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Home/ Questions/Q 872713
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Editorial Team
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Editorial Team
Asked: May 15, 20262026-05-15T10:49:32+00:00 2026-05-15T10:49:32+00:00

Assuming I have final Iterable<String> unsorted = asList(FOO, BAR, PREFA, ZOO, PREFZ, PREFOO); What

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Assuming I have

final Iterable<String> unsorted = asList("FOO", "BAR", "PREFA", "ZOO", "PREFZ", "PREFOO");

What can I do to transform this unsorted list into this:

[PREFZ, PREFA, BAR, FOO, PREFOO, ZOO]

(a list which begin with known values that must appears first (here “PREFA” and “PREFZ”) and the rest is alphabetically sorted)

I think there are some usefull classes in guava that can make the job (Ordering, Predicates…), but I have not yet found a solution…

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-15T10:49:32+00:00Added an answer on May 15, 2026 at 10:49 am

    You specifically mentioned guava; along with Sylvain M’s answer, here’s another way (more as an academic exercise and demonstration of guava’s flexibility than anything else)

    // List is not efficient here; for large problems, something like SkipList 
    // is more suitable
    private static final List<String> KNOWN_INDEXES = asList("PREFZ", "PREFA");
    
    private static final Function<Object, Integer> POSITION_IN_KNOWN_INDEXES 
        = new Function<Object, Integer>() {
      public Integer apply(Object in) {
         int index = KNOWN_INDEXES.indexOf(in);
         return index == -1 ? null : index;
      }     
    };
    
    
    ...
    
    
    List<String> values = asList("FOO", "BAR", "PREFA", "ZOO", "PREFZ", "PREFOO");
    
    Collections.sort(values,
      Ordering.natural().nullsLast().onResultOf(POSITION_IN_KNOWN_INDEXES).compound(Ordering.natural())
    );
    

    So, in other words, sort on natural order of the Integer returned by List.indexOf(), then break ties with natural order of the object itself.

    Messy, perhaps, but fun.

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