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How can I split a string without knowing the split characters a-priori?

For my project I have to read various input graphs. Unfortunately, the input edges have not the same format. Some of them are comma-separated, others are tab-separated, etc. For example:
File 1:
123,45
67,89
...
File 2
123 45
67 89
...
Rather than handling each case separately, I would like to automatically detect the split characters. Currently I have developed the following solution:
String str = "123,45";
String splitChars = "";
for(int i=0; i < str.length(); i++) {
if(!Character.isDigit(str.charAt(i))) {
splitChars += str.charAt(i);
}
}
String[] endpoints = str.split(splitChars);
Basically I pick the first row and select all the non-numeric characters, then I use the generated substring as split characters. Is there a cleaner way to perform this?

Split requires a regexp, so your code would fail for many reasons: If the separator has meaning in regexp (say, +), it'll fail. If there is more than 1 non-digit character, your code will also fail. If you code contains more than exactly 2 numbers, it will also fail. Imagine it contains hello, world - then your splitChars string becomes " , " - and your split would do nothing (that would split the string "test , abc" into two, nothing else).
Why not make a regexp to fetch digits, and then find all sequences of digits, instead of focussing on the separators?
You're using regexps whether you want to or not, so let's make it official and use Pattern, while we are at it.
private static final Pattern ALL_DIGITS = Pattern.compile("\\d+");
// then in your split method..
Matcher m = ALL_DIGITS.matcher(str);
List<Integer> numbers = new ArrayList<Integer>();
// dont use arrays, generally. List is better.
while (m.find()) {
numbers.add(Integer.parseInt(m.group(0)));
}
//d+ is: Any number of digits.
m.find() finds the next match (so, the next block of digits), returning false if there aren't any more.
m.group(0) retrieves the entire matched string.

Split the string on \\D+ which means one or more non-digit characters.
Demo:
import java.util.Arrays;
public class Main {
public static void main(String[] args) {
// Test strings
String[] arr = { "123,45", "67,89", "125 89", "678 129" };
for (String s : arr) {
System.out.println(Arrays.toString(s.split("\\D+")));
}
}
}
Output:
[123, 45]
[67, 89]
[125, 89]
[678, 129]

Why not split with [^\d]+ (every group of nondigfit) :
for (String n : "123,456 789".split("[^\\d]+")) {
System.out.println(n);
}
Result:
123
456
789

How to return only first n number of words in a sentence Java

Say i have a simple sentence as below.
For example, this is what have:
A simple sentence consists of only one clause. A compound sentence
consists of two or more independent clauses. A complex sentence has at
least one independent clause plus at least one dependent clause. A set
of words with no independent clause may be an incomplete sentence,
also called a sentence fragment.
I want only first 10 words in the sentence above.
I'm trying to produce the following string:
A simple sentence consists of only one clause. A compound
I tried this:
bigString.split(" " ,10).toString()
But it returns the same bigString wrapped with [] array.
Thanks in advance.

Assume bigString : String equals your text. First thing you want to do is split the string in single words.
String[] words = bigString.split(" ");
How many words do you like to extract?
int n = 10;
Put words together
String newString = "";
for (int i = 0; i < n; i++) { newString = newString + " " + words[i];}
System.out.println(newString);
Hope this is what you needed.
If you want to know more about regular expressions (i.e. to tell java where to split), see here: How to split a string in Java

If you use the split-Method with a limiter (yours is 10) it won't just give you the first 10 parts and stop but give you the first 9 parts and the 10th place of the array contains the rest of the input String. ToString concatenates all Strings from the array resulting in the whole input String. What you can do to achieve what you initially wanted is:
String[] myArray = bigString.split(" " ,11);
myArray[10] = ""; //setting the rest to an empty String
myArray.toString(); //This should give you now what you wanted but surrouned with array so just cut that off iterating the array instead of toString or something.

