Develop Reference

Deleting content of every string after first empty space

How can I delete everything after first empty space in a string which user selects? I was reading this how to remove some words from a string in java. Can this help me in my case?

You can use replaceAll with a regex \s.* which match every thing after space:
String str = "Hello java word!";
str = str.replaceAll("\\s.*", "");
output
Hello
regex demo
Like #Coffeehouse Coder mention in comment, This solution will replace every thing if the input start with space, so if you want to avoid this case, you can trim your input using string.trim() so it can remove the spaces in start and in end.

Assuming that there is no space in the beginning of the string.
Follow these steps-
Split the string at space. It will create an array.
Get the first element of that array.
Hope this helps.
str = "Example string"
String[] _arr = str.split("\\s");
String word = _arr[0];
You need to consider multiple white spaces and space in the beginning before considering the above code.
I am not native to JAVA Programming but have an idea that it has split function for string.
And the reference you cited in the question is bit complex, while you can achieve the desired thing very easily.
P.S. In future if you make a mind to get two words or three, splitting method is better (assuming you have already dealt with multiple white-spaces) else substring is better.

A simple way to do it can be:
System.out.println("Hello world!".split(" ")[0]);

// Taking 'str' as your string
// To remove the first space(s) of the string,
str = str.trim();
int index = str.indexOf(" ");
String word = str.substring(0, index);
This is just one method of many.
str = str.replaceAll("\\s+", " "); // This replaces one or more spaces with one space
String[] words = str.split("\\s");
String first = words[0];

The simplest solution in my opinion would be to just locate the index which the user wants it to be cut off at and then call the substring() method from 0 to the index they wanted. Set that = to a new string and you have the string they want.
If you want to replace the string then just set the original string = to the result of the substring() method.
Link to substring() method: https://docs.oracle.com/javase/7/docs/api/java/lang/String.html#substring(int,%20int)

There are already 5 perfectly good answers, so let me add a sixth one. Variety is the spice of life!
private static final Pattern FIRST_WORD = Pattern.compile("\\S+");
public static String firstWord(CharSequence text) {
Matcher m = FIRST_WORD.matcher(text);
return m.find() ? m.group() : "";
}
Advantages over the .split(...)[0]-type answers:
It directly does exactly what is being asked, i.e. "Find the first sequence of non-space characters." So the self-documentation is more explicit.
It is more efficient when called on multiple strings (e.g. for batch processing a large list of strings) because the regular expression is compiled only once.
It is more space-efficient because it avoids unnecessarily creating a whole array with references to each word when we only need the first.
It works without having to trim the string.
(I know this is probably too late to be of any use to the OP but I'm leaving it here as an alternative solution for future readers.)

This would be more efficient
String str = "Hello world!";
int spaceInd = str.indexOf(' ');
if(spaceInd != -1) {
str = str.substring(0, spaceInd);
}
System.out.println(String.format("[%s]", str));

Java : Specifying Case Insensitive Search on String into an ArrayList with Regex

Following are the intended output and the original output I got from using this line of code :- ArrayList<String> nodes = new ArrayList<String>
(Arrays.asList(str.split("(?i:"+word+")"+"[.,!?:;]?")));
on the input :-
input : "Cow shouts COW! other cows shout COWABUNGA! stupid cow."
The string will be split into an ArrayList at the acceptable "cow" versions.
Original Output(from line above) :
ArrayList nodes = {, shouts , other , s shout ,ABUNGA! stupid }
vs
Intended Output :
ArrayList nodes = {, shouts , other cows shout COWABUNGA! stupid }
What I'm trying to achieve :
Case insensitive search. (ACHIEVED)
Takes into account the possibilities of these punctuations ".,:;!?" behind the word that is to be split. hence "[.,!?:;]?" (ACHIEVED)
Only splits if it finds exact word lengths + "[.,!?:;]?". It will not split at "cows" nor "COWABUNGA!" (NOT ACHIEVED, need help)
Find a possible way to add the acceptable splitting-word versions {Cow,COW!,cow.} into another arrayList for future use later in the method. (IN PROGRESS)
As you can see, I have fulfilled 1. and 2. and I am pasting this question first whilst I work on 4.. I know this issue can be solved with more extra lines but I'd like to keep it minimal and efficient.
UPDATE : I found that "{"+input.length+"}" can limit the matches down to letter length but I don't know if it'll work or not.
All help will be appreciated. I apologize if this question is too trivial but alas, I am new. Thanks in advance!

