Going by the suggestion provided here
I tried using \\W as a delimiter for non word character in string.split function of java.
String str = "id-INT, name-STRING,";
This looks like a really simple string. I wanted to extract just the words from this string. The length of the array that I get is 5 whereas it should be 4. There is an empty string at position right after INT. I don't understand why the space in there is not being considered as non word
The , and the space are been treated as separate entities, try using \\W+ instead
String str = "id-INT, name-STRING,";
String[] parts = str.split("\\W+");
System.out.println(parts.length);
System.out.println(Arrays.toString(parts));
Which outputs
4
[id, INT, name, STRING]
Related
Going by the suggestion provided here
I tried using \\W as a delimiter for non word character in string.split function of java.
String str = "id-INT, name-STRING,";
This looks like a really simple string. I wanted to extract just the words from this string. The length of the array that I get is 5 whereas it should be 4. There is an empty string at position right after INT. I don't understand why the space in there is not being considered as non word
The , and the space are been treated as separate entities, try using \\W+ instead
String str = "id-INT, name-STRING,";
String[] parts = str.split("\\W+");
System.out.println(parts.length);
System.out.println(Arrays.toString(parts));
Which outputs
4
[id, INT, name, STRING]
So I want to split a string using by space and "-".
For example, a sample string : "White-spaces are necessary"
The above String should be split into parts like "White", "space", "are", "necessary"
is this possible using split() method in Java.
The split method takes a String in the form of a regex.
Using regex character classes, we can say split on either space OR dashes.
the regex would be the following [ -].
Each character inside the brackets is a possible character to split on.
However, this would split groups of 2 or more white space characters as separate split items, so you would end up with empty array elements. To avoid this you can use:
[ -]+
More reading on regex character classes.
https://www.regular-expressions.info/charclass.html
Given a string S, find the number of words in that string. For this problem a word is defined by a string of one or more English letters.
Note: Space or any of the special characters like ![,?.\_'#+] will act as a delimiter.
Input Format: The string will only contain lower case English letters, upper case English letters, spaces, and these special characters: ![,?._'#+].
Output Format: On the first line, print the number of words in the string. The words don't need to be unique. Then, print each word in a separate line.
My code:
Scanner sc = new Scanner(System.in);
String str = sc.nextLine();
String regex = "( |!|[|,|?|.|_|'|#|+|]|\\\\)+";
String[] arr = str.split(regex);
System.out.println(arr.length);
for(int i = 0; i < arr.length; i++)
System.out.println(arr[i]);
When I submit the code, it works for just over half of the test cases. I do not know what the test cases are. I'm asking for help with the Murphy's law. What are the situations where the regex I implemented won't work?
You don't escape some special characters in your regex. Let's start with []. Since you don't escape them, the part [|,|?|.|_|'|#|+|] is treated like a set of characters |,?._'#+. This means that your regex doesn't split on [ and ].
For example x..]y+[z is split to x, ]y and [z.
You can fix that by escaping those characters. That will force you to escape more of them and you end up with a proper definition:
String regex = "( |!|\\[|,|\\?|\\.|_|'|#|\\+|\\])+";
Note that instead of defining alternatives, you could use a set which will make your regex easier to read:
String regex = "[!\\[,?._'#+\\].]+";
In this case you only need to escape [ and ].
UPDATE:
There's also a problem with leading special character (like in your example ".Hi?there[broski.]#####"). You need to split on it but it produces an empty string in the results. I don't think there's a way to use split function without producing it but you can mitigate it by removing the first group before splitting using the same regex:
String[] arr = str.replaceFirst(regex, "").split(regex);
I have a string like
String myString = "hello world~~hello~~world"
I am using the split method like this
String[] temp = myString.split("~|~~|~~~");
I want the array temp to contain only the strings separated by ~, ~~ or ~~~.
However, the temp array thus created has length 5, the 2 additional 'strings' being empty strings.
I want it to ONLY contain my non-empty string. Please help. Thank you!
