Convert Set<Integer> to Set<String> in Java

Question

Is there a simple way to convert Set<Integer> to Set<String> without iterating through the entire set?

Solution 1

No. You have to format each integer and add it to your string set.

Solution 2

No. The best way is a loop.

HashSet<String> strs = new HashSet<String>(ints.size());
for(Integer integer : ints) {
  strs.add(integer.toString());
}

Something simple and relatively quick that is straightforward and expressive is probably best.

(Update:) In Java 8, the same thing can be done with a lambda expression if you'd like to hide the loop.

HashSet<String> strs = new HashSet<>(ints.size());
ints.forEach(i -> strs.add(i.toString()));

or, using Streams,

Set<String> strs = ints.stream().map(Integer::toString).collect(toSet());

Solution 3

use Java8 stream map and collect abilities:

 Set< String >  stringSet = 
   intSet.stream().map(e -> String.valueOf(e)).collect(Collectors.toSet());

Solution 4

You could use Commons Collections' TransformedSet or Guava's Collections2.transform(...)

In both cases, your functor would presumably simply call the Integer's toString().

Solution 5

You can use a decorator if you really don't want to iterate through the entire set

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Comments

[Cajunluke]

Can you provide an example? I'm not familiar with the concept of decorators.

[Oliver Charlesworth]

How would you apply the decorator pattern in this scenario?

[Vishy]

Perhaps you meant toString instead of stringValue ;)

[Kathy Van Stone]

If we just wanted a Set<Integer> to look like a Set<String> but not create another set (say if only a fraction of the set is likely to be checked) you would create a subclass of AbstractSet<String> that would wrap the Set<Integer> and do the conversions in individual methods.

[Cajunluke]

@Peter That indeed I did. I must have been thinking of Integer's .intValue().

[VirtualTroll]

Well it would depends on Mat needs. For a single need (function), i will probably create a simple object containing the set and implementing my required function. For more than one need, I will probably choose another solution :)

[Luke Hutteman]

Theoretically, your code could throw a NullPointerException in the call to toString() as a Set<T> can contain a null value.

[Isaac Truett]

Presumably, someone disagreed with the "no" answers and feels that a wrapper/decorator is a better answer.

[Cajunluke]

@Luke True, but, as this is an example, handling it would be unimportant and, honestly, I wouldn't handle it even in production code until it proved to be an issue. Why bother if you know the input set won't have any nulls?

[Vishy]

@Luke, For this reason I prefer the shorter ""+integer ;)

[Luke Hutteman]

@CajunLuke: I agree that if you know the input Set won't have any nulls, there's no need to bother with the null-check. The question does not mention whether nulls are possible though, and online code-samples do sometimes end up copied-and-pasted into production code (where I absolutely disagree with your approach of waiting "until it proved to be an issue").

[Luke Hutteman]

@Peter: that maps nulls to a String instance of "null" though, instead of the probably more accurate actual null value you would get with the longer integer == null ? null : integer.toString().

[Vishy]

@Luke, true, Another way to map null Integer is to an empty string, which works for loading text tables into a spreadsheet.

[Cajunluke]

@PeterLawrey Instead of dealing with a much of extra method calls and a new StringBuilder, String.valueOf(integer) is faster and has the same result as ""+integer.

[Vishy]

@CajunLuke While a little faster, it is more work for the developer. If you concerned about speed you wouldn't be adding Integer as a String to HashSet. You might be using a library which wraps an int[] to save the int as a primitive.

[Cajunluke]

@PeterLawrey I never thought of wrapping an int[] to keep the primitive. I'm going to have to keep that in mind.