The following code produces the output you specified given your input. I have broken the regular expression down into named components, so each bit should be self-explanatory.
import java.util.Arrays;
import java.util.List;
import java.util.regex.Pattern;
public class Moo {
public static void main(String[] args) {
String input = "Cow shouts COW! other cows shout COWABUNGA! stupid cow.";
System.out.println(splitter(input, "cow"));
}
public static List<String> splitter(String input, String word) {
String beginningOfInputOrWordBoundary = "(\\A|\\W)";
String caseInsensitiveWord = "(?i:"+Pattern.quote(word)+")";
String optionalPunctuation = "\\p{Punct}?";
String endOfInputOrWordBoundary = "(\\z|\\W)";
String regex =
beginningOfInputOrWordBoundary +
caseInsensitiveWord +
optionalPunctuation +
endOfInputOrWordBoundary;
return Arrays.asList(input.split(regex));
}
}
Resulting output:
[, shouts, other cows shout COWABUNGA! stupid]

A word is a sequence of letters. Any character that is not a letter implies the end of a word.
Thus, this should provide the desired result:
(?i:Cow)[^\\p{IsAlphabetic}]

Comparing parts of Arrays against each other?

I'm really really really not sure what is the best way to approach this. I've gotten as far as I can, but I basically want to scan a user response with an array of words and search for matches so that my AI can tell what mood someone is in based off the words they used. However, I've yet to find a clear or helpful answer. My code is pretty cluttered too because of how many different methods I've tried to use. I either need a way to compare sections of arrays to each other or portions of strings. I've found things for finding a part of an array. Like finding eggs in green eggs and ham, but I've found nothing that finds a section of an array in a section of another array.
public class MoodCompare extends Mood1 {
public static void MoodCompare(String inputMood){
int inputMoodLength = inputMood.length();
int HappyLength = Arrays.toString(Happy).length();
boolean itWorks = false;
String[] inputMoodArray = inputMood.split(" ");
if(Arrays.toString(Happy).contains(Arrays.toString(inputMoodArray)) == true)
System.out.println("Success!");
InputMood is the data the user has input that should have keywords lurking in them to their mood. Happy is an array of the class Mood1 that is being extended. This is only a small piece of the class, much less the program, but it should be all I need to make a valid comparison to complete the class.
If anyone can help me with this, you will save me hours of work. So THANK YOU!!!

Manipulating strings will be nicer when you do not use the relative primitive arrays, where you have to walk through yourself etcetera. A Dutch proverb says: not seeing the wood through the trees.
In this case it seems you check words of the input against a set of words for some mood.
Lets use java collections:
Turning an input string into a list of words:
String input = "...";
List<String> sentence = Arrays.asList(input.split("\\W+"));
sentence.remove("");
\\W+ is a sequence of one or more non-word characters. Mind "word" mean A-Za-z0-9_.
Now a mood would be a set of unique words:
Set<String> moodWords = new HashSet<>();
Collections.addAll(moodWords, "happy", "wow", "hurray", "great");
Evaluation could be:
int matches = 0;
for (String word : sentence) {
if (moodWords.contains(word)) {
++matches;
}
}
int percent = sentence.isEmpty() ? 0 : matches * 100 / sentence.size();
System.out.printf("Happiness: %d %%%n", percent);
In java 8 even compacter.
int matches = sentence.stream().filter(moodWords::contains).count();
Explanation:
The foreach-word-in-sentence takes every word. For every word it checks whether it is contained in moodWords, the set of all mood words.
The percentage is taken over the number of words in the sentence being moody. The boundary condition of an empty sentence is handled by the if-then-else expression ... ? ... : ... - an empty sentence given the arbitrary percentage 0%.
The printf format used %d for the integer, %% for the percent sign % (self-escaped) and %n for the line break character(s).

If I'm understanding your question correctly, you mean something like this?
String words[] = {"green", "eggs", "and", "ham"};
String response = "eggs or ham";
Mood mood = new Mood();
for(String foo : words)
{
if(response.contains(foo))
{
//Check if happy etc...
if(response.equals("green")
mood.sad++;
...
}
}
System.out.println("Success");
...
//CheckMood() etc... other methods.

