使用Java流排序时将数字解析为字符串的简单方法吗？

Question

I have a list of objects that contain values stored as strings but are actually numbers and I'm trying to find a way to parse strings to integers within a Java stream to sort them numerically. Here's what I got so far.

List<SearchResultObject> returnedValues = doTheSearchMethod();
returnedValues = returnedValues
.stream()
.sorted(Comparator.comparing(SearchResultObject::getNumber))
.collect(Collectors.toList());

As of right now this kind of works, but the problem is that because those numbers are actually strings, it sorts the numbers lexicographically, not numerically. Is there a way to incorporate parsing into the stream? All numbers are padded to be the same size.

3356001017053000000000000
3356001017002000000000000
3356001017004000000000000
3356001017026000000000000
5101004003020000000000000
4123000006002510000000000
4758000005010000000000000

Answer 1

use comparingLong and then parse long. the numbers are too large for integers.

returnedValues.sort(Comparator.comparingLong(sro->Long.parseLong(sro.getNumber())));

Answer 2

For just sorting you don't need stream, just use custom Comparator using lambda expression

Comparator<SearchResultObject> c = Comparator.comparingLong(s->Long.parseLong(s.getNumber()));

And then use Collections.sort method

Collections.sort(returnedValues,c);

Answer 3

You don't need to use Streams API for sorting the list, you can directly use sort() method from the List interface. You just need to ensure that you provide a proper comparator to the sort() method.

As you mentioned, you can directly use:

Comparator.comparing(SearchResultObject::getNumber)

comparator to compare the strings.

However, as you need the list to be sorted by comparing the Integer/Long, you can use the following:

1. For Integer:

   list.sort(Comparator.comparingInt(s ->Integer.parseInt(s.getNumber())));

2. For Long:

   list.sort(Comparator.comparingLong(s ->Long.parseLong(s.getNumber())));

You can check the output for the following code below:

Before sorting
SearchResultObject{number='100'}
SearchResultObject{number='2'}
SearchResultObject{number='10'}
After sorting
SearchResultObject{number='2'}
SearchResultObject{number='10'}
SearchResultObject{number='100'}

The above output is based on the assumption the SearchResultObject contains a number as the field.

Answer 4

If you can modify SearchResultObject add a method to it converting the String to BigInteger and use it in your Comparator

public SearchResultObject {
    ...
    public BigInteger getNumberAsBigInteger() {
        return new BigInteger(this.number);
    }
    ...
}

And the invocation is

returnedValues.stream()
              .sorted(
                  Comparator.comparing(
                      SearchResultObject::getNumberAsBigInteger
                  )
              )
              .collect(Collectors.toList());

---

If you can't modify the SearchResultObject class, then you'll have to customize your Comparator a bit, but nothing too complicated

returnedValues.stream()
              .sorted(
                  Comparator.comparing(
                      sro -> new BigInteger(sro.getNumber())
                  )
              )
              .collect(toList());