I am trying to split a given string using the java split method while the string should be devided by two different characters (+ and -) and I am willing to save the characters inside the array aswell in the same index the string has been saven.
for example :
input : String s = "4x^2+3x-2"
output :
arr[0] = 4x^2
arr[1] = +3x
arr[2] = -2
I know how to get the + or - characters in a different index between the numbers but it is not helping me,
any suggestions please?
You can face this problem in many ways. I´m sure there are clever and fancy ways to split this expression. I will show you the simplest problem-solving process that can help you.
State the problem you need to solve, the input and output
Problem: Split a math expression into subexpressions at + and - signals
Input: 4x^2+3x-2
Output: 4x^2,+3x,-2
Create a pseudo code with some logic you might think works
Given an expression string
Create an empty list of expressions
Create a subExpression string
For each character in the expression
Check if the character is + ou - then
add the subExpression in the list and create a new empty subexpression
otherwise, append the character in the subExpression
In the end, add the left subexpression in the list
Implement the pseudo-code in the programming language of your choice
String expression = "4x^2+3x-2";
List<String> expressions = new ArrayList();
StringBuilder subExpression = new StringBuilder();
for (int i = 0; i < expression.length(); i++) {
char character = expression.charAt(i);
if (character == '-' || character == '+') {
expressions.add(subExpression.toString());
subExpression = new StringBuilder(String.valueOf(character));
} else {
subExpression.append(String.valueOf(character));
}
}
expressions.add(subExpression.toString());
System.out.println(expressions);
Output
[4x^2, +3x, -2]
You will end with one algorithm that works for your problem. You can start to improve it.
Try this code:
String s = "4x^2+3x-2";
s = s.replace("+", "#+");
s = s.replace("-", "#-");
String[] ss = s.split("#");
for (int i = 0; i < ss.length; i++) {
Log.e("XOP",ss[i]);
}
This code replaces + and - with #+ and #- respectively and then splits the string with #. That way the + and - operators are not lost in the result.
If you require # as input character then you can use any other Unicode character instead of #.
Try this one:
String s = "4x^2+3x-2";
String[] arr = s.split("[\\+-]");
for(int i=0;i<arr.length;i++){
System.out.println(arr[i]);
}
Personally I like it better to have positive matches of patterns, especially if the split pattern itself is empty.
So for instance you could use a Pattern and Matcher like this:
Pattern p = Pattern.compile("(^|[+-])([^+-]*)");
Matcher m = p.matcher("4x^2+3x-2");
while (m.find()) {
System.out.printf("%s or %s %s%n", m.group(), m.group(1), m.group(2));
}
This matches the start of the string or a plus or minus: ^|[+-], followed by any amount of characters that are not a plus or minus: [^+-]*.
Do note that the ^ first matches the start of the string, and is then used to negate a character class when used between brackets. Regular expressions are tricky like that.
Bonus: you can also use the two groups (within the parenthesis in the pattern) to match the operators - if any.
All this is presuming that you want to use/test regular expressions; generally things like this require a parser rather than a regular expression.
A one-liner for persons thinking that this is too complex:
var expressions = Pattern.compile("^|[+-][^+-]*")
.matcher("4x^2+3x-2")
.results()
.map(r -> r.group())
.collect(Collectors.toList());
I have a string vaguely like this:
foo,bar,c;qual="baz,blurb",d;junk="quux,syzygy"
that I want to split by commas -- but I need to ignore commas in quotes. How can I do this? Seems like a regexp approach fails; I suppose I can manually scan and enter a different mode when I see a quote, but it would be nice to use preexisting libraries. (edit: I guess I meant libraries that are already part of the JDK or already part of a commonly-used libraries like Apache Commons.)
the above string should split into:
foo
bar
c;qual="baz,blurb"
d;junk="quux,syzygy"
note: this is NOT a CSV file, it's a single string contained in a file with a larger overall structure
Try:
public class Main {
public static void main(String[] args) {
String line = "foo,bar,c;qual=\"baz,blurb\",d;junk=\"quux,syzygy\"";
String[] tokens = line.split(",(?=(?:[^\"]*\"[^\"]*\")*[^\"]*$)", -1);
for(String t : tokens) {
System.out.println("> "+t);
}
}
}
Output:
> foo
> bar
> c;qual="baz,blurb"
> d;junk="quux,syzygy"
In other words: split on the comma only if that comma has zero, or an even number of quotes ahead of it.
