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    <p>EDIT: Here's some actual numbers! The <code>izip</code>, list comprehension, and <code>numpy</code> ways of doing this are all about the same speed.</p> <pre><code># zip &gt;&gt;&gt; timeit.timeit( "newlist = zip(*someList)[0]", setup = "someList = [range(1000000), range(1000000), range(1000000)]", number = 10 ) 1.4984046398561759 # izip &gt;&gt;&gt; timeit.timeit( "newlist = izip(*someList).next()", setup = "someList = range(1000000), range(1000000), range(1000000)]; from itertools import izip", number = 10 ) 2.2186223645803693e-05 # list comprehension &gt;&gt;&gt; timeit.timeit( "newlist = [li[0] for li in someList]", setup = "someList = [range(1000000), range(1000000), range(1000000)]", number = 10 ) 1.4677040212518477e-05 # numpy &gt;&gt;&gt; timeit.timeit( "newlist = someList[0,:]", setup = "import numpy as np; someList = np.array([range(1000000), range(1000000), range(1000000)])", number = 10 ) 6.6217344397045963e-05 &gt;&gt;&gt; </code></pre> <hr> <p>For large data structures like this you should use <a href="http://numpy.scipy.org/" rel="noreferrer"><code>numpy</code></a>, which implementes an array type in C and hence is significantly more efficient. It also provides all the matrix manipulation you will ever want.</p> <pre><code>&gt;&gt;&gt; import numpy as np &gt;&gt;&gt; foo = np.array([[0,1,2],[3,4,5],[6,7,8]]) &gt;&gt;&gt; foo[:,0] array([0, 3, 6]) </code></pre> <p>You can also transpose...</p> <pre><code>&gt;&gt;&gt; foo.transpose() array([[0, 3, 6], [1, 4, 7], [2, 5, 8]]) </code></pre> <p>...work with n-dimensional arrays...</p> <pre><code>&gt;&gt;&gt; foo = np.zeros((3,3,3)) &gt;&gt;&gt; foo array([[[ 0., 0., 0.], [ 0., 0., 0.], [ 0., 0., 0.]], [[ 0., 0., 0.], [ 0., 0., 0.], [ 0., 0., 0.]], [[ 0., 0., 0.], [ 0., 0., 0.], [ 0., 0., 0.]]]) &gt;&gt;&gt; foo[0,...] array([[ 0., 0., 0.], [ 0., 0., 0.], [ 0., 0., 0.]]) </code></pre> <p>..do efficient linear algebra...</p> <pre><code>&gt;&gt;&gt; foo = no.ones((3,3)) &gt;&gt;&gt; np.linalg.qr(foo) (array([[-0.57735027, 0.81649658, 0. ], [-0.57735027, -0.40824829, -0.70710678], [-0.57735027, -0.40824829, 0.70710678]]), array([[ -1.73205081e+00, -1. 73205081e+00, -1.73205081e+00], [ 0.00000000e+00, -1.57009246e-16, -1.57009246e-16], [ 0.00000000e+00, 0.00000000e+00, 0.00000000e+00]])) </code></pre> <p>...and basically do <a href="http://mathesaurus.sourceforge.net/matlab-numpy.html" rel="noreferrer">anything that Matlab can</a>.</p>
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    1. COWOW, thanks so much. This is SUPER FAST compared to other solutions. If I could +100000 I would
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    2. CO@Richard: That depends ... the larger the list is, the better the izip and list comprehension solutions perform against numpy, because they don't consider the vast majority of the matrix entries.
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    3. CO@jellybean: I doubt that... numpy is really quite well optimised. I think it's safe to say that a simple column-wise slice won't read the entire matrix. (Though I may of course be wrong, I don't know the C implementation details.) In fact, I think that `numpy` should perform *better* with a larger list, as the overhead of Python's native `list` type starts to add up. But of course, there's only one way to find out!
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