Sort Entry Set of String, Integer by more than one field

Question

Have a Map<String, Integer> and trying to sort on value and length of String. I am trying to compare two different things in the statement so don't know if I need two different statements. This is being used to compare digit root so the String length and then the digit root is the value and the value.

For example:

("103",4); (1+0+3 == 4)
("4",4); (4 ==4)
("11101",4); (1+1+1+0+1 == 4)
("5",5); (5 == 5 )
("1003",4); (1+0+0+3 == 4)

But ("103",4) > ("4",4) because the length of "103" > "4", and ("11101",4) > ("103",4);, length "11101" > "103"

Collections.sort(list, new Comparator<Map.Entry<String, Integer>>() {
public int compare(Map.Entry<String, Integer> o1, Map.Entry<String, Integer> o2) { int length = o1.getKey().length().compareTo(o2.getKey().length());
if(length != 0) {
 return length;
}
return (o1.getValue()).compareTo(o2.getValue());
}
});

Edit Answered to the above Question(Also, response given)

 Map<String,Integer> unsortMap = new.
                        TreeMap<String,Integer>();

unsortMap.put("103",4);
unsortMap.put("4",4);
unsortMap.put("11101",4);   
unsortMap.put("5",5);
unsortMap.put("1003",4); Map<String,

 Integer> result =unsortMap.entrySet().stream() .sorted(Map.Entry.comparingByKey(Comparator.comparingInt(String::length)) )
   .sorted(Map.Entry.comparingByValue()) .collect(Collectors.toMap
     (Map.Entry::getKey, Map.Entry::getValue, (oldValue, newValue) -> oldValue, LinkedHashMap::new)); 

 System.println(result);

`Collections.sort(List list)` is for list . You can use TreeMap instead — Haifeng Zhang, Nov 13 '17 at 18:58
I can guess this is what you are asking for https://stackoverflow.com/questions/25899929/in-java-sort-hash-map-by-its-key-length — Maytham Fahmi, Nov 13 '17 at 18:59
Are you sure that you want to sort by the lenght of the key on the map?? — Marcos Vasconcelos, Nov 13 '17 at 19:19
Added explanation in the problem statement. I need to have the values also ordered so though 4 and 22 have value of 4 I need 4 to come before 22. — April_Nara, Nov 13 '17 at 19:34
I am working with a linked list trying to sort the entrySet from the map. — April_Nara, Nov 13 '17 at 19:57

developer_hatch · Accepted Answer · 2017-11-13T20:58:29.603

2

If you already have a map, and you want to order it, by lenght key, and by value then:

Map<String,Integer> unsortMap = new TreeMap<String,Integer>();

unsortMap.put("103",4);
unsortMap.put("4",4);
unsortMap.put("11101",4);
unsortMap.put("5",5);
unsortMap.put("1003",4);

Map<String, Integer> result = unsortMap.entrySet().stream()
        .sorted(Map.Entry.comparingByKey(Comparator.comparingInt(String::length))

        ).sorted(Map.Entry.comparingByValue())
        .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue,
                (oldValue, newValue) -> oldValue, LinkedHashMap::new));


System.out.println(result);

out => {4=4, 103=4, 1003=4, 11101=4, 5=5}

edited Nov 13 '17 at 20:58

answered Nov 13 '17 at 19:00

developer_hatch

15,898
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I get that it can't find Function. I think this is what I am looking for but when I get to this sort I already have the treemap @Damian Lattenero – April_Nara Nov 13 '17 at 19:51
@April_Nara I updated the answer! don't forget to accept if it was usefull – developer_hatch Nov 13 '17 at 20:04
It doesn't do what I want it to now though. I still need it to sort the value. the second solution only orders by length. I do appreciate the help. – April_Nara Nov 13 '17 at 20:17
@April_Nara let me know if now I understood your question, and if that is your expected output – developer_hatch Nov 13 '17 at 20:31
That's an incorrect output. I want to also sort on value. First you'd never get 22=7 as the digit root is 4 so it would need to be {4=4, 22=4, 103=4, 23=5} would be what I need. @Damian Lattenero – April_Nara Nov 13 '17 at 20:38
@April_Nara but who is bigger? ("103", 1) or ("9", 3) ? because "103" is larger than "9", but 1 < 3... so... – developer_hatch Nov 13 '17 at 20:40
Let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/158900/discussion-between-april-nara-and-damian-lattenero). – April_Nara Nov 13 '17 at 20:43
Thank you. I was familiar with the sort in 8 but was still attempting in 7. – April_Nara Nov 13 '17 at 21:03
@April_Nara I have always loved to help :) – developer_hatch Nov 13 '17 at 21:12
@DamianLattenero this the nice way to sort :). – Simmant Nov 14 '17 at 10:25

score 1 · Answer 2 · answered Nov 13 '17 at 19:16

1

You can compare in this way, if you are aiming to sort with length.

Collections.sort(list, new Comparator<Map.Entry<String, Integer>>() {

    @Override
    public int compare(Entry<String, Integer> o1, Entry<String, Integer> o2) {
        // TODO Auto-generated method stub

        if (o1.getKey().length() == o2.getKey().length()) {
            return 0;
        }

        if (o1.getKey().length() > o2.getKey().length()) {
            return 1;
        } else {
            return -1;
        }

    }

});

answered Nov 13 '17 at 19:16

Simmant

1,477
25
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That's a good answer, and a plus one ;), you can do it shorter, take a look at my answer if you want – developer_hatch Nov 13 '17 at 19:18
Thank you :), I checked and I really like that way, but reason why I use multiple if statement because it make code bit debug friendly :P in general. – Simmant Nov 13 '17 at 19:25
Thanks but this doesn't solve the problem of sorting them by value as well. – April_Nara Nov 13 '17 at 20:24
Sorry for delay in response, but now I have one question to @April_Nara, and its basis how much I understand your question and please correct me if I m wrong. You wanted to sort it with key and sorting is based on sum of individual key. If this the case then why you are using String as key?? It can easily be done with Integer as sum of key. – Simmant Nov 14 '17 at 10:24

score 0 · Answer 3 · answered Nov 13 '17 at 19:21

You haven't said why it doesn't work. What happens when you try?

In Java, a Map is not meant to be sorted, it is meant for fast access. However, implementations like TreeMap maintain a deterministic ordering, and can be sorted. Your code suggests you are trying to sort the Map's EntrySet. A Java Set also has no order, and cannot be sorted.

I suggest you either utilise a NavigableMap like a TreeMap, or work with a List rather than a Set. A Java List has an order, and can be sorted using Collections.sort(...).

There are a number of reasons why your existing code may not be working. For one thing, you are attempting to call compareToon an int (at o1.getKey().length().compareTo(...)). This is not possible. You could instead use:

Ints.compare(o1.getKey().length(), o2.getKey().length());

The question isn't really being addressed because, I am not sure if I can but I think I should, be able to compare two things in the compare statement. — April_Nara, Nov 13 '17 at 19:31

Sort Entry Set of String, Integer by more than one field

3 Answers3