Or, a bit friendlier for the eyes:
public class Main {
public static void main(String[] args) {
String line = "foo,bar,c;qual=\"baz,blurb\",d;junk=\"quux,syzygy\"";
String otherThanQuote = " [^\"] ";
String quotedString = String.format(" \" %s* \" ", otherThanQuote);
String regex = String.format("(?x) "+ // enable comments, ignore white spaces
", "+ // match a comma
"(?= "+ // start positive look ahead
" (?: "+ // start non-capturing group 1
" %s* "+ // match 'otherThanQuote' zero or more times
" %s "+ // match 'quotedString'
" )* "+ // end group 1 and repeat it zero or more times
" %s* "+ // match 'otherThanQuote'
" $ "+ // match the end of the string
") ", // stop positive look ahead
otherThanQuote, quotedString, otherThanQuote);
String[] tokens = line.split(regex, -1);
for(String t : tokens) {
System.out.println("> "+t);
}
}
}
which produces the same as the first example.
EDIT
As mentioned by #MikeFHay in the comments:
I prefer using Guava's Splitter, as it has saner defaults (see discussion above about empty matches being trimmed by String#split(), so I did:
Splitter.on(Pattern.compile(",(?=(?:[^\"]*\"[^\"]*\")*[^\"]*$)"))
While I do like regular expressions in general, for this kind of state-dependent tokenization I believe a simple parser (which in this case is much simpler than that word might make it sound) is probably a cleaner solution, in particular with regards to maintainability, e.g.:
String input = "foo,bar,c;qual=\"baz,blurb\",d;junk=\"quux,syzygy\"";
List<String> result = new ArrayList<String>();
int start = 0;
boolean inQuotes = false;
for (int current = 0; current < input.length(); current++) {
if (input.charAt(current) == '\"') inQuotes = !inQuotes; // toggle state
else if (input.charAt(current) == ',' && !inQuotes) {
result.add(input.substring(start, current));
start = current + 1;
}
}
result.add(input.substring(start));
If you don't care about preserving the commas inside the quotes you could simplify this approach (no handling of start index, no last character special case) by replacing your commas in quotes by something else and then split at commas:
String input = "foo,bar,c;qual=\"baz,blurb\",d;junk=\"quux,syzygy\"";
StringBuilder builder = new StringBuilder(input);
boolean inQuotes = false;
for (int currentIndex = 0; currentIndex < builder.length(); currentIndex++) {
char currentChar = builder.charAt(currentIndex);
if (currentChar == '\"') inQuotes = !inQuotes; // toggle state
if (currentChar == ',' && inQuotes) {
builder.setCharAt(currentIndex, ';'); // or '♡', and replace later
}
}
List<String> result = Arrays.asList(builder.toString().split(","));
http://sourceforge.net/projects/javacsv/
https://github.com/pupi1985/JavaCSV-Reloaded
(fork of the previous library that will allow the generated output to have Windows line terminators \r\n when not running Windows)
http://opencsv.sourceforge.net/
CSV API for Java
Can you recommend a Java library for reading (and possibly writing) CSV files?
Java lib or app to convert CSV to XML file?
I would not advise a regex answer from Bart, I find parsing solution better in this particular case (as Fabian proposed). I've tried regex solution and own parsing implementation I have found that:
Parsing is much faster than splitting with regex with backreferences - ~20 times faster for short strings, ~40 times faster for long strings.
Regex fails to find empty string after last comma. That was not in original question though, it was mine requirement.
My solution and test below.
