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I have a complex strings like the following:

$company=>Company(code->MyCompany)
$brand=>Brand(code->Brand_1, company->$company)

and then I prepare objects out of them as follows:

Macro [key=$company, type=Company, map={code=MyCompany}]
Macro [key=$brand, type=Brand, map={ company=$company, code=Brand_1}]

using the following code for parsing. Is there a better way?

public class Macro {

    private String key;

    private String type;

    private Map<String, String> map;
}

public static Macro getMacro(String str) {
    final Macro macro = new Macro();
    final Map<String, String> map = new HashMap<String, String>();
    final Matcher matcher = Pattern.compile("\\(([^)]+)\\)").matcher(str);
    while (matcher.find()) {
        final List<String> list = Arrays.asList(matcher.group(1).split(",")).stream()
                .collect(Collectors.toList());
        list.stream().forEach(a ->
        {
            final String[] ar = a.split("->");
            map.put(ar[0], ar[1]);
        });
    }
    str = str.replaceAll("\\(.*\\)", "");
    final String[] s = str.split("=>");
    macro.setMap(map);
    macro.setKey(s[0]);
    macro.setType(s[1]);
    return macro;
}
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3 Answers 3

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Pattern.compile("\\(([^)]+)\\)");

The Pattern is always the same so you may just as well make it a private static final constant. This will avoid having to recompile the pattern each time.


Arrays.asList(/*..*/).stream().collect(Collectors.toList());

Why this dance? Arrays.asList returns a List. You only do a stream on the return result.

    final List<String> list = Arrays.asList(matcher.group(1).split(","));
    list.stream().forEach(a ->
    {
        final String[] ar = a.split("->");
        map.put(ar[0], ar[1]);
    });

However a Pattern an return a Stream split according to the pattern directly:

Pattern COMMA_PATTERN = Pattern.compile("\\s*,\\s*");


COMMA_PATTERN.splitAsStream(matcher.group(3)).forEach(a ->
{
    //...
}

You can use a single match of a Pattern to find both the leading values and the string inside the parenthesis:

private static final Pattern FULL_PATTERN = Pattern.compile("^(\\$\\w+)\\s*=>\\s*(\\w+)\\s*\\(([^)]+)\\)$");
private static final Pattern KEY_VALUE_PATTERN = Pattern.compile("(\\w+)\\s*->\\s*([\\$\\w]+)");
private static final Pattern COMMA_PATTERN = Pattern.compile("\\s*,\\s*");

public static Macro getMacro(String str) {
    final Macro macro = new Macro();
    final Map<String, String> map = new HashMap<String, String>();
    final Matcher matcher = FULL_PATTERN.matcher(str);
    if(!matcher.matches()) {
        throw new IllegalArgumentException("...");
    }

    COMMA_PATTERN.splitAsStream(matcher.group(3)).forEach(a ->
    {
        final Matcher kvMatcher = KEY_VALUE_PATTERN.matcher(a);
        if(!kvMatcher.matches()){
             throw new IllegalArgumentException("...");
        }

        map.put(kvMatcher.group(1), kvMatcher.group(2));
    });
    macro.setMap(map);
    macro.setKey(matcher.group(1));
    macro.setType(matcher.group(2));
    return macro;

}

You'll note I added \\s* to various points in each pattern, This lets you ignore the whitespace that may surround the operators.

The \\w in a Pattern means a word character. If you only want alphanumeric +undescore then you'llneed to replace each instance of it with [\\p{Alnum}_]

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  • \$\begingroup\$ Thank you sir. When i check your Full pattern ^(\\$\\w+)\\s*=>\\s*(\\w+)\\s*\(([^)]+)\)$ to the following regex tool regex101.com, it is not matching anything. Will it be possible for you to have a look \$\endgroup\$ Commented Aug 3, 2017 at 10:19
  • \$\begingroup\$ @SaurabhKumar if I insert ^(\$\w+)\s*=>\s*(\w+)\s*\(([^)]+)\)$ into regex101 it properly matches your test strings: regex101.com/r/E7uvML/1 \$\endgroup\$ Commented Aug 3, 2017 at 10:25
  • \$\begingroup\$ Correct but for (\w+)\s*->\s*(\w+) regex sir it's not matching the following str for example company->$company. Its goes into IllegalArgument. \$\endgroup\$ Commented Aug 3, 2017 at 10:37
  • \$\begingroup\$ @SaurabhKumar Ah I see it was missing the $, should have been (\w+)\s*->\s*([\$\w]+) fixed in the answer \$\endgroup\$ Commented Aug 3, 2017 at 10:41
  • \$\begingroup\$ Yup. Thank you very much for helping me out. I learnt new things about regex from you :) \$\endgroup\$ Commented Aug 3, 2017 at 10:44
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Simplifications:

