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copied!<p>What you are attempting to do in this code is not what you think (i suspect) for instance:<br> <code>x[::2]</code> will return a slice containing every <code>odd</code> element of x, since x is of size 20,<br> the slice will be of size 10, but you are trying to assign a non-iterable of size 1 to it.</p> <p>to successfully use the code you have you will need to do:</p> <pre><code>x = [True]*20 x[::2] = [False]*10 </code></pre> <p>wich will assign an iterable of size 10 to a slice of size 10.</p> <p>Why work in the dark with the number of elements? use </p> <pre><code>len(x[::2]) </code></pre> <p>which would be equal to 10, and then use </p> <pre><code>x[::2] = [False]*len(x[::2]) </code></pre> <p>you could also do something like:</p> <pre><code>x = [True if (index & 0x1 == 0) else False for index, element in enumerate(x)] </code></pre> <hr> <p>EDIT: Due to OP edit</p> <p>The <a href="http://docs.python.org/library/itertools.html#itertools.cycle" rel="nofollow">documentation</a> on cycle says it <code>Repeats indefinitely.</code> which means it will continuously 'cycle' through the iterator it has been given.</p> <p>Repeat has a similar implementation, however <a href="http://docs.python.org/library/itertools.html#itertools.repeat" rel="nofollow">documentation</a> states that it<br> <code>Runs indefinitely unless the times argument is specified.</code><br> which has not been done in the questions code. Thus both will lead to infinite loops.</p> <p>About the <code>itertools</code> being faster comment. Yes itertools are generally faster than other implementations because they are optimised to be as fast as the creators could make them.</p> <p>However if you do not want to recreate a list you can use <code>generator expressions</code> such as the following:</p> <pre><code>x = (True if (index & 0x1 == 0) else False for index, element in enumerate(x)) </code></pre> <p>which do not store all of their elements in memory but produce them as they are needed, however, generator functions can be used up.</p> <p>for instance:</p> <pre><code>x = [True]*20 print(x) y = (True if (index & 0x1 == 0) else False for index, element in enumerate(x)) print ([a for a in y]) print ([a for a in y]) </code></pre> <p>will print <code>x</code> then the elements in the generator <code>y</code>, then a null list, because the generator has been used up.</p>

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