String tested = "foo,bar,c;qual=\"baz,blurb\",d;junk=\"quux,syzygy\",";
long start = System.nanoTime();
String[] tokens = tested.split(",(?=([^\"]*\"[^\"]*\")*[^\"]*$)");
long timeWithSplitting = System.nanoTime() - start;
start = System.nanoTime();
List<String> tokensList = new ArrayList<String>();
boolean inQuotes = false;
StringBuilder b = new StringBuilder();
for (char c : tested.toCharArray()) {
switch (c) {
case ',':
if (inQuotes) {
b.append(c);
} else {
tokensList.add(b.toString());
b = new StringBuilder();
}
break;
case '\"':
inQuotes = !inQuotes;
default:
b.append(c);
break;
}
}
tokensList.add(b.toString());
long timeWithParsing = System.nanoTime() - start;
System.out.println(Arrays.toString(tokens));
System.out.println(tokensList.toString());
System.out.printf("Time with splitting:\t%10d\n",timeWithSplitting);
System.out.printf("Time with parsing:\t%10d\n",timeWithParsing);
Of course you are free to change switch to else-ifs in this snippet if you feel uncomfortable with its ugliness. Note then lack of break after switch with separator. StringBuilder was chosen instead to StringBuffer by design to increase speed, where thread safety is irrelevant.
You're in that annoying boundary area where regexps almost won't do (as has been pointed out by Bart, escaping the quotes would make life hard) , and yet a full-blown parser seems like overkill.
If you are likely to need greater complexity any time soon I would go looking for a parser library. For example this one
I was impatient and chose not to wait for answers... for reference it doesn't look that hard to do something like this (which works for my application, I don't need to worry about escaped quotes, as the stuff in quotes is limited to a few constrained forms):
final static private Pattern splitSearchPattern = Pattern.compile("[\",]");
private List<String> splitByCommasNotInQuotes(String s) {
if (s == null)
return Collections.emptyList();
List<String> list = new ArrayList<String>();
Matcher m = splitSearchPattern.matcher(s);
int pos = 0;
boolean quoteMode = false;
while (m.find())
{
String sep = m.group();
if ("\"".equals(sep))
{
quoteMode = !quoteMode;
}
else if (!quoteMode && ",".equals(sep))
{
int toPos = m.start();
list.add(s.substring(pos, toPos));
pos = m.end();
}
}
if (pos < s.length())
list.add(s.substring(pos));
return list;
}
(exercise for the reader: extend to handling escaped quotes by looking for backslashes also.)
Try a lookaround like (?!\"),(?!\"). This should match , that are not surrounded by ".
The simplest approach is not to match delimiters, i.e. commas, with a complex additional logic to match what is actually intended (the data which might be quoted strings), just to exclude false delimiters, but rather match the intended data in the first place.
The pattern consists of two alternatives, a quoted string ("[^"]*" or ".*?") or everything up to the next comma ([^,]+). To support empty cells, we have to allow the unquoted item to be empty and to consume the next comma, if any, and use the \\G anchor:
Pattern p = Pattern.compile("\\G\"(.*?)\",?|([^,]*),?");
The pattern also contains two capturing groups to get either, the quoted string’s content or the plain content.
Then, with Java 9, we can get an array as
String[] a = p.matcher(input).results()
.map(m -> m.group(m.start(1)<0? 2: 1))
.toArray(String[]::new);
whereas older Java versions need a loop like
for(Matcher m = p.matcher(input); m.find(); ) {
String token = m.group(m.start(1)<0? 2: 1);
System.out.println("found: "+token);
}
Adding the items to a List or an array is left as an excise to the reader.
For Java 8, you can use the results() implementation of this answer, to do it like the Java 9 solution.
For mixed content with embedded strings, like in the question, you can simply use
Pattern p = Pattern.compile("\\G((\"(.*?)\"|[^,])*),?");
But then, the strings are kept in their quoted form.
what about a one-liner using String.split()?
