CODEWITHSUNDEEP
javajava-8java-stream
mat_boy
mat_boy

Reputation: 13698

Using streams to convert a list of objects into a string obtained from the toString method

There are a lot of useful new things in Java 8. E.g., I can iterate with a stream over a list of objects and then sum the values from a specific field of the Object's instances. E.g.

public class AClass {
  private int value;
  public int getValue() { return value; }
}

Integer sum = list.stream().mapToInt(AClass::getValue).sum();

Thus, I'm asking if there is any way to build a String that concatenates the output of the toString() method from the instances in a single line.

List<Integer> list = ...

String concatenated = list.stream().... //concatenate here with toString() method from java.lang.Integer class

Suppose that list contains integers 1, 2 and 3, I expect that concatenated is "123" or "1,2,3".

Upvotes: 288

Views: 588573

Answers (15)

Karim Omaya
Karim Omaya

Reputation: 1039

Using Stream

    List<Integer> list = Arrays.asList(1, 2, 3);
    String s = list.stream().map((i)->i.toString()).collect(Collectors.joining());

Upvotes: 4

Shinu Mathew
Shinu Mathew

Reputation: 47

A clean way to do this is by mapping the elements of the list to string and then using the joining operation in Collectors class.

List<Integer> ls = new ArrayList<Integer>();
ls.add(1);
ls.add(2);
ls.add(3);
String s = ls.stream().map(Object::toString).collect(Collectors.joining(","));

Upvotes: 2

Shailesh Aswal
Shailesh Aswal

Reputation: 6802

One simple way is to append your list items in a StringBuilder

List<Integer> list = new ArrayList<>();
list.add(1);
list.add(2);
list.add(3);

StringBuilder b = new StringBuilder();
list.forEach(b::append);

System.out.println(b);

you can also try:

String s = list.stream().map(e -> e.toString()).reduce("", String::concat);

Explanation: map converts Integer stream to String stream, then its reduced as concatenation of all the elements.

Note: This is normal reduction which performs in O(n2)

for better performance use a StringBuilder or mutable reduction similar to F. Böller's answer.

String s = list.stream().map(Object::toString).collect(Collectors.joining(","));

Ref: Stream Reduction

Upvotes: 507

Larry Brown
Larry Brown

Reputation: 1

I'm going to use the streams api to convert a stream of integers into a single string. The problem with some of the provided answers is that they produce a O(n^2) runtime because of String building. A better solution is to use a StringBuilder, and then join the strings together as the final step.

//              Create a stream of integers 
    String result = Arrays.stream(new int[]{1,2,3,4,5,6 })                
            // collect into a single StringBuilder
            .collect(StringBuilder::new, // supplier function
                    // accumulator - converts cur integer into a string and appends it to the string builder
                    (builder, cur) -> builder.append(Integer.toString(cur)),
                    // combiner - combines two string builders if running in parallel
                    StringBuilder::append) 
            // convert StringBuilder into a single string
            .toString();

You can take this process a step further by converting the collection of object to a single string.

// Start with a class definition
public static class AClass {
    private int value;
    public int getValue() { return value; }
    public AClass(int value) { this.value = value; }

    @Override
    public String toString() {
        return Integer.toString(value);
    }
}

// Create a stream of AClass objects
        String resultTwo = Arrays.stream(new AClass[]{
                new AClass(1),
                new AClass(2),
                new AClass(3),
                new AClass(4)
        })
                // transform stream of objects into a single string
                .collect(StringBuilder::new,
                        (builder, curObj) -> builder.append(curObj.toString()),
                        StringBuilder::append
                )
            // finally transform string builder into a single string
            .toString();

Upvotes: 0

bachrc
bachrc

Reputation: 1196

There is a method in the String API for those "joining list of string" usecases, you don't even need Stream.

List<String> myStringIterable = Arrays.asList("baguette", "bonjour");

String myReducedString = String.join(",", myStringIterable);

// And here you obtain "baguette,bonjour" in your myReducedString variable

Upvotes: 10

F. B&#246;ller
F. B&#246;ller

Reputation: 4304

There is a collector joining in the API. It's a static method in Collectors.

list.stream().map(Object::toString).collect(Collectors.joining(","))

Not perfect because of the necessary call of toString, but works. Different delimiters are possible.

Upvotes: 252

Rekha Ghanate
Rekha Ghanate

Reputation: 39

StringListName = ObjectListName.stream().map( m -> m.toString() ).collect( Collectors.toList() );

Upvotes: 4

Lorenzo Cipriani
Lorenzo Cipriani

Reputation: 11

Testing both approaches suggested in Shail016 and bpedroso answer (https://stackoverflow.com/a/24883180/2832140), the simple StringBuilder + append(String) within a for loop, seems to execute much faster than list.stream().map([...].

