How can I find the last word of a string? I am not trying to find a fixed word, in other words, I would not know what the last word is, however I want to retrieve it.
Here is my code:
myString = myString.trim();
String[] wordList = myString.split("\\s+");
System.out.println(wordList[wordList.length-1]);
Providing you consider words in a sentence to be delimited by whitespace and punctuation (particularly commas, spaces, new lines, brackets, and so on), which means punctuation can appear at the end of the sentence, and you want to include non-ASCII characters in the words, then the following will find you the last word in a string without the punctuation included:
static String lastWord(String sentence) {
Pattern p = Pattern.compile("([\\p{Alpha}]+)(?=\\p{Punct}*$)", Pattern.UNICODE_CHARACTER_CLASS);
Matcher m = p.matcher(sentence);
if (m.find()) {
return m.group();
}
return ""; // or null
}
The regular expression uses look-ahead to find zero-or-more punctuations at the end of the string and matches the alphabetical word before it.
If you want to also allow numbers in the word, change {Alpha} to {Alnum}.
Read the String API for various methods you might use.
For example you could:
Use the lastIndexOf(...) method to find where the start of the word is
Then use the substring(...) method to get the word
Use the StringTokenizer for this
StringTokenizer st = new StringTokenizer("this is a test");//Take any String
int count = st.countTokens();//it will count the number of token in that particular String
String[] myStringArray = new String[count];
for (int i = 0; i < count; i++) {
`myStringArray[i] = st.nextToken();`//insert the words/Tken in to the string array
}
`System.out.println("Last Word is--" + myStringArray[myStringArray.length - 1])`;//get the last words of the given String
System.out.println("enter the string");
Scanner input = new Scanner(System.in);
String instr = input.nextLine();
instr = instr.trim();
int index = instr.lastIndexOf(" ");
int l = instr.length();
System.out.println(l);
String lastStr = instr.substring(index+1,l);
System.out.println("last string .."+lastStr);
Related
I have a question about replacing words. I have some strings, each of which looks like this:
String string = "today is a (happy) day, I would like to (explore) more about Java."
I need to replace the words that have parentheses. I want to replace "(happy)" with "good", and "(explore)" with "learn".
I have some ideas, but I don't know how.
for (int i = 0; i <= string.length(), i++) {
for (int j = 0; j <= string.length(), j++
if ((string.charAt(i)== '(') && (string.charAt(j) == ')')) {
String w1 = line.substring(i+1,j);
string.replace(w1, w2)
}
}
}
My problem is that I can only replace one word with one new word...
I am thinking of using a scanner to prompt me to give a new word and then replace it, how can I do this?
The appendReplacement and appendTail methods of Matcher are designed for this purpose. You can use a regex to scan for your pattern--a pair of parentheses with a word in the middle--then do whatever you need to do to determine the string to replace it with. See the javadoc.
An example, based on the example in the javadoc. I'm assuming you have two methods, replacement(word) that tells what you want to replace the word with (so that replacement("happy") will equal "good" in your example), and hasReplacement(word) that tells whether the word has a replacement or not.
Pattern p = Pattern.compile("\\((.*?)\\)");
Matcher m = p.matcher(source);
StringBuffer sb = new StringBuffer();
while (m.find()) {
String word = m.group(1);
String newWord = hasReplacement(word) ? replacement(word) : m.group(0);
m.appendReplacement(sb, newWord); // appends the replacement, plus any not-yet-used text that comes before the match
}
m.appendTail(sb); // appends any text left over after the last match
String result = sb.toString();
Use below code for replacing the string.
String string = "today is a (happy) day, I would like to (explore) more about Java.";
string = string.replaceAll("\\(happy\\)", "good");
string = string.replaceAll("\\(explore\\)", "learn");
System.out.println(string);`
What you can do is run a loop from 0 to length-1 and if loop encounters a ( then assign its index to a temp1 variable. Now go on further as long as you encounter ).Assign its index to temp2 .Now you can replace that substring using string.replace(string.substring(temp1+1,temp2),"Your desired string")).
