How to extract multiple integers from String in Java?

Question

I got a series of lines like "(123, 234; 345, 456) (567, 788; 899, 900)" . "(123, 234; 345, 456) (567, 788; 899, 900)" How to extract these numbers into an array like aArray[0]=123, aArray=[234], ....aArray[8]=900;

thanks

Answer 1

This is probably too confusing, but the hay ...

The first thing we need to do is remove all the crap we don't need ...

 String[] crap = {"(", ")", ",", ";"}; String text = "(123, 234; 345, 456) (567, 788; 899, 900)"; for (String replace : crap) { text = text.replace(replace, " ").trim(); } // This replaces any multiple spaces with a single space while (text.contains(" ")) { text = text.replace(" ", " "); } 

Next, we need to separate the individual elements of the string into a more manageable form.

 String[] values = text.split(" "); 

Then we need to convert each String value to an int value

 int[] iValues = new int[values.length]; for (int index = 0; index < values.length; index++) { String sValue = values[index]; iValues[index] = Integer.parseInt(values[index].trim()); } 

Then we print the values ​​...

 for (int value : iValues) { System.out.println(value); } 

Answer 2

Strategy: Find one or more numbers that are together, through a regular expression to be added to the list.

Code:

  LinkedList<String> list = new LinkedList<>(); Matcher matcher = Pattern.compile("\\d+").matcher("(123, 234; 345, 456) (567, 788; 899, 900)"); while (matcher.find()) { list.add(matcher.group()); } String[] array = list.toArray(new String[list.size()]); System.out.println(Arrays.toString(array)); 

Output:

 [123, 234, 345, 456, 567, 788, 899, 900] 

Answer 3

You almost certainly saw a quote:

Some people, faced with a problem, think: "I know, I will use regular expressions." Now they have two problems.

But regular expressions really are your friend for these kinds of things.

 import java.util.ArrayList; import java.util.List; import java.util.regex.Matcher; import java.util.regex.Pattern; class Numbers { public static void main(String[] args) { String s = "(123, 234; 345, 456) (567, 788; 899, 900)"; Matcher m = Pattern.compile("\\d+").matcher(s); List<Integer> numbers = new ArrayList<Integer>(); while(m.find()) { numbers.add(Integer.parseInt(m.group())); } System.out.println(numbers); } } 

Outputs:

 [123, 234, 345, 456, 567, 788, 899, 900] 

Answer 4

Swipe through each character and store the numbers in a temporary array until you find a character (e.g.,,, ; ), then save the data from the temporary array to your array and then delete this temporary array for next use.

Answer 5

Since your numbers are separated by a specific character set, you can take a look at the .split(String regex) method.

Answer 6

Since there may be several different delimiters in your string, you can go through it and replace all the obscure characters with spaces . Then you can use split("\\s") to split your string into an array of substrings with numbers. And finally, convert them to numbers.

Answer 7

This method will extract integers from the given string. It also processes strings in which other characters are used to separate numbers, and not just those specified in your example:

 public static Integer[] extractIntegers( final String source ) { final int length = source.length(); final char[] chars = source.toCharArray(); final List< Integer > list = new ArrayList< Integer >(); for ( int i = 0; i < length; i++ ) { // Find the start of an integer: it must be a digit or a sign character if ( chars[ i ] == '-' || chars[ i ] == '+' || Character.isDigit( chars[ i ] ) ) { final int start = i; // Find the end of the integer: for ( i++; i < length && Character.isDigit( chars[ i ] ); i++ ) ; // Now extract this integer: list.add( Integer.valueOf( source.substring( start, i ) ) ); } } return list.toArray( new Integer[ list.size() ] ); } 

Note: Since the inner positions of the for loop are after the integer, and the outer for loop increments the variable i when searching for the next integer, the algorithm will require at least one character to separate the integers, but I think this is desirable. For example, the source "-23-12" will produce the numbers [ -23, 12 ] and NOT [ -23, -12 ] (but the "-23 -12" will generate [-23, -12] as expected).

Answer 8

The easiest way is to use a combination of String.indexOf() (or something like this) and NumberFormat.parse(ParsePosition) . The algorithm will follow:

• start at the beginning of the line
• find a number starting at this position
• analyze the NumberFormat method used, which will stop at a non-digital character and return a value
• repeat 2), starting from this position (to the end of the line)

At the same time, the string has a certain structure, so IMHO some parser will be better suited, since it will also check the correctness of the format (if necessary, I don’t know). There are many tools for generating Java code from a grammar description (e.g. ANTLR, etc.). But this may be too complicated a solution for the occasion.

Answer 9

I think you can use regex to get the result. Something like this is possible:

 String string = "(123, 234; 345, 456) (567, 788; 899, 900)"; String[] split = string.split("[^\\d]+"); int number; ArrayList<Integer> numberList = new ArrayList<Integer>(); for(int index = 0; index < split.length; index++){ try{ number = Integer.parseInt(split[index]); numberList.add(number); }catch(Exception exe){ } } Integer[] numberArray = numberList.toArray(new Integer[numberList.size()]); for(int index = 0; index < numberArray.length; index++){ System.out.println(numberArray[index]); } 

Answer 10

Another way. Maybe it’s good if you want to write less code, it can be bad if you can’t add libraries to your project.

 import com.google.common.base.Splitter; import com.google.common.collect.Iterables; public static void main(String[] args) throws IOException { String text = "(123, 234; 345, 456) (567, 788; 899, 900)"; Splitter splitter = Splitter.onPattern("[,;\\)\\(]").omitEmptyStrings(); String[] cleanString = Iterables.toArray(splitter.split(text), String.class); System.out.println(Arrays.toString(cleanString)); } 

AM sure gurus can cleanse it further.

Answer 11

  for (int i = 0; i < faces.total(); i++) { CvRect r = new CvRect(cvGetSeqElem("(123, 234; 345, 456)", i)); String x=""+Integer.toString(rx()); String y=""+Integer.toString(ry()); String w=""+Integer.toString(r.width()); String h=""+Integer.toString(r.height()); for(int j=0;j<(4-Integer.toString(rx()).length());j++) x="0"+x; for(int j=0;j<(4-Integer.toString(ry()).length());j++) y="0"+y; for(int j=0;j<(4-Integer.toString(r.width()).length());j++) w="0"+w; for(int j=0;j<(4-Integer.toString(r.height()).length());j++) h="0"+h; r_return=""+x+y+w+h; } 

Above the code will return the string "0123023403540456"

 int[] rectArray = new int[rectInfo.length()/4]; for(int i=0;i<rectInfo.length()/4; i++) { rectArray[i]=Integer.valueOf(rectInfo.substring(i*4, i*4+4)); } 

and he will receive [123, 234, 345, 456]