Example: This code walks through a Map<Long, List<Long>> builds a json string, using list.stream().map([...]:

if (mapSize > 0) {
    StringBuilder sb = new StringBuilder("[");

    for (Map.Entry<Long, List<Long>> entry : threadsMap.entrySet()) {

        sb.append("{\"" + entry.getKey().toString() + "\":[");
        sb.append(entry.getValue().stream().map(Object::toString).collect(Collectors.joining(",")));
    }
    sb.delete(sb.length()-2, sb.length());
    sb.append("]");
    System.out.println(sb.toString());
}

On my dev VM, junit usually takes between 0.35 and 1.2 seconds to execute the test. While, using this following code, it takes between 0.15 and 0.33 seconds:

if (mapSize > 0) {
    StringBuilder sb = new StringBuilder("[");

    for (Map.Entry<Long, List<Long>> entry : threadsMap.entrySet()) {

        sb.append("{\"" + entry.getKey().toString() + "\":[");

        for (Long tid : entry.getValue()) {
            sb.append(tid.toString() + ", ");
        }
        sb.delete(sb.length()-2, sb.length());
        sb.append("]}, ");
    }
    sb.delete(sb.length()-2, sb.length());
    sb.append("]");
    System.out.println(sb.toString());
}

Upvotes: 1

0vint0
0vint0

Reputation: 31

List<String> list = Arrays.asList("One", "Two", "Three");
    list.stream()
            .reduce("", org.apache.commons.lang3.StringUtils::join);

Or 

List<String> list = Arrays.asList("One", "Two", "Three");
        list.stream()
                .reduce("", (s1,s2)->s1+s2);

This approach allows you also build a string result from a list of objects Example 

List<Wrapper> list = Arrays.asList(w1, w2, w2);
        list.stream()
                .map(w->w.getStringValue)
                .reduce("", org.apache.commons.lang3.StringUtils::join);

Here the reduce function allows you to have some initial value to which you want to append new string Example:

 List<String> errors = Arrays.asList("er1", "er2", "er3");
            list.stream()
                    .reduce("Found next errors:", (s1,s2)->s1+s2);

Upvotes: 2

Bill K
Bill K

Reputation: 62789

Just in case anyone is trying to do this without java 8, there is a pretty good trick. List.toString() already returns a collection that looks like this:

[1,2,3]

Depending on your specific requirements, this can be post-processed to whatever you want as long as your list items don't contain [] or , .

For instance:

list.toString().replace("[","").replace("]","")

or if your data might contain square brackets this:

String s=list.toString();
s = s.substring(1,s.length()-1)

will get you a pretty reasonable output.

One array item on each line can be created like this:

list.toString().replace("[","").replace("]","").replaceAll(",","\r\n")

I used this technique to make html tooltips from a list in a small app, with something like:

list.toString().replace("[","<html>").replace("]","</html>").replaceAll(",","<br>")

If you have an array then start with Arrays.asList(list).toString() instead

I'll totally own the fact that this is not optimal, but it's not as inefficient as you might think and is pretty straightforward to read and understand. It is, however, quite inflexible--in particular don't try to separate the elements with replaceAll if your data might contain commas and use the substring version if you have square brackets in your data, but for an array of numbers it's pretty much perfect.

Upvotes: 11

James Wierzba
James Wierzba

Reputation: 17568

With Java 8+

String s = Arrays.toString(list.stream().toArray(AClass[]::new));

Not the most efficient, but it is a solution with a small amount of code.

Upvotes: -2

Evgenii
Evgenii

Reputation: 447

Also, you can do like this.

    List<String> list = Arrays.asList("One", "Two", "Three");
    String result = String.join(", ", list);
    System.out.println(result);

Upvotes: -2

Joe Almore
Joe Almore

Reputation: 4329

String actual = list.stream().reduce((t, u) -> t + "," + u).get();

Upvotes: 0

Kumar Abhishek
Kumar Abhishek

Reputation: 3134

Can we try this.

public static void main(String []args){
        List<String> stringList = new ArrayList<>();
        for(int i=0;i< 10;i++){
            stringList.add(""+i);
        }
        String stringConcated = String.join(",", stringList);
        System.out.println(stringConcated);

    }

Upvotes: -1

Edward J Beckett
Edward J Beckett

Reputation: 5140

The other answers are fine. However, you can also pass Collectors.toList() as parameter to Stream.collect() to return the elements as an ArrayList.

System.out.println( list.stream().map( e -> e.toString() ).collect( toList() ) );

Upvotes: 6

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