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.

(doseq [i (interleave items1 items2)]
  (println i))

using System.Collections.Generic;
using System.Math;

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]);
}

import std.range;
import std.algorithm.iteration;
import std.stdio;

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])
	}
}

Demo

import "fmt"

for i := range min(len(items1), len(items2)) {
	fmt.Println(items1[i])
	fmt.Println(items2[i])

}

Demo

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

mapM_ print . concat 
	$ zipWith (\a b -> [a,b]) items1 items2

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]);
});

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()
  }
}

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.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 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);

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();
}

# 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";

uses Math;

for i := 0 to Min(items1.Count-1,items2.Count-1) do 
  writeln(items1[i],', ',items2[i]);

use List::SomeUtils qw(zip);

print for zip @items1, @items2;

for pair in zip(item1, item2): print(pair)

items1.zip(items2){|pair| puts pair}

extern crate itertools;

for pair in izip!(&items1, &items2) {
    println!("{}", pair.0);
    println!("{}", pair.1);
}

Demo
Doc

(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].

Do you know the best way to do this in your language ? New implementation...

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Idiom created by BBaz

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Idiom #143 Iterate alternatively over two lists

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