You should use quantifier with your character:
String[] temp = myString.split("~+");
String#split() takes a regex. ~+ will match 1 or more ~, so it will split on ~, or ~~, or ~~~, and so on.
Also, if you just want to split on ~, ~~, or ~~~, then you can limit the repetition by using {m,n} quantifier, which matches a pattern from m to n times:
String[] temp = myString.split("~{1,3}");
When you split it the way you are doing, it will split a~~b twice on ~, and thus the middle element will be an empty string.
You could also have solved the problem by reversing the order of your delimiter like this:
String[] temp = myString.split("~~~|~~|~");
That will first try to split on ~~, before splitting on ~ and will work fine. But you should use the first approach.
Just turn the pattern around:
String myString = "hello world~~hello~~world";
String[] temp = myString.split("~~~|~~|~");
Try This :
myString.split("~~~|~~|~");
It will definitely works. In your code, what actually happens that when ~ occurs for the first time,it count as a first separator and split the string from that point. So it doesn't get ~~ or ~~~ anywhere in your string though it is there. Like :
[hello world]~[]~[hello]~[]~[world]
Square brackets are split-ed in to 5 different string values.
I need to split a java string into an array of words. Let's say the string is:
"Hi!! I need to split this string, into a serie's of words?!"
At the moment I'm tried using this String[] strs = str.split("(?!\\w)") however it keeps symbols such as ! in the array and it also keeps strings like "Hi!" in the array as well. The string I am splitting will always be lowercase. What I would like is for an array to be produced that looks like:
{"hi", "i", "need", "to", "split", "this", "string", "into", "a", "serie's", "of", "words"} - Note the apostrophe is kept.
How could I change my regex to not include the symbols in the array?
Apologies, I would define a word as a sequence of alphanumeric characters only but with the ' character inclusive if it is in the above context such as "it's", not if it is used to a quote a word such as "'its'". Also, in this context "hi," or "hi-person" are not words but "hi" and "person" are. I hope that clarifies the question.
You can remove all ?! symbols and then split into words
str = str.replaceAll("[!?,]", "");
String[] words = str.split("\\s+");
Result:
Hi, I, need, to, split, this, string, into, a, serie's, of, words
Should work for what you want.
String line = "Hi!! I need to split this string, into a serie's of words?! but not '' or ''' word";
String regex = "([^a-zA-Z']+)'*\\1*";
String[] split = line.split(regex);
System.out.println(Arrays.asList(split));
Gives
[Hi, I, need, to, split, this, string, into, a, serie's, of, words, but, not, or, word]
If you define a word as a sequence of non-whitespace characters (whitespace character as defined by \s), then you can split along space characters:
str.split("\\s+")
Note that ";.';.##$>?>#4", "very,bad,punctuation", and "'goodbye'" are words under the definition above.
Then the other approach is to define a word as a sequence of characters from a set of allowed characters. If you want to allow a-z, A-Z, and ' as part of a word, you can split along everything else:
str.split("[^a-zA-Z']+")
This will still allow "''''''" to be defined as a word, though.
So what you want is to split on anything that is not a wordcharacter [a-zA-Z] and is not a '
This regex will do that "[^a-zA-Z']\s"
There will be a problem if the string contains a quote that is quoted in '
I usually use this page for testing my regex'
http://www.regexplanet.com/advanced/java/index.html
I would use str.split("[\\s,?!]+"). You can add whatever character you want to split with inside the brackets [].
You could filter out the characters you deem as "non-word" characters:
String[] strs = str.split("[,!? ]+");
myString.replaceAll("[^a-zA-Z'\\s]","").toLowerCase().split("\\s+");
replaceAll("[^a-zA-Z'\\s]","") method replaces all the characters which are not a-z or A-Z or ' or a whitespace with nothing ("") and then toLowerCase method make all the chars returned from replaceAll method lower case. Finally we are splitting the string in terms of whitespace char. more readable one;
myString = myString.replaceAll("[^a-zA-Z'\\s]","");
myString = myString.toLowerCase();
String[] strArr = myString.split("\\s+");