This will help you
String[] strings = Arrays.stream(bigstring.split(" "))
.limit(10)
.toArray(String[]::new);

Here is exactly what you want:
String[] result = new String[10];
// regex \s matches a whitespace character: [ \t\n\x0B\f\r]
String[] raw = bigString.split("\\s", 11);
// the last entry of raw array is the whole sentence, need to be trimmed.
System.arraycopy(raw, 0, result , 0, 10);
System.out.println(Arrays.toString(result));

split a string in java into equal length substrings while maintaining word boundaries

How to split a string into equal parts of maximum character length while maintaining word boundaries?
Say, for example, if I want to split a string "hello world" into equal substrings of maximum 7 characters it should return me
"hello "
and
"world"
But my current implementation returns
"hello w"
and
"orld "
I am using the following code taken from Split string to equal length substrings in Java to split the input string into equal parts
public static List<String> splitEqually(String text, int size) {
// Give the list the right capacity to start with. You could use an array
// instead if you wanted.
List<String> ret = new ArrayList<String>((text.length() + size - 1) / size);
for (int start = 0; start < text.length(); start += size) {
ret.add(text.substring(start, Math.min(text.length(), start + size)));
}
return ret;
}
Will it be possible to maintain word boundaries while splitting the string into substring?
To be more specific I need the string splitting algorithm to take into account the word boundary provided by spaces and not solely rely on character length while splitting the string although that also needs to be taken into account but more like a max range of characters rather than a hardcoded length of characters.

If I understand your problem correctly then this code should do what you need (but it assumes that maxLenght is equal or greater than longest word)
String data = "Hello there, my name is not importnant right now."
+ " I am just simple sentecne used to test few things.";
int maxLenght = 10;
Pattern p = Pattern.compile("\\G\\s*(.{1,"+maxLenght+"})(?=\\s|$)", Pattern.DOTALL);
Matcher m = p.matcher(data);
while (m.find())
System.out.println(m.group(1));
Output:
Hello
there, my
name is
not
importnant
right now.
I am just
simple
sentecne
used to
test few
things.
Short (or not) explanation of "\\G\\s*(.{1,"+maxLenght+"})(?=\\s|$)" regex:
(lets just remember that in Java \ is not only special in regex, but also in String literals, so to use predefined character sets like \d we need to write it as "\\d" because we needed to escape that \ also in string literal)
\G - is anchor representing end of previously founded match, or if there is no match yet (when we just started searching) beginning of string (same as ^ does)
\s* - represents zero or more whitespaces (\s represents whitespace, * "zero-or-more" quantifier)
(.{1,"+maxLenght+"}) - lets split it in more parts (at runtime :maxLenght will hold some numeric value like 10 so regex will see it as .{1,10})
. represents any character (actually by default it may represent any character except line separators like \n or \r, but thanks to Pattern.DOTALL flag it can now represent any character - you may get rid of this method argument if you want to start splitting each sentence separately since its start will be printed in new line anyway)
{1,10} - this is quantifier which lets previously described element appear 1 to 10 times (by default will try to find maximal amout of matching repetitions),
.{1,10} - so based on what we said just now, it simply represents "1 to 10 of any characters"
( ) - parenthesis create groups, structures which allow us to hold specific parts of match (here we added parenthesis after \\s* because we will want to use only part after whitespaces)
(?=\\s|$) - is look-ahead mechanism which will make sure that text matched by .{1,10} will have after it:
space (\\s)
OR (written as |)
end of the string $ after it.
So thanks to .{1,10} we can match up to 10 characters. But with (?=\\s|$) after it we require that last character matched by .{1,10} is not part of unfinished word (there must be space or end of string after it).

Non-regex solution, just in case someone is more comfortable (?) not using regular expressions:
private String justify(String s, int limit) {
StringBuilder justifiedText = new StringBuilder();
StringBuilder justifiedLine = new StringBuilder();
String[] words = s.split(" ");
for (int i = 0; i < words.length; i++) {
justifiedLine.append(words[i]).append(" ");
if (i+1 == words.length || justifiedLine.length() + words[i+1].length() > limit) {
justifiedLine.deleteCharAt(justifiedLine.length() - 1);
justifiedText.append(justifiedLine.toString()).append(System.lineSeparator());
justifiedLine = new StringBuilder();
}
}
return justifiedText.toString();
}
Test:
String text = "Long sentence with spaces, and punctuation too. And supercalifragilisticexpialidocious words. No carriage returns, tho -- since it would seem weird to count the words in a new line as part of the previous paragraph's length.";
System.out.println(justify(text, 15));
Output:
Long sentence
with spaces,
and punctuation
too. And
supercalifragilisticexpialidocious
words. No
carriage
returns, tho --
since it would
seem weird to
count the words
in a new line
as part of the
previous
paragraph's
length.
It takes into account words that are longer than the set limit, so it doesn't skip them (unlike the regex version which just stops processing when it finds supercalifragilisticexpialidosus).
PS: The comment about all input words being expected to be shorter than the set limit, was made after I came up with this solution ;)

