There are two maps and I am trying to merge them into a single map (finalResp
).
Map<String, String[]> map1 = new HashMap<>();
Map<String, String> map2 = new HashMap<>();
HashMap<String, String> finalResp = new HashMap<String, String>();
Solution - pre Java 8 - achieved like below:
for (Map.Entry<String, String[]> entry : map1.entrySet()) {
if (map2.containsKey(entry.getKey())) {
String newValue = changetoAnother(map1.get(entry.getKey()), map2.get(entry.getKey()));
finalResp.put(entry.getKey(), newValue);
}
}
Using Java 8, I am stuck at this:
HashMap<String, String> map3 = new HashMap<>(map2);
map1.forEach((k, v) -> map3.merge(k, v, (i, j) -> mergeValue(i, j) ));
How can I check if a map 2 key is not present in map 1 and modify the values?
One possible way is to filter the unwanted elements (not contained in map2
) and collect the result into a new Map:
Map<String, String> finalResp =
map1.entrySet().stream().filter(e -> map2.containsKey(e.getKey()))
.collect(Collectors.toMap(
Entry::getKey,
e -> changetoAnother(e.getValue(), map2.get(e.getKey()))
));
Another way would be to create a copy of map2
, retain all the keys of this Map
that are also contained in map1
keys and finally replace all the values by applying the function changetoAnother
.
Map<String, String> result = new HashMap<>(map2);
result.keySet().retainAll(map1.keySet());
result.replaceAll((k, v) -> changetoAnother(map1.get(k), v));
Note that the advantage of the first solution is that it can be easily generalized to work for any two Maps:
private <K, V, V1, V2> Map<K, V> merge(Map<K, V1> map1, Map<K, V2> map2, BiFunction<V1, V2, V> mergeFunction) {
return map1.entrySet().stream()
.filter(e -> map2.containsKey(e.getKey()))
.collect(Collectors.toMap(
Entry::getKey,
e -> mergeFunction.apply(e.getValue(), map2.get(e.getKey()))
));
}
with
Map<String, String> finalResp = merge(map1, map2, (v1, v2) -> changetoAnother(v1, v2));
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