How to get the difference of two maps based on the key set?

Question

I have two maps:

Map<String, Sample> newMap = convertJSONObjectToSampleMap(newMapStr);
Map<String, Sample> oldMap = convertJSONObjectToSampleMap(oldMapStr);

the Sample is some custom class

The newMap has keys: [1,2,3,4,5] The oldMap has keys: [2,3,4,5,8]

What is the best way to get difference between them, .e, get Samples with keys: 1 and 8?

I thought to use Collections and extract Set<>:

Set<String> newSet = newMap.keySet();
Set<String> oldSet = oldMap.keySet();

Thank you,

see http://stackoverflow.com/questions/8064570/what-is-the-best-way-get-the-symmetric-difference-between-two-sets-in-java — Arturo Volpe, May 03 '14 at 14:33

Alexis C. · Answer 1 · 2014-05-03T14:44:07.907

6

What you want is called the symmetric difference.

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Guava provides such a method.

Set<String> diff = Sets.symmetricDifference(newSet, oldSet);

Then just iterate through the set get the samples.

List<Sample> samples = new ArrayList<>();
for(String key : diff){
    if(oldMap.containsKey(key)){
        samples.add(oldMap.get(key));
    } else {
        samples.add(newMap.get(key));
    }
}

You can also do it with the official API, basically the symmetric difference is the union minus the intersection of both sets, but why reinvent the wheel?

Using an external dependency can be bad if you're using only one method, but Guava provides a lot of useful features that are a must.

edited May 03 '14 at 14:44

answered May 03 '14 at 14:26

Alexis C.

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-1 This answer introduces an external dependancy on your project. – Erick Robertson May 03 '14 at 14:30
2

@ErickRobertson Surely you can do it with the official JDK, but Guava is really a must and I don't see any reasons for not using an external library for this sort of things. – Alexis C. May 03 '14 at 14:35
The question is how to obtain a the `values` of the map, not only the `keys`, you need to add a step to obtain these keys. – Arturo Volpe May 03 '14 at 14:37
@AVolpe You're right, fixed. – Alexis C. May 03 '14 at 14:40
It's not gaining you anything here. You still have to have a loop around a `containsKey` in your code, and it's less efficient. I disagree that "Guava is really a must." – Erick Robertson May 03 '14 at 19:18

score 2 · Answer 2 · answered May 03 '14 at 14:28

2

Loop through one map and compare.

It's only O(n) to loop through one of the maps. Consider this code:

for (String key: oldMap.keySet()) {
    if (newMap.containsKey(key))
        newMap.remove(key);
    else
        newMap.put(key, oldMap.get(key));
}

And the newMap will now contain only the unique entries from both sets. It gives you the keys and the values, so all the data is in one place. It's fast, because you only need to loop through the keys in one of the maps and you don't have to create sets.

answered May 03 '14 at 14:28

Erick Robertson

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1

`newMap.keySet().removeAll(oldMap.keySet())` I guess this is significantly simpler, isn't it? – allprog May 03 '14 at 14:37
@allprog then you would have keys from `newMap` not found in `oldMap`, but you won't have the keys that are only in `oldMap`. – Erick Robertson May 03 '14 at 19:15
Ah, Ok. I didn't consider the `else` branch. – allprog May 03 '14 at 19:24

Kostas Kryptos · Answer 3 · 2014-05-03T14:53:51.063

Use a function like this if you're interested only on the keys:

public static Set<String> getDiff(Map<String,Object> mapA, Map<String,Object> mapB) {
    Set<String> diff = mapA.keySet();
    for (String s: mapB.keySet()) {
        if (diff.contains(s))
            diff.remove(s);
        else
            diff.add(s);
    }
    return diff;
}

Use the following function for the full diff Map (without affecting your existing Map objects):

public static Map<String,Object> getDiff(Map<String,Object> mapA, Map<String,Object> mapB) {
   Map<String,Object> diff = new HashMap<String,Object>();
   diff.putAll(mapA);
   for (String s: mapB.keySet()) {
       if (diff.containsKey(s))
           diff.remove(s);
       else
           diff.put(s, mapB.get(s));
   }
   return diff;
}

How to get the difference of two maps based on the key set?

3 Answers3