CODEWITHSUNDEEP
javajava-8java-stream
student
student

Reputation: 289

Use Java 8 stream to transform nested collection

I have the following nested list:

List<List<String>> nestedList = new ArrayList<>();
nestedList.add(Arrays.asList("aM0", "b"));
nestedList.add(Arrays.asList("c", "aM0"));

I want to remove "M0" from all the strings. I want to challenge myself using Java 8 stream. The following is what I wrote:

nestedList.stream()
  .map(t -> t.stream()
  .map(s -> s.substring(0, s.indexOf('M')))
  .collect(Collectors.toCollection(ArrayList::new)))
  .collect(Collectors.toList());

But JDK throws a lot of exceptions. Can anyone tell me how to fix? Or we can not use streams in this special situation? Thank you.

Upvotes: 0

Views: 664

Answers (3)

Hadi
Hadi

Reputation: 17289

The other simple way is use replaceAll(UnaryOperator operator)

nestedList.forEach(list->list.replaceAll(s -> s.replace("M0", "")));

Upvotes: 1

student
student

Reputation: 289

Thank you for quickly spotting the issue. Since after marker 'M', it may not be just '0', I change the code as follows, and it now works:

    nestedList.stream().map(t->t.stream().map(s->s.contains("M") ? s.substring(0, s.indexOf('M')) : s).collect(Collectors.toCollection(ArrayList<String>::new))).collect(Collectors.toList());

Thank you very much for your time to answer my question!

Upvotes: 0

Youcef LAIDANI
Youcef LAIDANI

Reputation: 59950

Your real problem is with String::substring so when you use for example :

String str = "b";
str = str.substring(0, str.indexOf('M'));

this will throws :

java.lang.StringIndexOutOfBoundsException: begin 0, end -1, length 1

because str.indexOf('M') return -1if there is no such occurrence (String::indexOf) so str.substring(0, -1) will fail with that error

Instead in such situations its better to use String::replace like so :

nestedList = nestedList.stream()
        .map(t -> t.stream()
                .map(s -> s.replace("M0", ""))
                .collect(Collectors.toList())
        ).collect(Collectors.toList());

Outputs

[[a, b], [c, a]]

Upvotes: 4

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