project\s)))
Following is a simple program to demonstrate it. In this example you just put the words that don't need separation in the array unseparated.
class NoSeparation {
private static String[][] unseparated = {{"test", "case"}, {"in", "project"}};
private static String getRegex() {
String regex = "\\s(?<!";
for (int i = 0; i < unseparated.length; i++)
regex += "(\\s" + separated[i][0] + "\\s(?=" + separated[i][1] + "\\s))|";
// Remove the last |
regex = regex.substring(0, regex.length() - 1);
return (regex + ")");
}
public static void main(String[] args) {
String input = "Give me list of test case in project X";
String []sentence = input.split(getRegex());
for (String i: sentence)
System.out.println(i);
}
}
EDIT 3: Following is a very dirty way to handle strings with more than 2 words. It works, but I am pretty sure that you can do it in a more efficient way. It will work fine in short inputs, but in longer it will probably be slow.
You have to put the words that should not be splitted in a 2d array, as in unseparated. You should also choose a separator if you don't want to use %% for some reason (e.g. if there is a chance your input contains it).
class NoSeparation {
private static final String SEPARATOR = "%%";
private static String[][] unseparated = {{"of", "test", "case"}, {"in", "project"}};
private static String[] splitString(String in) {
String[] splitted;
for (int i = 0; i < unseparated.length; i++) {
String toReplace = "";
String replaceWith = "";
for (int j = 0; j < unseparated[i].length; j++) {
toReplace += unseparated[i][j] + ((j < unseparated[i].length - 1)? " " : "");
replaceWith += unseparated[i][j] + ((j < unseparated[i].length - 1)? SEPARATOR : "");
}
in = in.replaceAll(toReplace, replaceWith);
}
splitted = in.split(" ");
for (int i = 0; i < splitted.length; i++)
splitted[i] = splitted[i].replaceAll(SEPARATOR, " ");
return splitted;
}
public static void main(String[] args) {
String input = "Give me list of test case in project X";
// Uncomment this if there is a chance to have multiple spaces/tabs
// input = input.replaceAll("[\\s\\t]+", " ");
for (String str: splitString(input))
System.out.println(str);
}
}

Java: Split string by number of characters but with guarantee that string will be split only after whitespace

I want to achieve something like this.
String str = "This is just a sample string";
List<String> strChunks = splitString(str,8);
and strChunks should should be like:
"This is ","just a ","sample ","string."
Please note that string like "sample " have only 7 characters as with 8 characters it will be "sample s" which will break down my next word "string".
Also we can go with the assumption that a word will never be larger than second argument of method (which is 8 in example) because in my use case second argument is always static with value 32000.
The obvious approach that I can think of is looping thru the given string, breaking the string after 8 chars and than searching the next white space from the end. And then repeating same thing again for remaining string.
Is there any more elegant way to achieve the same. Is there any utility method already available in some standard third libraries like Guava, Apache Commons.

Splitting on "(?<=\\G.{7,}\\s)" produces the result that you need (demo).
\\G means the end of previous match; .{7,} means seven or more of any characters; \\s means a space character.

Not a standard method, but this might suit your needs
See it on http://ideone.com/2RFIZd
public static List<String> splitString(String str, int chunksize) {
char[] chars = str.toCharArray();
ArrayList<String> list = new ArrayList<String>();
StringBuilder builder = new StringBuilder();
int count = 0;
for(char character : chars) {
if(count < chunksize - 1) {
builder.append(character);
count++;
}
else {
if(character == ' ') {
builder.append(character);
list.add(builder.toString());
count = 0;
builder.setLength(0);
}
else {
builder.append(character);
count++;
}
}
}
list.add(builder.toString());
builder.setLength(0);
return list;
}
Please note, I used the human notation for string length, because that's what your sample reflects( 8 = postion 7 in string). that's why the chunksize - 1 is there.
This method takes 3 milliseconds on a text the size of http://catdir.loc.gov/catdir/enhancements/fy0711/2006051179-s.html

Splitting String using method 1.
String text="This is just a sample string";
List<String> strings = new ArrayList<String>();
int index = 0;
while (index < text.length()) {
strings.add(text.substring(index, Math.min(index + 8,text.length())));
index += 8;
}
for(String s : strings){
System.out.println("["+s+"]");
}
Splitting String using Method 2
String[] s=text.split("(?<=\\G.{"+8+"})");
for (int i = 0; i < s.length; i++) {
System.out.println("["+s[i]+"]");
}

This uses a hacked reduction to get it done without much code:
String str = "This is just a sample string";
List<String> parts = new ArrayList<>();
parts.add(Arrays.stream(str.split("(?<= )"))
.reduce((a, b) -> {
if (a.length() + b.length() <= 8)
return a + b;
parts.add(a);
return b;
}).get());
See demo using edge case input (that breaks some other answers!)
This splits after each space, then either joins up parts or adds to the list depending on the length of the pair.

