Converting a string into a double[][] - Java

Question

Since Java doesn't have an eval() module , and I want to write my own regex to parse strings into a double[][] , eg

[in]:

`{{1.23,8.4},{92.12,-0.57212}}`
`{{1.23,-8.4}, {-92.12,-0.57212}}`

[code]:

 double[][] xArr;
 // Somehow read the string into something like this:
 xArr = new double[][] {{1.23,8.4},{92.12,-0.57212}};
 System.out.println(xArr[0][0] + " " + xArr[0][1]);

[out]:

 1.23 -8.4

Currently, I'm doing it as such:

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

import java.util.ArrayList;
import java.util.List;

public class HelloWorld {

    public static void main(String[] args) {
        String s = "{{1.23,8.4}, {92.12,-0.57212}}";

        Pattern regex = Pattern.compile("((-)?\\d+(?:\\.\\d+)?)");
        Matcher matcher = regex.matcher(s);
        List<double[]> locations = new ArrayList<double[]>();
        int i = 0;
        while(matcher.find()){
            double d1 = Double.parseDouble(matcher.group(1));
            matcher.find();
            double d2 = Double.parseDouble(matcher.group(1));
            locations.add(new double[] {d1, d2});
            i++;        
        }; 


    }

}

Is there a better way to do this? Ie:

• Now, the code is sort of cheating by know that my inner size of the double[][] is 2 and during iteration through match.find() . It does 2 passes to skip to the next pair, is there a way to change the regex such that it extracts 2 groups at a time?

• Currently it's reading into the d1 and d2 variable before create a new double[] to add to the List , is there a way to do it directly without creating d1 and d2 ?

Answer 1

Use jackson but you will have to replace the braces with boxes/parenthesis.

With this you don't need to specify the dimensions of the expected array

public static void main(String[] args) throws IOException {
    String jsonString = "{{1.23,8.4}, {92.12,-0.57212}}";
    jsonString = jsonString.replace("{", "[").replace("}", "]");
    Double[][] doubles = new ObjectMapper().readValue(jsonString, Double[][].class);
    System.out.println(Arrays.deepToString(doubles));
}

Answer 2

Here is a Java 8 solution I came up with:

String test = "{{1.23,8.4},{92.12,-0.57212}}";

double[][] vals = Arrays.stream(test.replaceAll(" ", "").split("},\\{"))
                        .map(str -> str.replaceAll("[{}]", "").split(","))
                        .map(Arrays::stream)
                        .map(stream -> stream.mapToDouble(Double::parseDouble)
                                             .toArray())
                        .toArray(double[][]::new);

System.out.println(Arrays.deepToString(vals));

Output:

[[1.23, 8.4], [92.12, -0.57212]]

Or if you want a Double[][] :

Double[][] vals = Arrays.stream(test.replaceAll(" ", "").split("},\\{"))
                        .map(str -> str.replaceAll("[{}]", "").split(","))
                        .map(Arrays::stream)
                        .map(stream -> stream.map(Double::parseDouble)
                                             .toArray(Double[]::new))
                        .toArray(Double[][]::new);

Explanation:

First any whitespace is removed and the string is split on the pattern },\\\\{ which will result in a String[] where each String is one of the double[] s with some excess curly braces:

["{{1.23,8.4", "92.12,-0.57212}}"]

Then, for each String the curly braces are removed, and the String is split again. So each String becomes a String[] where each value is the String representation of a " double ":

[["1.23", "8.4"],["92.12", "-0.57212"]]

These strings are then parsed into doubles, and everything is collected into a double[] and then a double[][] :

[[1.23, 8.4], [92.12, -0.57212]]

Answer 3

1. Question:

   is there a way to change the regex such that it extracts 2 groups at a time

Yes. You could extract a whole line with one pattern, by using your existing one. You have to add the paranthesis and the komma and a the + operator (for multiple values). When you have a single line, you can extract the values with your current pattern, until there are no more found (with a while loop).

2. Question:

   is there a way to do it directly without creating d1 and d2?

I don't know exactly what you mean. Where is the problem with creating them? With the suggested way on question 1, you could store every value directly in a list as well and hen create a array from it.

I like the solution of Olayinka, too if your format dont change.