  • Arrays.asList(...).stream() can be shorter (and faster) as Arrays.stream(...)

  • In the same vein, the code collects the stream into a list, just to restream the list. Additionally using an Action (as in forEach) to collect results is semantically questionable. The whole while-loop should be written like this:

    while (matcher.find()) {
        map = Arrays.stream(matcher.group(1).split(","))
            .map(content -> content.split("->")
            .collect(Collectors.toMap(a -> a[0], a -> a[1]));
    }
    
  • Since the Pattern instance you have does not change, I'd extract it into a static variable

  • Instead of making this an external static method, I'd put it into the Macro class. This enables calling it as follows: Macro.fromString(...).

Last but not least this is a pretty nicely usable place for regex as follows:

static final Pattern macroPattern = 
    Pattern.compile("(\\$[^=]+)=>([A-Z][^\\(]+)\\((([^-]+)->([^\\)]+)(?:\\s*,\\s*)?\\)");

Now if java had named capture groups this would suddenly become much easier (~hint, hint, nudge nudge). How to use this pattern is something left as exercise to the reader

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  • 1
    \$\begingroup\$ But Java does have named-capturing groups! (?<key>\$.+?)=> etc. Also, your regex does not compile. You have to use double backslashes in string literals to represent a single backslash in the regular expression. \$\endgroup\$
    – Stingy
    Commented Aug 3, 2017 at 22:42
  • \$\begingroup\$ thanks for catching the backslashes and the apparently needed refresher on named capture groups... \$\endgroup\$
    – Vogel612
    Commented Aug 3, 2017 at 23:14
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You are calling Stream.forEach(Consumer) with a Consumer whose action accesses shared state (namely the local and final variable map) without taking care of synchronization (a HashMap is not thread-safe). This is bad. Stream.forEach(Consumer) might look like a simple for-each loop, but it is more than that, because unlike the latter, it might process multiple stream elements simultaneously in multiple threads, so accessing shared state without synchronization is dangerous. A safer approach would have been to use forEachOrdered(Consumer) instead of forEach(Consumer).

The solution proposed by Vogel612 for this problem is even more elegant, but, technically, it is not equivalent to your procedure, because your procedure doesn't create a new Map for every iteration of while (matcher.find()), as opposed to Vogel612's code, which assigns a new Map to map for every loop iteration. However, the while (matcher.find()) loop is probably a design flaw anyway, because as far as I understand your question, the input string is not supposed to contain multiple parentheses-enclosed "maps" (not to be confused with key-value-"mappings"). This can be rectified by using a single regular expression for the whole input string, as the other answers have already suggested.

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  • \$\begingroup\$ FWIW making "synchronization" an issue when referring to a local variable is a separate level of overkill. You're of course correct in principle, but it doesn't matter that much, given that the stream isn't a parallelStream in the first place. Unless there is a parallel in the pipeline, Java / JIT is highly unlikely if not disallowed from making the Stream parallel \$\endgroup\$
    – Vogel612
    Commented Aug 3, 2017 at 22:44
  • \$\begingroup\$ @Vogel612 It is irrelevant whether the variable is local or a field. The point is that it can be accessed concurrently and therefore behave in an unexpected manner. However, you're right, the documentation of Collection.stream() says explicitly that it returns a sequential stream. Thank you for pointing that out, I wasn't aware of that. Still, I think it is better to point out a practice that has the potential of causing trouble, even if, in this particular case, it is unproblematic. \$\endgroup\$
    – Stingy
    Commented Aug 3, 2017 at 23:07

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