String s = "foo,bar,c;qual=\"baz,blurb\",d;junk=\"quux,syzygy\"";
String[] split = s.split( "(?<!\".{0,255}[^\"]),|,(?![^\"].*\")" );
A regular expression is not capable of handling escaped characters. For my application, I needed the ability to escape quotes and spaces (my separator is spaces, but the code is the same).
Here is my solution in Kotlin (the language from this particular application), based on the one from Fabian Steeg:
fun parseString(input: String): List<String> {
val result = mutableListOf<String>()
var inQuotes = false
var inEscape = false
val current = StringBuilder()
for (i in input.indices) {
// If this character is escaped, add it without looking
if (inEscape) {
inEscape = false
current.append(input[i])
continue
}
when (val c = input[i]) {
'\\' -> inEscape = true // escape the next character, \ isn't added to result
',' -> if (inQuotes) {
current.append(c)
} else {
result += current.toString()
current.clear()
}
'"' -> inQuotes = !inQuotes
else -> current.append(c)
}
}
if (current.isNotEmpty()) {
result += current.toString()
}
return result
}
I think this is not a place to use regular expressions. Contrary to other opinions, I don't think a parser is overkill. It's about 20 lines and fairly easy to test.
Rather than use lookahead and other crazy regex, just pull out the quotes first. That is, for every quote grouping, replace that grouping with __IDENTIFIER_1 or some other indicator, and map that grouping to a map of string,string.
After you split on comma, replace all mapped identifiers with the original string values.
I would do something like this:
boolean foundQuote = false;
if(charAtIndex(currentStringIndex) == '"')
{
foundQuote = true;
}
if(foundQuote == true)
{
//do nothing
}
else
{
string[] split = currentString.split(',');
}
I'm trying to convert a string array that is itself a part of another array fed into Java from an external file.
There are two parts to this question:
How do I convert the string's substring elements to doubles or ints?
How do I skip the header which is itself a part of the string?
I have the following piece of code that is NOT giving me an error but neither is it giving me output. The data is arranged in columns, so as far as the split, I'm not sure what delimiter to use as the argument for that method. I've tried \r, \n, ",", " " and nothing works.
str0 = year.split(",");
year = year.trim();
int[] yearData = new int[str0.length-1];
for(i = 0; i < str0.length-1; i++) {
yearData[i] = Integer.parseInt(str0[i]);
System.out.println(yearData[i]);
}
The code you have provided is not working. Anyway consider the given example which is using Regular Expression, where you found all the numbers in the string, so our regular expression works well. By changing the Regular Expression you can get the substring as well as you can skip the head part. I hope it would help.
String regEx = "[+|-]?(\\d+(\\.\\d*)?)|(\\.\\d+)";
String str = "256 is the square of 16 and -2.5 squared is 6.25 “ + “and -.243 is less than 0.1234.";
Pattern pattern = Pattern.compile(regEx);
Matcher m = pattern.matcher(str);
int i = 0;
String subStr = null;
while(m.find()) {
System.out.println(m.group());
Try something like this:
year = year.trim(); // This should come before the split()...
str0 = year.split("[\\s,;]+"); // split() uses RegEx...
int[] yearData = new int[str0.length-1];
for(i = 0; i < str0.length-1; i++) {
yearData[i] = Integer.parseInt(str0[i]);
System.out.println(yearData[i]);
}
If I have a piece of text around 3000 characters. I want search for strings with certain characteristics for example strings like [*].
That is, I want to get [a] and [bc] from
sjfhshdkfjhskdhfksdf[a]sfdsgfsdf[bc]
I know there is an algorithm called KMP that guarantee a linear time searching operation through a text, but here I don't have a fixed string to be found, maybe I have to use some regular expression at some place.
How can I do this better than O(n^2)? Is there any light libraries for this if I'm using java?
No libraries needed, you've effectively described a use case for regex! They are highly optimized for searching, and in this case will be O(n).
