Best way to concatenate List of String objects?
JavaStringListJava Problem Overview
What is the best way to concatenate a list of String objects? I am thinking of doing this way:
List<String> sList = new ArrayList<String>();
// add elements
if (sList != null)
{
String listString = sList.toString();
listString = listString.subString(1, listString.length() - 1);
}
I somehow found this to be neater than using the StringBuilder/StringBuffer approach.
Any thoughts/comments?
Java Solutions
Solution 1 - Java
Use one of the the StringUtils.join methods in Apache Commons Lang.
import org.apache.commons.lang3.StringUtils;
String result = StringUtils.join(list, ", ");
If you are fortunate enough to be using Java 8, then it's even easier...just use String.join
String result = String.join(", ", list);
Solution 2 - Java
Using Java 8+
String str = list.stream().collect(Collectors.joining())
or even
String str = String.join("", list);
Solution 3 - Java
Your approach is dependent on Java's ArrayList#toString() implementation.
While the implementation is documented in the Java API and very unlikely to change, there's a chance it could. It's far more reliable to implement this yourself (loops, StringBuilders, recursion whatever you like better).
Sure this approach may seem "neater" or more "too sweet" or "money" but it is, in my opinion, a worse approach.
Solution 4 - Java
A variation on codefin's answer
public static String concatStringsWSep(Iterable<String> strings, String separator) {
StringBuilder sb = new StringBuilder();
String sep = "";
for(String s: strings) {
sb.append(sep).append(s);
sep = separator;
}
return sb.toString();
}
Solution 5 - Java
If you are developing for Android, there is TextUtils.join
provided by the SDK.
Solution 6 - Java
This is the most elegant and clean way I've found so far:
list.stream().collect(Collectors.joining(delimiter));
Solution 7 - Java
Guava is a pretty neat library from Google:
Joiner joiner = Joiner.on(", ");
joiner.join(sList);
Solution 8 - Java
I prefer String.join(list) in Java 8
Solution 9 - Java
Have you seen this Coding Horror blog entry?
The Sad Tragedy of Micro-Optimization Theater
I am not shure whether or not it is "neater", but from a performance-standpoint it probably won't matter much.
Solution 10 - Java
ArrayList
inherits its toString()
-method from AbstractCollection
, ie:
public String toString() {
Iterator<E> i = iterator();
if (! i.hasNext())
return "[]";
StringBuilder sb = new StringBuilder();
sb.append('[');
for (;;) {
E e = i.next();
sb.append(e == this ? "(this Collection)" : e);
if (! i.hasNext())
return sb.append(']').toString();
sb.append(", ");
}
}
Building the string yourself will be far more efficient.
If you really want to aggregate the strings beforehand in some sort of List, you should provide your own method to efficiently join them, e.g. like this:
static String join(Collection<?> items, String sep) {
if(items.size() == 0)
return "";
String[] strings = new String[items.size()];
int length = sep.length() * (items.size() - 1);
int idx = 0;
for(Object item : items) {
String str = item.toString();
strings[idx++] = str;
length += str.length();
}
char[] chars = new char[length];
int pos = 0;
for(String str : strings) {
str.getChars(0, str.length(), chars, pos);
pos += str.length();
if(pos < length) {
sep.getChars(0, sep.length(), chars, pos);
pos += sep.length();
}
}
return new String(chars);
}
Solution 11 - Java
It seems to me that the StringBuilder will be quick and efficient.
The basic form would look something like this:
public static String concatStrings(List<String> strings)
{
StringBuilder sb = new StringBuilder();
for(String s: strings)
{
sb.append(s);
}
return sb.toString();
}
If that's too simplistic (and it probably is), you can use a similar approach and add a separator like this:
public static String concatStringsWSep(List<String> strings, String separator)
{
StringBuilder sb = new StringBuilder();
for(int i = 0; i < strings.size(); i++)
{
sb.append(strings.get(i));
if(i < strings.size() - 1)
sb.append(separator);
}
return sb.toString();
}
I agree with the others who have responded to this question when they say that you should not rely on the toString() method of Java's ArrayList.
Solution 12 - Java
> I somehow found this to be neater than > using the StringBuilder/StringBuffer > approach.
I guess it depends on what approach you took.
The AbstractCollection#toString() method simply iterates over all the elements and appends them to a StringBuilder. So your method may be saving a few lines of code but at the cost of extra String manipulation. Whether that tradeoff is a good one is up to you.
Solution 13 - Java
Rather than depending on ArrayList.toString()
implementation, you could write a one-liner, if you are using java 8:
String result = sList.stream()
.reduce("", String::concat);
If you prefer using StringBuffer
instead of String since String::concat
has a runtime of O(n^2)
, you could convert every String
to StringBuffer
first.
StringBuffer result = sList.stream()
.map(StringBuffer::new)
.reduce(new StringBuffer(""), StringBuffer::append);
Solution 14 - Java
Next variation on Peter Lawrey's answer without initialization of a new string every loop turn
String concatList(List<String> sList, String separator)
{
Iterator<String> iter = sList.iterator();
StringBuilder sb = new StringBuilder();
while (iter.hasNext())
{
sb.append(iter.next()).append( iter.hasNext() ? separator : "");
}
return sb.toString();
}
Solution 15 - Java
Assuming it's faster to just move a pointer / set a byte to null (or however Java implements StringBuilder#setLength), rather than check a condition each time through the loop to see when to append the delimiter, you could use this method:
public static String Intersperse (Collection<?> collection, String delimiter) { StringBuilder sb = new StringBuilder (); for (Object item : collection) { if (item == null) continue; sb.append (item).append (delimiter); } sb.setLength (sb.length () - delimiter.length ()); return sb.toString (); }
Solution 16 - Java
Depending on the need for performance and amount of elements to be added, this might be an ok solution. If the amount of elements are high, the Arraylist
s reallocation of memory might be a bit slower than StringBuilder
.
Solution 17 - Java
Using the Functional Java library, import these:
import static fj.pre.Monoid.stringMonoid;
import static fj.data.List.list;
import fj.data.List;
... then you can do this:
List<String> ss = list("foo", "bar", "baz");
String s = stringMonoid.join(ss, ", ");
Or, the generic way, if you don't have a list of Strings:
public static <A> String showList(List<A> l, Show<A> s) {
return stringMonoid.join(l.map(s.showS_()), ", ");
}
Solution 18 - Java
if you have json in your dependencies.you can use new JSONArray(list).toString()
Solution 19 - Java
In java 8 you can also use a reducer, something like:
public static String join(List<String> strings, String joinStr) {
return strings.stream().reduce("", (prev, cur) -> prev += (cur + joinStr));
}