Remove characters at certain position at string

I want to remove certain characters at specific positions of the String. I have the positions, but I am facing problems removing the characters.
what i am doing is:
if (string.subSequence(k, k + 4).equals("\n\t\t\t")){
string = string.subSequence(0, k) + "" + s.subSequence(k, s.length());
}
I need to remove "\n\t\t\t" from string

Use StringBuilder:
StringBuilder sb = new StringBuilder(str);
sb.delete(start, end);
sb.deleteCharAt(index);
String result = sb.toString();

Use StringBuilder
String str=" ab a acd";
StringBuilder sb = new StringBuilder(str);
sb.delete(0,3);
sb.deleteCharAt(0);
String result = sb.toString();
System.out.println(result);

public static String remove(int postion, String stringName) {
char [] charArray = stringName.toCharArray();
char [] resultArray = new char[charArray.length];
int count = 0;
for (int i=0; i< charArray.length; i++) {
if (i != postion-1) {
resultArray[count] = charArray[i];
count++;
}
}
return String.valueOf(resultArray);
}

Use String.ReplaceAll() instead of this.
But if you only want to remove specific element only you can use substring().
Now you want to know position which you already know.

Put your points in a HashSet called set
StringBuilder sb=new StringBuilder();
for(int i=0;i<string.length();i++){
if(!set.contains(string.charAt(i)))
sb.append(string.charAt(i));
}
String reformattedString=sb.toString();

First you have to put \ in front of the special characters in order to do the matching of the two string, thus you will have .equals("\"\\n\\t\\t\\t\""), otherwise the substring is not going to be recognized inside the string. Then the other thing which you have to fix is the position of the index begin and end inside .subSequence(k,k+10) since the first and the last character are 10 positions apart and not 4. Note also that when you patch the string you go from position 0 to k and from k+10 to str.length(). If you go from 0 --> k and k --> length() you just join the old string together :).
Your code should work like this, I have tested it already
if(str.substring(k, k+10).equals("\"\\n\\t\\t\\t\""))
{
newstr = str.substring(0,k)+str.substring(k+10,(str.length()));
}
also you don't need +" "+ since you are adding strings. Whoever wants to see the effect of this can run this simple code:
public class ReplaceChars_20354310_part2 {
/**
* #param args the command line arguments
*/
public static void main(String[] args) {
String str = "This is a weird string containg balndbfhr frfrf br brbfbrf b\"\\n\\t\\t\\t\"";
System.out.println(str); //print str
System.out.println(ReplaceChars(str)); //then print after you replace the substring
System.out.println("\n"); //skip line
String str2 = "Whatever\"\\n\\t\\t\\t\"you want to put here"; //print str
System.out.println(str2); //then print after you replace the substring
System.out.println(ReplaceChars(str2));
}
//Method ReplaceChars
public static String ReplaceChars (String str) {
String newstr ="";
int k;
k = str.indexOf("\"\\n\\t\\t\\t\""); //position were the string starts within the larger string
if(str.substring(k, k+10).equals("\"\\n\\t\\t\\t\""))
{
newstr = str.substring(0,k)+str.substring(k+10,(str.length())); //or just str
}
return newstr;
}//end method
}

how to count the number of words of the same(PALINDROME) in java

I'm a newbie Java Developer. I want to write code to count the number of palindrome words in the paragraph using Java.
The assumptions are : User can enter a paragraph containing as many sentences as possible. Each word is separated by a whitespace, and each sentence is separated by a period and The punctuation right before or after the word will be ignored, while the punctuation inside the word will be counted.
Sample Input : Otto goes to school. Otto sees a lot of animals at the pets store.
Sample output : Otto = 2 a = 1 Sees = 1

Read the file into your program, split the entries at every space and enter those into an arraylist. Afterwards, apply your palindrome algorithm onto each value in your arraylist and keep track of the words that were a palindrome and their occurences (for example a 2D array, or an arraylist with an object that holds both values).
When you've followed these steps, you should pretty much be there. More specific help will probably be given once you've shown attempts of your own.

Using Collections in java will reduce the programming effort
Algorithm :
Read the paragraph to a String variable
Split the String using StringTokenizer using token as ' '(space) and add each word to ArrayList (Set wont allow duplicates)
Write a method which return boolean (TRUE/ FALSE) value based on whether a given String is palindrome or not.
Define a Map to hold the values of palindrome String and number of times it is repeated.
If yes
add the String to Map with key as palindrome String and value as number of times
else
dont add the String to Map
Repeat the same logic until all the words are finished
Sample Code:
` public class StringPalindromeCalculator {
private Map<String, int> wordsMap = new HashMap<>();
private List<String> wordsList = new ArrayLiat<>();
private boolean isPalindrome(String inputString) {
// write String palindrome logic here
}
public Map<String, int> findPalindromeWords(String completeString) {
StringTokenizer wordTokenizer = new StringTokenizer(completeString, ' ');
while(wordTokenizer.hasMoreTokens()) {
wordsList.add(wordTokenizer.nextToken());
}
for(String word : wordsList) {
if(isPalindrome(word)) {
if(wordsMap.containsKey(word)) {
// increment the value of word
}
} else {
// put the word into Map and return the map value
}
}
return wordsMap;
}
}`
Hope this Helps :)

public class Palindrome {
int count = 0;
public static void main(String[] args) {
String a = "malayalammadyoydaraarasdasdkfjasdsjhtj";
Palindrome palindrome = new Palindrome();
palindrome.countPalin(a);
}
private int countPalin(String str) {
for (int i = 0; i < str.length() - 1; i++) {
char start = str.charAt(i);
String st = "";
st += start;
for (int j = i + 1; j < str.length(); j++) {
st += str.charAt(j);
StringBuffer rev = new StringBuffer(st).reverse();
if (st.equals(rev.toString()) && st.length() > 1) {
System.out.println(st.toString());
count++;
}
}
st = "";
}
System.out.println("Total Count : " + count);
return count;
}
}