Try to use tokens.
Every time that the program needs to compare the contents of a row from one array to the other array, just tokenize the contents in parallel and compare them.
Visit the following Java Doc page for farther reference: http://docs.oracle.com/javase/7/docs/api/java/util/StringTokenizer.html
or even view the following web pages:
http://introcs.cs.princeton.edu/java/72regular/Tokenizer.java.html

How to delete duplicated characters in a string?

Okay, I'm a huge newbie in the world of java and I can't seem to get this program right. I am suppose to delete the duplicated characters in a 2 worded string and printing the non duplicated characters.
for example:I input the words "computer program." the output should be "cute" because these are the only char's that are not repeated.
I made it until here:
public static void main(String[] args) {
System.out.print("Input two words: ");
String str1 = Keyboard.readString();
String words[] = str1.split(" ");
String str2 = words[0] + " ";
String str3 = words[words.length - 1] ;
}
but i don't know how to output the characters. Could someone help me?
I don't know if I should use if, switch, for, do, or do-while...... I'm confused.

what you need is to build up logic for your problem. First break the problem statement and start finding solution for that. Here you go for steps,
Read every character from a string.
Add it to a collection, but before adding that, just check whether it exists.
If it exists just remove it and continue the reading of characteer.
Once you are done with reading the characters, just print the contents of collection to console using System.out.println.
I will recommend you to refer books like "Think like A Programmer". This will help you to get started with logic building.

Just a hint: use a hash map (http://docs.oracle.com/javase/6/docs/api/java/util/HashMap.html).

Adding following code after last line of your main program will resolve your issue.
char[] strChars = str2.toCharArray();
String newStr="";
for (char c : strChars) {
String charStr = ""+c;
if(!str3.contains(charStr.toLowerCase()) && !str3.contains(charStr.toUpperCase())){
newStr+=c;
}
}
System.out.println(newStr);
This code loops through all the characters of the first word and check if the second string contains that character (In any form of case Lower or Upper). If it is not containing, adding it to output string and at the end printing it.
Hope this will work in your case.

How about doing it in just 1 line?
str = str.replaceAll("(.)(?=.*\\1)", "");

Python split semantics in Java

When I split a string in python, adjacent space delimiters are merged:
>>> str = "hi there"
>>> str.split()
['hi', 'there']
In Java, the delimiters are not merged:
$ cat Split.java
class Split {
public static void main(String args[]) {
String str = "hi there";
String result = "";
for (String tok : str.split(" "))
result += tok + ",";
System.out.println(result);
}
}
$ javac Split.java ; java Split
hi,,,,,,,,,,,,,,there,
Is there a straightforward way to get python space split semantics in java?

String.split accepts a regular expression, so provide it with one that matches adjacent whitespace:
str.split("\\s+")
If you want to emulate the exact behaviour of Python's str.split(), you'd need to trim as well:
str.trim().split("\\s+")
Quote from the Python docs on str.split():
If sep is not specified or is None, a different splitting algorithm is applied: runs of consecutive whitespace are regarded as a single separator, and the result will contain no empty strings at the start or end if the string has leading or trailing whitespace. Consequently, splitting an empty string or a string consisting of just whitespace with a None separator returns [].
So the above is still not an exact equivalent, because it will return [''] for the empty string, but it's probably okay for your purposes :)

Use str.split("\\s+") instead. This will do what you need.

Java uses Regex to split.
so splitting on a single space will absolutely give you many array elements.
Python split, ltrims and rtrims and then takes runs of spaces into a single space when no parameter has been passed.
So it would more properly be
"my string".trim().split("\\s+");

The problem with Niklas B.'s answer is that trim has its own definition of whitespace, i.e., anything with code up to '\u0020'. The following should get close enough to the Python version, including the fix for the empty string:
class TestSplit {
private static final String[] EMPTY = {};
private static String[] pySplit(String s) {
s = s.replaceAll("^\\s+", "").replaceAll("\\s+$", "");
if (s.isEmpty()) return EMPTY;
return s.split("\\s+");
}
}

In java, String.split takes a regex. So you can do str.split(" +") to get python semantics.