CODEWITHSUNDEEP
javaregexfileparsing
user3025403
user3025403

Reputation: 1080

Java - Ideas to turn strings into numbers (RegEx/Parsing/etc.)?

I'm reading in from a file, and the input is like this:

Description (1.0,2.0) (2,7.6) (2.1,3.0)
Description2 (4,1)
...
Description_n (4,18) (8, 7.20)

I want to be able to take the numbers inside parentheses and use turn them from strings into numbers so that I can do mathematical operations of them. Right now, to simplify things, my code only reads in the first line and then splits it based on spaces:

BufferedReader reader = null;

try {
    reader = new BufferedReader(new FileReader(new File("filename.txt")));

    //reader reads in the first line
    String firstLine = reader.readLine();

    //splits into an array of ["Description","(1.0,2.0)","(2,7.6)","(2.1,3.0)"]
    String[] parts = first.split(" "); 

    //now I want to store 1.0, 2, and 2.1 in one array as ints and 2.0, 7.6, and 3.0 in another int array

} catch (Exception e) {
    System.exit(0); 
}

What are some ways I can store the numbers inside parentheses into two separate arrays of ints (see comment above)? Should I use regular expressions to somehow capture something of the form "( [1-9.] , [1-9.] )" and then pass those into another function that will then separate the first number in the pair from the second and then convert them both into integers? I'm new to regular expression parsing in Java, so I'm not sure how to implement this.

Or is there a simply, better way to do this?

Upvotes: 0

Views: 122

Answers (4)

tgkprog
tgkprog

Reputation: 4598

I would use a String Tokenizer.But need more information and thought for full impl.

This is your line : "Description (1.0,2.0) (2,7.6) (2.1,3.0)"

First thing - can there be cases without parenthesis? Will there always be sets f 2,2,2 numbers ?

Do you want to take care of errors at each number or just skip the line or skip processing if there is an error (like number of numbers does not match?).

Now you need a data structure to hold numbers. You could make a class to hold each individual element in a seperate property if each number has a distinct meaning in the domain or have an array list or simple array if you want to treat them as a simple list of numbers. If a class one sample (incopmplete):

class LineItem{

}

Now to actually break up the string there are many ways to do it. Really depends on the quality of data and how you want to deal with possible errors

One way is find the first opening parenthesis( take rest of string and parse out using a String Tokenizer.

Something like:

int i = str.indexOf("(");
String s2 = str.substring(i);
StringTokenizer st = new StringTokenizer(s2, "() ,";//parenthesis, comma and space

ArrayList<Double> lineVals1 = new ArrayList<Double>();
ArrayList<Double> lineVals1 = new ArrayList<Double>();
int cnt = 0;
while(st.hasMoreTokens()){
    cnt++;//use this to keep count of how many numbers you got in line and raise error if need be
    String stemp = st.nextToken();
    if(isNumeric(stemo)){
         if(cnt % 2 == 1){
             lineVals1.add(Double.parseDouble(stemp));
         }else{
            lineVals2.add(Double.parseDouble(stemp));    
         }    
    }else{
         /raise error if not numberic
    }
}


public static boolean isNumeric(String str)
{
  NumberFormat formatter = NumberFormat.getInstance();
  ParsePosition pos = new ParsePosition(0);
  formatter.parse(str, pos);
  return str.length() == pos.getIndex();
}

Upvotes: 0

aliteralmind
aliteralmind

Reputation: 20163

This stores the numbers into Double-arrays (not two-dimensional arrays, arrays of Double objects), since some have .#. int-arrays would eliminate the post decimal part.

It uses the regex \b([\d.]+)\b to find each number within each paren-group, adding each to an ArrayList<Double>. Note that it assumes all input is perfect (nothing like (bogus,3.2). The list is then translated into an array of Double objects.

This should give you a good start towards your goal.

import  java.util.Arrays;
import  java.util.ArrayList;
import  java.util.regex.Matcher;
import  java.util.regex.Pattern;

/**
   <P>{@code java DoubleInParenStringsToArrays}</P>
 **/
public class DoubleInParenStringsToArrays  {
   public static final void main(String[] ignored)  {
      String input = "(1.0,2.0) (2,7.6) (2.1,3.0)";
      String[] inputs = input.split(" ");

      //"": Dummy string, to reuse matcher
      Matcher mtchrGetNums = Pattern.compile("\\b([\\d.]+)\\b").matcher("");
      for(String s : inputs)  {

         ArrayList<Double> doubleList = new ArrayList<Double>();
         mtchrGetNums.reset(s);
         while(mtchrGetNums.find())  {
            //TODO: Crash if it's not a number!
            doubleList.add(Double.parseDouble(mtchrGetNums.group(0)));
         }

         Double[] doubles = doubleList.toArray(new Double[doubleList.size()]);
         System.out.println(Arrays.toString(doubles));
      }
   }
}

Output:

[C:\java_code\]java DoubleInParenStringsToArrays
[1.0, 2.0]
[2.0, 7.6]
[2.1, 3.0]

Upvotes: 1

sn0wman
sn0wman

Reputation: 75

Just do Integer.parseInt(string), or Double.parseDouble(string), then add those to the array. I'm not really 100% sure what you're asking, though.

Upvotes: 1

Abiel Paltao
Abiel Paltao

Reputation: 315

How to parse per item:

Double.parseDouble("your string here");

As for the storing, I didnt get the pattern you want to store your values. What's the reason why you want 1.0, 2, and 2.1 in 1 array and 2.0, 7.6, and 3.0 to another?

Upvotes: 1

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