How can I convert List<Integer> to int[] in Java?
JavaArraysCollectionsJava Problem Overview
I'm new to Java. How can I convert a List<Integer>
to int[]
in Java?
I'm confused because List.toArray()
actually returns an Object[]
, which can be cast to neither Integer[]
nor int[]
.
Right now I'm using a loop to do so:
int[] toIntArray(List<Integer> list) {
int[] ret = new int[list.size()];
for(int i = 0; i < ret.length; i++)
ret[i] = list.get(i);
return ret;
}
Is there's a better way to do this?
This is similar to the question *https://stackoverflow.com/questions/880581/java-convert-int-to-integer*.</sub>
Java Solutions
Solution 1 - Java
With streams added in Java 8 we can write code like:
int[] example1 = list.stream().mapToInt(i->i).toArray();
// OR
int[] example2 = list.stream().mapToInt(Integer::intValue).toArray();
Thought process:
-
The simple
Stream#toArray
returns anObject[]
array, so it is not what we want. Also,Stream#toArray(IntFunction<A[]> generator)
doesn't do what we want, because the generic typeA
can't represent the primitive typeint
-
So it would be nice to have some stream which could handle the primitive type
int
instead of the wrapperInteger
, because itstoArray
method will most likely also return anint[]
array (returning something else likeObject[]
or even boxedInteger[]
would be unnatural here). And fortunately Java 8 has such a stream which isIntStream
-
So now the only thing we need to figure out is how to convert our
Stream<Integer>
(which will be returned fromlist.stream()
) to that shinyIntStream
.Quick searching in documentation of
Stream
while looking for methods which returnIntStream
points us to our solution which ismapToInt(ToIntFunction<? super T> mapper)
method. All we need to do is provide a mapping fromInteger
toint
.Since
ToIntFunction
is functional interface we can provide its instance via lambda or method reference.Anyway to convert Integer to int we can use
Integer#intValue
so insidemapToInt
we can write:mapToInt( (Integer i) -> i.intValue() )
(or some may prefer:
mapToInt(Integer::intValue)
.)But similar code can be generated using unboxing, since the compiler knows that the result of this lambda must be of type
int
(the lambda used inmapToInt
is an implementation of theToIntFunction
interface which expects as body a method of type:int applyAsInt(T value)
which is expected to return anint
).So we can simply write:
mapToInt((Integer i)->i)
Also, since the
Integer
type in(Integer i)
can be inferred by the compiler becauseList<Integer>#stream()
returns aStream<Integer>
, we can also skip it which leaves us withmapToInt(i -> i)
Solution 2 - Java
Unfortunately, I don't believe there really is a better way of doing this due to the nature of Java's handling of primitive types, boxing, arrays and generics. In particular:
List<T>.toArray
won't work because there's no conversion fromInteger
toint
- You can't use
int
as a type argument for generics, so it would have to be anint
-specific method (or one which used reflection to do nasty trickery).
I believe there are libraries which have autogenerated versions of this kind of method for all the primitive types (i.e. there's a template which is copied for each type). It's ugly, but that's the way it is I'm afraid :(
Even though the Arrays
class came out before generics arrived in Java, it would still have to include all the horrible overloads if it were introduced today (assuming you want to use primitive arrays).
Solution 3 - Java
In addition to Commons Lang, you can do this with Guava's method Ints.toArray(Collection<Integer> collection)
:
List<Integer> list = ...
int[] ints = Ints.toArray(list);
This saves you having to do the intermediate array conversion that the Commons Lang equivalent requires yourself.
Solution 4 - Java
The easiest way to do this is to make use of Apache Commons Lang. It has a handy ArrayUtils class that can do what you want. Use the toPrimitive
method with the overload for an array of Integer
s.
List<Integer> myList;
... assign and fill the list
int[] intArray = ArrayUtils.toPrimitive(myList.toArray(new Integer[myList.size()]));
This way you don't reinvent the wheel. Commons Lang has a great many useful things that Java left out. Above, I chose to create an Integer list of the right size. You can also use a 0-length static Integer array and let Java allocate an array of the right size:
static final Integer[] NO_INTS = new Integer[0];
....
int[] intArray2 = ArrayUtils.toPrimitive(myList.toArray(NO_INTS));
Solution 5 - Java
Java 8 has given us an easy way to do this via streams...
Using the collections stream()
function and then mapping to ints, you'll get an IntStream. With the IntStream
we can call toArray() which gives us int []
int [] ints = list.stream().mapToInt(Integer::intValue).toArray();
Solution 6 - Java
Use:
int[] toIntArray(List<Integer> list) {
int[] ret = new int[list.size()];
int i = 0;
for (Integer e : list)
ret[i++] = e;
return ret;
}
This slight change to your code is to avoid expensive list indexing (since a List is not necessarily an ArrayList, but it could be a linked list, for which random access is expensive).