Java String Regex Divide - Always the Same Pattern

I never understood how to make properly regex to divide my Strings.
I have this types of Strings example = "on[?a, ?b, ?c]";
Sometimes I have this, Strings example2 = "not clear[?c]";
For the first Example I would like to divide into this:
[on, a, b, c]
or
String name = "on";
String [] vars = [a,b,c];
And for the second example I would like to divide into this type:
[not clear, c]
or
String name = "not clear";
String [] vars = [c];
Thanks alot in advance guys ;)

If you know the character set of your identifiers, you can simply do a split on all of the text that isn't in that set. For example, if your identifiers only consist of word characters ([a-zA-Z_0-9]) you can use:
String[] parts = "on[?a, ?b, ?c]".split("[\\W]+");
String name = parts[0];
String[] vars = Arrays.copyOfRange(parts, 1, parts.length);
If your identifiers only have A-Z (upper and lower) you could replace \\W above with ^A-Za-z.
I feel that this is more elegant than using a complex regular expression.
Edit: I realize that this will have issues with your second example "not clear". If you have no option of using something like an underscore instead of a space there, you could do one split on [? (or substring) to get the "name", and another split on the remainder, like so:
String s = "not clear[?a, ?b, ?c]";
String[] parts = s.split("\\[\\?"); //need the '?' so we don't get an extra empty array element in the next split
String name = parts[0];
String[] vars = parts[1].split("[\\W]+");

This comes close, but the problem is the third remembered group is actually repeated so it only captures the last match.
(.*?)\[(?:\s*(?:\?(.*?)(?:\s*,\s*\?(.*?))*)\s*)?]
For example, the first one you list on[?a, ?b, ?c] would give group 1 as on, 2 as a 3 as c. If you are using perl, you could the g flag to apply a regex to a line multiple times and use this:
my #tokens;
while ( my $line =~ /\s*(.*?)\s*[[,\]]/g ) {
push( #tokens, $1 );
}
Note, i did not actually test the perl code, just off the top of my head. It should give you the idea though

String[] parts = example.split("[^\\w ]");
List<String> x = new ArrayList<String>();
for (int i = 0; i < parts.length; i++) {
if (!"".equals(parts[i]) && !" ".equals(parts[i])) {
x.add(parts[i]);
}
}
This will work as long as you don't have more than one space separating your non-space characters. There's probably a cleverer way of filtering out the null and " " strings.

Need to split a string into two parts in java

I have a string which contains a contiguous chunk of digits and then a contiguous chunk of characters. I need to split them into two parts (one integer part, and one string).
I tried using String.split("\\D", 1), but it is eating up first character.
I checked all the String API and didn't find a suitable method.
Is there any method for doing this thing?

Use lookarounds: str.split("(?<=\\d)(?=\\D)")
String[] parts = "123XYZ".split("(?<=\\d)(?=\\D)");
System.out.println(parts[0] + "-" + parts[1]);
// prints "123-XYZ"
\d is the character class for digits; \D is its negation. So this zero-matching assertion matches the position where the preceding character is a digit (?<=\d), and the following character is a non-digit (?=\D).
References
regular-expressions.info/Lookarounds and Character Class
Related questions
Java split is eating my characters.
Is there a way to split strings with String.split() and include the delimiters?
Alternate solution using limited split
The following also works:
String[] parts = "123XYZ".split("(?=\\D)", 2);
System.out.println(parts[0] + "-" + parts[1]);
This splits just before we see a non-digit. This is much closer to your original solution, except that since it doesn't actually match the non-digit character, it doesn't "eat it up". Also, it uses limit of 2, which is really what you want here.
API links
String.split(String regex, int limit)
If the limit n is greater than zero then the pattern will be applied at most n - 1 times, the array's length will be no greater than n, and the array's last entry will contain all input beyond the last matched delimiter.

There's always an old-fashioned way:
private String[] split(String in) {
int indexOfFirstChar = 0;
for (char c : in.toCharArray()) {
if (Character.isDigit(c)) {
indexOfFirstChar++;
} else {
break;
}
}
return new String[]{in.substring(0,indexOfFirstChar), in.substring(indexOfFirstChar)};
}
(hope it works with digit-only or char-only Strings too - can't test it here - if not, take it as a general idea)