String str = "sjfhshdkfjhskdhfksdf[a]sfdsgfsdf[bc]";
List<String> allMatches = new ArrayList<>();
Matcher m = Pattern.compile("\\[[^\\]]*]").matcher(str);
while (m.find()) {
allMatches.add(m.group());
}
Regex Demo
If you have any doubts though and really want some O(n) that you can see, here's an algorithm:
String str = "sjfhshdkfjhskdhfksdf[a]sfdsgfsdf[bc]";
List<String> allMatches = new ArrayList<>();
for (int i = str.indexOf('['), j; i != -1; i = str.indexOf('[', j + 1)) {
j = str.indexOf(']', i + 1);
// if `j` is -1, the brackets are unbalanced. Perhaps throw an Exception?
allMatches.add(str.substring(i, j + 1));
}
Here's how to do it in one line:
String[] hits = str.replaceAll("^.*?\\[|][^\\]]*$", "").split("].*?\\[");
This works by stripping off leading and trailing chars up to and including the first/last opening/closing square bracket, then splits on a close bracket to the next opening bracket (inclusive).
I have a code that works but is extremely slow. This code determines whether a string contains a keyword. The requirements I have need to be efficient for hundreds of keywords that I will search for in thousands of documents.
What can I do to make finding the keywords (without falsely returning a word that contains the keyword) efficiently?
For example:
String keyword="ac";
String document"..." //few page long file
If i use :
if(document.contains(keyword) ){
//do something
}
It will also return true if document contains a word like "account";
so I tried to use regular expression as follows:
String pattern = "(.*)([^A-Za-z]"+ keyword +"[^A-Za-z])(.*)";
Pattern r = Pattern.compile(pattern);
Matcher m = r.matcher(document);
if(m.find()){
//do something
}
Summary:
This is the summary: Hopefully it will be useful to some one else:
My regular expression would work but extremely impractical while
working with big data. (it didn't terminate)
#anubhava perfected the regular expression. it was easy to
understand and implement. It managed to terminate which is a big
thing. but it was still a bit slow. (Roughly about 240 seconds)
#Tomalak solution is abit complex to implement and understand but it
was the fastest solution. so hats off mate.(18 seconds)
so #Tomalak solution was ~15 times faster than #anubhava.
Don't think you need to have .* in your regex.
Try this regex:
String pattern = "\\b"+ Pattern.quote(keyword) + "\\b";
Here \\b is used for word boundary. If the keyword can contain special characters, make sure they are not at the start or end of the word, or the word boundaries will fail to match.
Also you must be using Pattern.quote if your keyword contains special regex characters.
EDIT: You might use this regex if your keywords are separated by space.
String pattern = "(?<=\\s|^)"+ Pattern.quote(keyword) + "(?=\\s|$)";
The fastest-possible way to find substrings in Java is to use String.indexOf().
To achieve "entire-word-only" matches, you would need to add a little bit of logic to check the characters before and after a possible match to make sure they are non-word characters:
public class IndexOfWordSample {
public static void main(String[] args) {
String input = "There are longer strings than this not very long one.";
String search = "long";
int index = indexOfWord(input, search);
if (index > -1) {
System.out.println("Hit for \"" + search + "\" at position " + index + ".");
} else {
System.out.println("No hit for \"" + search + "\".");
}
}
public static int indexOfWord(String input, String word) {
String nonWord = "^\\W?$", before, after;
int index, before_i, after_i = 0;
while (true) {
index = input.indexOf(word, after_i);
if (index == -1 || word.isEmpty()) break;
before_i = index - 1;
after_i = index + word.length();
before = "" + (before_i > -1 ? input.charAt(before_i) : "");
after = "" + (after_i < input.length() ? input.charAt(after_i) : "");
if (before.matches(nonWord) && after.matches(nonWord)) {
return index;
}
}
return -1;
}
}
This would print:
Hit for "long" at position 44.
This should perform better than a pure regular expressions approach.
Think if ^\W?$ already matches your expectation of a "non-word" character. The regular expression is a compromise here and may cost performance if your input string contains many "almost"-matches.
For extra speed, ditch the regex and work with the Character class, checking a combination of the many properties it provides (like isAlphabetic, etc.) for before and after.
I've created a Gist with an alternative implementation that does that.