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copied!<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>>>> from itertools import zip_longest >>> a = [1, 2, 3] >>> b = ["hi", "bye", True, False, 33] >>> [y for x in zip_longest(a, b) for y in x] [1, 'hi', 2, 'bye', 3, True, None, False, None, 33] >>> [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] >>> </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>>>> a = [1, 2, 3] >>> b = ["hi", "bye", True, False, 33] >>> items = sorted((a,b), key=len) >>> out = [] >>> 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 ... >>> out ['hi', 1, 'bye', 2, True, 3, False, 33] >>> </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>>>> a = [1, 2, 3] >>> b = ["hi", "bye", True, False, 33] >>> out = [] >>> items = sorted((a,b), key=len) >>> 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 ... >>> out ['hi', 1, 'bye', 2, True, 3, False, 33] >>> </code></pre>

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