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Programming-Idioms

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  • Java

Idiom #143 Iterate alternatively over two lists

Iterate alternatively over the elements of the lists items1 and items2. For each iteration, print the element.

Explain what happens if items1 and items2 have different size.

import java.util.Iterator;
Iterator<String> iter1 = items1.iterator();
Iterator<String> iter2 = items2.iterator();
while (iter1.hasNext() || iter2.hasNext()) {
	if (iter1.hasNext()) {
		System.out.println(iter1.next());
	}
	if (iter2.hasNext()) {
		System.out.println(iter2.next());
	}
}
import static java.lang.System.out;
import java.util.Iterator;
Iterator<T> A = a.iterator(), B = b.iterator();
boolean x = A.hasNext(), y = B.hasNext();
while (x || y) {
    if (x) out.println(A.next());
    if (y) out.println(B.next());
    x = A.hasNext();
    y = B.hasNext();
}
import static java.lang.Math.max;
import static java.lang.System.out;
int i, A, B, n = max(A = a.size(), B = b.size());
for (i = 0; i < n; ++i) {
    if (i < A) out.println(a.get(i));
    if (i < B) out.println(b.get(i));
}
import java.util.stream.IntStream;
import java.util.stream.Stream;
IntStream.range(0, Math.max(items1.size(), items2.size()))
	.boxed()
	.flatMap(idx -> Stream.of(
		items1.size() > idx ? items1.get(idx) : null,
		items2.size() > idx ? items2.get(idx) : null
	))
	.filter(Objects::nonNull)
	.forEach(System.out::println);

Get a stream of the list indices.
For each index, map to the elements from the lists at that index
Filter out null values which occur when the index is after the end of the list
Print the result
const iterator1 = items1[Symbol.iterator]()
const iterator2 = items2[Symbol.iterator]()

let result1 = iterator1.next()
let result2 = iterator2.next()

while(!(result1.done && result2.done)) {
  if (!result1.done) {
    console.log(result1.value)
    result1 = iterator1.next()
  }
  if (!result2.done) {
    console.log(result2.value)
    result2 = iterator2.next()
  }
}

Approach that purely uses Iterators, similar to the Java Iterator example
const shorter = _items1.length > _items2.length ? _items2 : _items1;
const longer = _items1.length <= _items2.length ? _items2 : _items1;
shorter.map((m, i) => {
  console.log(m);
  console.log(longer[i]);
});

will limit each array to the length of the shortest array
(doseq [i (interleave items1 items2)]
  (println i))

interleave can receive any number of items, but truncates when any of them is exhausted
New implementation...
< > ⌨
BBaz 
Iterator<String> iter1 = items1.iterator();
Iterator<String> iter2 = items2.iterator();
while (iter1.hasNext() || iter2.hasNext()) {
	if (iter1.hasNext()) {
		System.out.println(iter1.next());
	}
	if (iter2.hasNext()) {
		System.out.println(iter2.next());
	}
}
Iterator<T> A = a.iterator(), B = b.iterator();
boolean x = A.hasNext(), y = B.hasNext();
while (x || y) {
    if (x) out.println(A.next());
    if (y) out.println(B.next());
    x = A.hasNext();
    y = B.hasNext();
}
int i, A, B, n = max(A = a.size(), B = b.size());
for (i = 0; i < n; ++i) {
    if (i < A) out.println(a.get(i));
    if (i < B) out.println(b.get(i));
}
IntStream.range(0, Math.max(items1.size(), items2.size()))
	.boxed()
	.flatMap(idx -> Stream.of(
		items1.size() > idx ? items1.get(idx) : null,
		items2.size() > idx ? items2.get(idx) : null
	))
	.filter(Objects::nonNull)
	.forEach(System.out::println);
const iterator1 = items1[Symbol.iterator]()
const iterator2 = items2[Symbol.iterator]()

let result1 = iterator1.next()
let result2 = iterator2.next()

while(!(result1.done && result2.done)) {
  if (!result1.done) {
    console.log(result1.value)
    result1 = iterator1.next()
  }
  if (!result2.done) {
    console.log(result2.value)
    result2 = iterator2.next()
  }
}
const shorter = _items1.length > _items2.length ? _items2 : _items1;
const longer = _items1.length <= _items2.length ? _items2 : _items1;
shorter.map((m, i) => {
  console.log(m);
  console.log(longer[i]);
});
(doseq [i (interleave items1 items2)]
  (println i))
for(int i = 0; i < Math.Max(items1.Count, items2.Count); i++)
{
  if (i < items1.Count) Console.WriteLine(items1[i]);
  if (i < items2.Count) Console.WriteLine(items2[i]);
}

roundRobin(items1, items2).each!writeln;
do i=1, min(size(items1), size(items2))
  print *,items1(i), items2(i)
end do
for i := 0; i < len(items1) || i < len(items2); i++ {
	if i < len(items1) {
		fmt.Println(items1[i])
	}
	if i < len(items2) {
		fmt.Println(items2[i])
	}
}
for i := range min(len(items1), len(items2)) {
	fmt.Println(items1[i])
	fmt.Println(items2[i])

}
mapM_ print . concat 
	$ zipWith (\a b -> [a,b]) items1 items2
interweave :: [a] -> [a] -> [a]   -- optional signature
interweave [] ys = ys
interweave xs [] = xs
interweave (x:xs) (y:ys) = x : y : interweave xs ys

main = mapM_ print $ interweave items1 items2

# php 5.3+
print array_reduce( array_map( null, $items1, $items2 ), function($c,$d){return $c.($c?"\n":'').implode("\n", $d); }, "")."\n" ;

#php 5.6+
print implode("\n", array_merge( ...array_map( null, $items1, $items2 )))."\n";
for i := 0 to Min(items1.Count-1,items2.Count-1) do 
  writeln(items1[i],', ',items2[i]);
print for zip @items1, @items2;
for pair in zip(item1, item2): print(pair)
print(*zip_longest(items1, items2))
a, b = iter(items1), iter(items2)
print(*zip_longest(a, b))
a, b = len(items1), len(items2)
for i in range(max(a, b)):
    if i < a: print(items1[i])
    if i < b: print(items2[i])
items1.zip(items2){|pair| puts pair}
for pair in izip!(&items1, &items2) {
    println!("{}", pair.0);
    println!("{}", pair.1);
}
(map (lambda (x)
       (display x)
       (newline))
     (foldr append '()
            (map list items1 items2)))
stream1 := items1 readStream.
stream2 := items2 readStream.
[stream1 atEnd ifFalse: [Transcript showln: stream1 next].
 stream2 atEnd ifFalse: [Transcript showln: stream2 next]]
	doWhileFalse: [stream1 atEnd or: [stream2 atEnd]].
items1 with: items2 do: [:item1 :item2 |
  Transcript 
	showln: item1;
	showln: item2].