Converting List<Integer> to List<String>

Converting ListInteger to ListString

As far as I know, iterate and instantiate is the only way to do this. Something like (for others potential help, since I'm sure you know how to do this):

List<Integer> oldList = ...
/* Specify the size of the list up front to prevent resizing. */
List<String> newList = new ArrayList<>(oldList.size());
for (Integer myInt : oldList) { 
  newList.add(String.valueOf(myInt)); 
}

Cast Listint to Liststring in .NET 2.0

.NET 2.0 has the ConvertAll method where you can pass in a converter function:

List<int>    l1 = new List<int>(new int[] { 1, 2, 3 } );
List<string> l2 = l1.ConvertAll<string>(delegate(int i) { return i.ToString(); });

Convert all strings in a list to int

Given:

xs = ['1', '2', '3']

Use map then list to obtain a list of integers:

list(map(int, xs))

In Python 2, list was unnecessary since map returned a list:

map(int, xs)

Convert ListString to ListInteger directly

No, you need to loop over the array

for(String s : strList) intList.add(Integer.valueOf(s));

Java: Convert ListInteger to String

One way would be:

Iterate over list, add each item to StringBuffer (or) StringBuilder and do toString() at end.

Example:

StringBuilder strbul  = new StringBuilder();
     Iterator<Integer> iter = list.iterator();
     while(iter.hasNext())
     {
         strbul.append(iter.next());
        if(iter.hasNext()){
         strbul.append(",");
        }
     }
 strbul.toString();

Convert a list of integers into a comma-separated string

Yes. However, there is no Collectors.joining for a Stream<Integer>; you need a Stream<String> so you should map before collecting. Something like,

System.out.println(i.stream().map(String::valueOf)
        .collect(Collectors.joining(",")));

Which outputs

1,2,3,4,5

Also, you could generate Stream<Integer> in a number of ways.

System.out.println(
        IntStream.range(1, 6).boxed().map(String::valueOf)
               .collect(Collectors.joining(","))
);

How to convert Liststring to Listint?

listofIDs.Select(int.Parse).ToList()

Convert ListString to ListInteger for only valid Integer

In order to remove all non-numeric characters from each string, you need to apply map() before converting the stream of String into stream of Integer.

List<Integer> nums =
     strList.stream()
            .map(s -> s.replaceAll("[^\\d]", ""))
            .map(Integer::valueOf)
            .collect(Collectors.toList());

System.out.println(nums);

Output for input {"464 mm", "80 mm", "1000", "1220", "48 mm", "1020"}

[464, 80, 1000, 1220, 48, 1020]

Note:

• trim() after replaceAll() is redundant.
• Since you are adding all the contents of the list generated by the stream pipeline into another list, you might substitute collect(Collectors.toList()) with toList() which is accessible with Java 16 onwards. The difference is that toList() returns an unmodifiable list and it more performant than collector.
• Iterative solutions almost always perform better than stream-based. Lambdas and streams were introduced in Java to provide a way of organizing the code that more readable and concise.

Convert array of Strings to list of Integers?

You only need to stream once.

Instead of using int Integer::parseInt(String s), you should use Integer Integer::valueOf(String s), so you don't have to call boxed() or rely on auto-boxing.

Then use collect(Collectors.toList()) directly, instead of creating intermediate array first.

List<Integer> allAnswerList = Arrays.stream(allAnswers)    // stream of String
                                    .map(Integer::valueOf) // stream of Integer
                                    .collect(Collectors.toList());

Convert a list of integers to string

There's definitely a slicker way to do this, but here's a very straight forward way:

mystring = ""

for digit in new:
    mystring += str(digit)

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