Collect the result with only a stream (without external loop use)

Is there a way to do the following code only with lambdas ?

// translate someList1 to someList3
// .. get sublist
List<String> someList2 = someList1.stream()
.map(i -> i.getField())
.collect(Collectors.toList());
// .. create new (target) list
List<SomeClass> someList3 = new ArrayList<>();
for (String item : someList2) {
SomeClass someObj = new SomeClass();
someObj.setField(item);
someList3.add(someObj);
}


3 Answers
3

Just one collect is required :

List<SomeClass> someList2 =
someList1.stream()
.map(i -> {
SomeClass someObj = new SomeClass();
someObj.setField(i.getField());
return someObj;
}
)
.collect(Collectors.toList());


But note that with a constructor in SomeClass that accepts the value of getField(), it would be really more neat :

SomeClass

getField()

List<SomeClass> someList2 =
someList1.stream()
.map(i-> new SomeClass(i.getField())
.collect(Collectors.toList());


Or by spliting the map() operation in two distinct transformations, you can use method references, which improves the readability :

map()

List<SomeClass> someList2 =
someList1.stream()
.map(OneClass::getField)
.map(SomeClass::new)
.collect(Collectors.toList());


You can achieve that with a multiline lambda and another map operation:

List<SomeClass> someList3 = someList1.stream()
.map(i -> i.getField())
.map(f -> {
SomeClass someObj = new SomeClass();
someObj.setField(f);
return someObj;
})
.collect(Collectors.toList());


No need to map it to a String then back to SomeClass, do it in the one map:

map

String

SomeClass

List<SomeClass> someList2 = list.stream()
.map(i -> new SomeClass(i.getField())) //in case you have such constructor
.collect(Collectors.toList());


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