How can I use Lambda to convert a String array to Integer array?

Question

I am trying to use lambda expressions to convert a String array to an Integer array.

I have provided my code below, along with a brief description of what I have tried so far:

String [] inputData = userInput.split("\\s+");
Integer [] toSort = new Integer [] {};
try {
    toSort = Arrays.asList(inputData).stream().mapToInt(Integer::parseInt).toArray();
}catch(NumberFormatException e) {
    System.out.println("Error. Invalid input!\n" + e.getMessage());
}

The lamba expression I have above is one which maps to an int array, which is not what I want, upon compiling this code I get the following error message:

BubbleSort.java:13: error: incompatible types: int[] cannot be converted to Integer[]
            toSort = Arrays.asList(inputData).stream().mapToInt(Integer::parseIn
t).toArray();

          ^
1 error

Is there a simple way which allows me to use lambdas, or other means, to get from a String array to an Integer array?

just use a for loop ...

ZhongYu
– ZhongYu

2015-09-11 17:48:31 +00:00
Commented Sep 11, 2015 at 17:48 — ZhongYu
– ZhongYu, Commented Sep 11, 2015 at 17:48

Holger · Accepted Answer · 2015-09-11 17:56:30Z

6

As already pointed out by others, mapToInt returns an IntStream whose toArray method will return int[] rather than Integer[]. Besides that, there are some other things to improve:

Integer [] toSort = new Integer [] {};

is an unnecessarily complicated way to initialized an array. Use either

Integer[] toSort = {};

or

Integer[] toSort = new Integer[0];

but you should not initialize it at all, if you are going to overwrite it anyway. If you want to have a fallback value for the exceptional case, do the assignment inside the exception handler:

String[] inputData = userInput.split("\\s+");
Integer[] toSort;
try {
    toSort = Arrays.stream(inputData).map(Integer::parseInt).toArray(Integer[]::new);
}catch(NumberFormatException e) {
    System.out.println("Error. Invalid input!\n" + e.getMessage());
    toSort=new Integer[0];
}

Further, note that you don’t need the String[] array in your case:

Integer[] toSort;
try {
    toSort = Pattern.compile("\\s+").splitAsStream(userInput)
        .map(Integer::parseInt).toArray(Integer[]::new);
}catch(NumberFormatException e) {
    System.out.println("Error. Invalid input!\n" + e.getMessage());
    toSort=new Integer[0];
}

Pattern refers to java.util.regex.Pattern which is the same class which String.split uses internally.

answered Sep 11, 2015 at 17:56

Holger

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1 Comment

Holger Over a year ago

But if you don’t mind sort an int[] instead of Integer[], you could just live with the result of .mapToInt(Integer::parseInt).toArray() and change the type of toSort to int[].

Pshemo · Accepted Answer · 2015-09-11 17:49:17Z

6

mapToInt(Integer::parseInt).toArray() returns int[] array since matToInt produces IntStream but int[] array can't be used by Integer[] reference (boxing works only on primitive types, which arrays are not).

What you could use is

import java.util.stream.Stream;
//...
toSort = Stream.of(inputData)
               .map(Integer::valueOf) //value of returns Integer, parseInt returns int
               .toArray(Integer[]::new); // to specify type of returned array

edited Sep 11, 2015 at 17:49

answered Sep 11, 2015 at 17:42

Pshemo

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6 Comments

James Over a year ago

This has returned an error on the variable Stream: BubbleSort.java:13: error: cannot find symbol ... symbol: variable Stream Is there something I need to import for this method to work?

Pshemo Over a year ago

@JamesZafar Yes, you need to add import to java.util.stream.Stream;. You should probably start using IDE which will allow you to add imports automatically.

ZhongYu Over a year ago

or IntStream.boxed()

Holger Over a year ago

@bayou.io: sure, but it would be a bit nonsensical to use mapToInt just to be followed by boxed() when you can use map in the first place.

Holger Over a year ago

@Pshemo: it doesn’t matter whether you use .map(Integer::valueOf) or .map(Integer::parseInt); both will first parse to an int and then wrap it to an Integer. The only thing that matters is whether you use map or mapToInt.

|

Eran · Accepted Answer · 2015-09-11 17:45:39Z

2

If you want an Integer array, don't map to an IntStream, map to a Stream<Integer> :

toSort = Arrays.asList(inputData).stream().map(Integer::valueOf).toArray(Integer[]::new);

edited Sep 11, 2015 at 17:45

answered Sep 11, 2015 at 17:40

Eran

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2 Comments

James Over a year ago

This returns anotherBubbleSort.java:13: error: incompatible types: Object[] cannot be converted to I nteger[] error for me, similar to the first one: BubbleSort.java:13: error: incompatible types: Object[] cannot be converted to I nteger[]

Eran Over a year ago

@JamesZafar Sorry. I used the wrong toArray method. Fixed now.

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