Solution 7 - Java
Here is a Java 8 single line code for this:
public int[] toIntArray(List<Integer> intList){
return intList.stream().mapToInt(Integer::intValue).toArray();
}
Solution 8 - Java
If you are simply mapping an Integer
to an int
then you should consider using parallelism, since your mapping logic does not rely on any variables outside its scope.
int[] arr = list.parallelStream().mapToInt(Integer::intValue).toArray();
Just be aware of this
> Note that parallelism is not automatically faster than performing operations serially, although it can be if you have enough data and processor cores. While aggregate operations enable you to more easily implement parallelism, it is still your responsibility to determine if your application is suitable for parallelism.
There are two ways to map Integers to their primitive form:
-
Via a
ToIntFunction
.mapToInt(Integer::intValue)
-
Via explicit unboxing with lambda expression.
mapToInt(i -> i.intValue())
-
Via implicit (auto-) unboxing with lambda expression.
mapToInt(i -> i)
Given a list with a null
value
List<Integer> list = Arrays.asList(1, 2, null, 4, 5);
Here are three options to handle null
:
-
Filter out the
null
values before mapping.int[] arr = list.parallelStream().filter(Objects::nonNull).mapToInt(Integer::intValue).toArray();
-
Map the
null
values to a default value.int[] arr = list.parallelStream().map(i -> i == null ? -1 : i).mapToInt(Integer::intValue).toArray();
-
Handle
null
inside the lambda expression.int[] arr = list.parallelStream().mapToInt(i -> i == null ? -1 : i.intValue()).toArray();
Solution 9 - Java
This simple loop is always correct! no bugs
int[] integers = new int[myList.size()];
for (int i = 0; i < integers.length; i++) {
integers[i] = myList.get(i);
}
Solution 10 - Java
I've noticed several uses of for loops, but you don't even need anything inside the loop. I mention this only because the original question was trying to find less verbose code.
int[] toArray(List<Integer> list) {
int[] ret = new int[ list.size() ];
int i = 0;
for( Iterator<Integer> it = list.iterator();
it.hasNext();
ret[i++] = it.next() );
return ret;
}
If Java allowed multiple declarations in a for loop the way C++ does, we could go a step further and do for(int i = 0, Iterator
In the end though (this part is just my opinion), if you are going to have a helping function or method to do something for you, just set it up and forget about it. It can be a one-liner or ten; if you'll never look at it again you won't know the difference.
Solution 11 - Java
There is really no way of "one-lining" what you are trying to do, because toArray returns an Object[] and you cannot cast from Object[] to int[] or Integer[] to int[].
Solution 12 - Java
int[] ret = new int[list.size()];
Iterator<Integer> iter = list.iterator();
for (int i=0; iter.hasNext(); i++) {
ret[i] = iter.next();
}
return ret;
Solution 13 - Java
Also try Dollar (check this revision):
import static com.humaorie.dollar.Dollar.*
...
List<Integer> source = ...;
int[] ints = $(source).convert().toIntArray();
Solution 14 - Java
I would recommend you to use the List<?>
skeletal implementation from the Java collections API. It appears to be quite helpful in this particular case:
package mypackage;
import java.util.AbstractList;
import java.util.Arrays;
import java.util.Collections;
import java.util.List;
public class Test {
// Helper method to convert int arrays into Lists
static List<Integer> intArrayAsList(final int[] a) {
if(a == null)
throw new NullPointerException();
return new AbstractList<Integer>() {
@Override
public Integer get(int i) {
return a[i]; // Autoboxing
}
@Override
public Integer set(int i, Integer val) {
final int old = a[i];
a[i] = val; // Auto-unboxing
return old; // Autoboxing
}
@Override
public int size() {
return a.length;
}
};
}
public static void main(final String[] args) {
int[] a = {1, 2, 3, 4, 5};
Collections.reverse(intArrayAsList(a));
System.out.println(Arrays.toString(a));
}
}
Beware of boxing/unboxing drawbacks.
Solution 15 - Java
With Eclipse Collections, you can do the following if you have a list of type java.util.List<Integer>
:
List<Integer> integers = Lists.mutable.with(1, 2, 3, 4, 5);
int[] ints = LazyIterate.adapt(integers).collectInt(i -> i).toArray();
Assert.assertArrayEquals(new int[]{1, 2, 3, 4, 5}, ints);
If you already have an Eclipse Collections type like MutableList
, you can do the following:
MutableList<Integer> integers = Lists.mutable.with(1, 2, 3, 4, 5);
int[] ints = integers.asLazy().collectInt(i -> i).toArray();
Assert.assertArrayEquals(new int[]{1, 2, 3, 4, 5}, ints);
Note: I am a committer for Eclipse Collections
Solution 16 - Java
Using a lambda you could do this (compiles in JDK lambda):
public static void main(String ars[]) {
TransformService transformService = (inputs) -> {
int[] ints = new int[inputs.size()];
int i = 0;
for (Integer element : inputs) {
ints[ i++ ] = element;
}
return ints;
};
List<Integer> inputs = new ArrayList<Integer>(5) { {add(10); add(10);} };
int[] results = transformService.transform(inputs);
}
public interface TransformService {
int[] transform(List<Integer> inputs);
}