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    <p>Here is a solution using <a href="http://docs.python.org/3.2/library/itertools.html#itertools.zip_longest" rel="nofollow"><code>itertools.zip_longest</code></a> and a <a href="http://docs.python.org/2/tutorial/datastructures.html#list-comprehensions" rel="nofollow">list comprehension</a>:</p> <pre><code>&gt;&gt;&gt; from itertools import zip_longest &gt;&gt;&gt; a = [1, 2, 3] &gt;&gt;&gt; b = ["hi", "bye", True, False, 33] &gt;&gt;&gt; [y for x in zip_longest(a, b) for y in x] [1, 'hi', 2, 'bye', 3, True, None, False, None, 33] &gt;&gt;&gt; [y for x in zip_longest(a, b) for y in x if y is not None] [1, 'hi', 2, 'bye', 3, True, False, 33] &gt;&gt;&gt; </code></pre> <hr> <p><strong>Edit:</strong></p> <p>If you don't want to use <code>itertools</code>, I guess you could do:</p> <pre><code>&gt;&gt;&gt; a = [1, 2, 3] &gt;&gt;&gt; b = ["hi", "bye", True, False, 33] &gt;&gt;&gt; items = sorted((a,b), key=len) &gt;&gt;&gt; out = [] &gt;&gt;&gt; for x,y in enumerate(items[1]): ... out.append(y) ... try: ... out.append(items[0][x]) ... except IndexError: ... out.extend(items[1][x+1:]) ... break ... &gt;&gt;&gt; out ['hi', 1, 'bye', 2, True, 3, False, 33] &gt;&gt;&gt; </code></pre> <p>But I must say this method isn't as efficient as my first solution.</p> <hr> <p><strong>Edit 2:</strong></p> <p>Changing <code>enumerate</code> to <code>zip</code> reduces the efficiency again. However, if you <em>must</em>, then you can do this:</p> <pre><code>&gt;&gt;&gt; a = [1, 2, 3] &gt;&gt;&gt; b = ["hi", "bye", True, False, 33] &gt;&gt;&gt; out = [] &gt;&gt;&gt; items = sorted((a,b), key=len) &gt;&gt;&gt; for x,y in zip(range(len(items[1])), items[1]): ... out.append(y) ... try: ... out.append(items[0][x]) ... except IndexError: ... out.extend(items[1][x+1:]) ... break ... &gt;&gt;&gt; out ['hi', 1, 'bye', 2, True, 3, False, 33] &gt;&gt;&gt; </code></pre>
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    1. POhow to interleaving lists
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    1. COIs there a way to do it in the simplest form?
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    2. CO@user2829177 - What do you mean by "simplest form"? My solution is small and efficient. Plus, `itertools` is a standard library, so it will always be there.
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    3. COI don't want to use any functions that I don't know yet
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