Does it exist a better way to parse String to Integer using stream than this :
String line = "1 2 3 4 5";
List<Integer> elements = Arrays.stream(line.split(" ")).mapToInt(x -> Integer.parseInt(x))
.boxed().collect(Collectors.toList());
2 Answers
You can eliminate one step if you parse the String directly to Integer:
String line = "1 2 3 4 5";
List<Integer> elements = Arrays.stream(line.split(" ")).map(Integer::valueOf)
.collect(Collectors.toList());
Or you can stick to primitive types, which give better performance, by creating an int array instead of a List<Integer>:
int[] elements = Arrays.stream(line.split(" ")).mapToInt(Integer::parseInt).toArray ();
You can also replace
Arrays.stream(line.split(" "))
with
Pattern.compile(" ").splitAsStream(line)
I'm not sure which is more efficient, though.
Comments
There's one more way to do it that will be available since java-9 via Scanner#findAll:
int[] result = scan.findAll(Pattern.compile("\\d+"))
.map(MatchResult::group)
.mapToInt(Integer::parseInt)
.toArray();
7 Comments
fps
Hi Eugene! I've seen you've answered at least one similar question using this same approach. Could you please kindly explain why you need to use
map(MatchResult::group) in the returned stream? I mean, I know I can go to the docs and find out, but maybe it's better for future readers if you explain this approach a little bit. Thanks!
tucuxi
I fail to see where the line containing the space-separated ints is parsed. Can you point scanners to read from strings?
Eugene
@FedericoPeraltaSchaffner it's the first matched group in the regex... with index
0. that's what you get when you use group(). if you need another group, you just use group(n). And you are always welcomeEugene
@tucuxi that's a different way to get those numbers, not by splitting, but by getting all the continuous digits
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int? What if one fails? What should happen then?Integeris wanted,notint, I would useInteger::valueOfto avoidboxed())