No need to use the nested loops. Better use one loop and store the index when you find opening parenthesis and also for close parenthesis and replace it with the word. Continue the same loop and store next index. As you are replacing the words in same string it changes the length of string you need to maintain copy of string and perform loop and replace on different,
Do not use nested for loop. Search for occurrences of ( and ). Get the substring between these two characters and then replace it with the user entered value. Do it till there are not more ( and ) combinations left.
import java.util.Scanner;
public class ReplaceWords {
public static String replaceWords(String s){
while(s.contains(""+"(") && s.contains(""+")")){
Scanner keyboard = new Scanner(System.in);
String toBeReplaced = s.substring(s.indexOf("("), s.indexOf(")")+1);
System.out.println("Enter the word with which you want to replace "+toBeReplaced+" : ");
String replaceWith = keyboard.nextLine();
s = s.replace(toBeReplaced, replaceWith);
}
return s;
}
public static void main(String[] args) {
String myString ="today is a (happy) day, I would like to (explore) more about Java.";
myString = replaceWords(myString);
System.out.println(myString);
}
}
This snippet works for me, just load the HashMap up with replacements and iterate through:
import java.util.*;
public class Test
{
public static void main(String[] args) {
String string = "today is a (happy) day, I would like to (explore) more about Java.";
HashMap<String, String> hm = new HashMap<String, String>();
hm.put("\\(happy\\)", "good");
hm.put("\\(explore\\)", "learn");
for (Map.Entry<String, String> entry : hm.entrySet()) {
String key = entry.getKey();
String value = entry.getValue();
string = string.replaceAll(key, value);
}
System.out.println(string);
}
}
Remember, replaceAll takes a regex, so you want it to display "\(word\)", which means the slashes themselves must be escaped.
There's a string
String str = "ggg;ggg;nnn;nnn;aaa;aaa;xxx;xxx;";
How do I split it into strings like this
"ggg;ggg;"
"nnn;nnn;"
"aaa;aaa;"
"xxx;xxx;"
???????
Using Regex
String input = "ggg;ggg;nnn;nnn;aaa;aaa;xxx;xxx;";
Pattern p = Pattern.compile("([a-z]{3});\\1;");
Matcher m = p.matcher(input);
while (m.find())
// m.group(0) is the result
System.out.println(m.group(0));
Will output
ggg;ggg;
nnn;nnn;
aaa;aaa;
xxx;xxx;
I assume that the you only want to check if the last segment is similar and not every segment that has been read.
If that is not the case then you would probably have to use an ArrayList instead of a Stack.
I also assumed that each segment has the format /([a-z])\1\1/.
If that is not the case either then you should change the if statement with:
(stack.peek().substring(0,index).equals(temp))
public static Stack<String> splitString(String text, char split) {
Stack<String> stack = new Stack<String>();
int index = text.indexOf(split);
while (index != -1) {
String temp = text.substring(0, index);
if (!stack.isEmpty()) {
if (stack.peek().charAt(0) == temp.charAt(0)) {
temp = stack.pop() + split + temp;
}
}
stack.push(temp);
text = text.substring(index + 1);
index = text.indexOf(split);
}
return stack;
}
Split and join them.
public static void main(String[] args) throws Exception {
String data = "ggg;ggg;nnn;nnn;aaa;aaa;xxx;xxx;";
String del = ";";
int splitSize = 2;
StringBuilder sb = new StringBuilder();
for (Iterable<String> iterable : Iterables.partition(Splitter.on(del).split(data), splitSize)) {
sb.append("\"").append(Joiner.on(del).join(iterable)).append(";\"");
}
sb.delete(sb.length()-3, sb.length());
System.out.println(sb.toString());
}
Ref : Split a String at every 3rd comma in Java
Use split with a regex:
String data="ggg;ggg;nnn;nnn;aaa;aaa;xxx;xxx;";
String [] array=data.split("(?<=\\G\\S\\S\\S;\\S\\S\\S);");
S: A non-whitespace character
G: last match/start of string, think of it of a way to skip delimiting if the
previous string matches current one.
?<=:positive look-behind will match semicolon which has string behind it.
Some other answer, that only works given your specific example input.
You see, in your example, there are two similarities:
All patterns seem to have exactly three characters
All patterns occur exactly twice
In other words: if those two properties are really met for all your input, you could avoid splitting - as you know exactly what to find in each position of your string.
Of course, following the other answers for "real" splitting are more flexible; but (theoretically), you could just go forward and do a bunch of substring calls in order to directly access all elements.
I have requirement in which I have to find the no. of times a particular word appears in a file.
For eg.
String str = "Hi hello how are you. hell and heaven. hell, gjh, hello,sdnc ";
Now in this string I want to count no. of times the word "hell" appeared. The count should include "hell" , "hell," all these words but not "hello".
So according to the given string I want the count to be 2.
I used following approaches
1st:
int match = StringUtils.countMatches(str, "hell");
StringUtils is of org.apache.commons.lang3 library
2nd:
int count = 0;
Pattern p = Pattern.compile("hell");
Matcher m = p.matcher(str);
while (m.find()) {
count++;
}
3rd
int count =0;
String[] s = str.split(" ");
for(String word: s)
if(word.equals("hell")
count++;
the 1st two approaches gave 4 as answer and the 3rd approach gave 1 as answer.
Please suggest anyway in which I can get 2 as answer and fullfill my requirement.
You should use word boundary matchers in regex:
Pattern.compile("\\bhell\\b");
You can use a regular expression with the "\\b" word boundaries as follows:
int matches = 0;
Matcher matcher = Pattern.compile("\\bhell\\b", Pattern.CASE_SENSITIVE).matcher(str);
while (matcher.find()) matches++;
Give this a try
String str = "put the string to be searched here";
Scanner sc = new Scanner(str);
String search = "put the string you are searching here";
int counter = 0; //this will count the number of occurences
while (sc.hasNext())
{
if (sc.next() == search)
counter++;
}
Since sc.next() reads complete next token it will hell and hello will not trouble you.
So I wrote the following code:
String text = "This is a string. I want to break it into sentences";
String[] sentences = text.split("\\.");
for (int i = 0; i < sentences.length; i++)
System.out.println(sentences[i]);
The output of this code is:
This is a string
I want to break it into sentences
How do I change this code so that
Each new sentence will be created not only after ".", but also after "!" or "?".
There won't be any spaces in the beginning of sentence.
For example, if we have the following string
String text = "This is a string! Is this a string? I want to break it into sentences";
then the output should be:
This is a string
Is this a string
I want to break it into sentences
Put the delimiters inside a character class and add \\s* next to the char class so that it would consume also the following zero or more spaces.
String[] sentences = text.split("[?!.]\\s*");
Example:
String text = "This is a string! Is this a string? I want to break it into sentences";
String[] parts = text.split("[?!.]\\s*");
for(String i: parts)
{
System.out.println(i);
}
Output:
This is a string
Is this a string
I want to break it into sentences
You can use a character class to split around either one of the dot (.), ? or ! characters. To remove the space at the beginning (and possibly at the end) of the sentence, you can simply trim the resulting string:
String[] sentences = text.split("[.!?]");
for (int i = 0; i < sentences.length; i++) {
System.out.println(sentences[i].trim());
}
I am having a hard time figuring with out. Say I have String like this
String s could equal
s = "{1,4,204,3}"
at another time it could equal
s = "&5,3,5,20&"
or it could equal at another time
s = "/4,2,41,23/"
Is there any way I could just extract the numbers out of this string and make a char array for example?
You can use regex for this sample:
String s = "&5,3,5,20&";
System.out.println(s.replaceAll("[^0-9,]", ""));
result:
5,3,5,20
It will replace all the non word except numbers and commas. If you want to extract all the number you can just call split method -> String [] sArray = s.split(","); and iterate to all the array to extract all the number between commas.
You can use RegEx and extract all the digits from the string.
stringWithOnlyNumbers = str.replaceAll("[^\\d,]+","");
After this you can use split() using deliminator ',' to get the numbers in an array.
I think split() with replace() must help you with that
Use regular expressions
String a = "asdf4sdf5323ki";
String regex = "([0-9]*)";
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(a);
while (matcher.find())
{
String group = matcher.group(1);
if (group.length() > 0)
{
System.out.println(group);
}
}
from your cases, if the pattern of string is same in all cases, then something like below would work, check for any exceptions, not mentioned here :
String[] sArr= s.split(",");
sArr[0] = sArr[0].substring(1);
sArr[sArr.length()-1] =sArr[sArr.length()-1].substring(0,sArr[sArr.length()-1].length()-1);
then convert the String[] to char[] , here is an example converter method
You can use Scanner class with , delimiter
String s = "{1,4,204,3}";
Scanner in = new Scanner(s.substring(1, s.length() - 1)); // Will scan the 1,4,204,3 part
in.useDelimiter(",");
while(in.hasNextInt()){
int x = in.nextInt();
System.out.print(x + " ");
// do something with x
}
The above will print:
1 4 204 3