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    <p>Expanding upon Tobias' answer, using a <code>for</code> loop:</p> <pre><code>comp = [1, 2, 3] m = max(comp) for index in range(len(comp)): if comp[index] == m: # some action </code></pre> <p>Since indexing starts at 0, you <em>do not</em> need to do <code>len(comp) + 1</code>. I prefer using indexing in a <code>for</code> loop instead of the actual element, because it speeds things up considerably. Some times in a process, you may need the index of a specific element. Then, using <code>l.index(obj)</code> will waste time (even if only insignificant amounts --- for longer processes, this becomes tedious).</p> <p>This also assumes that every process (for the <code>comp[index]</code>) is very similar: same process but with different variables. This wouldn't work if you have significantly different processes for each index.</p> <p>However, by using <code>for index in range(len(l)):</code>, you already have the index and the item can easily be accessed with <code>l[index]</code> (along with the index, which is given by the loop).</p> <p>Oddly, it seems that Tobias' implementation is faster (I thought otherwise):</p> <pre><code>comp = [1, 2, 3] m = max(comp) from timeit import timeit def test1(): if comp[0] == m: return m if comp[1] == m: return m if comp[2] == m: return m def test2(): for index in range(len(comp)): if comp[index] == m: return m print 'test1:', timeit(test1, number = 1000) print 'test2:', timeit(test2, number = 1000) </code></pre> <p>Returns: </p> <pre><code>test1: 0.00121262329299 test2: 0.00469034990534 </code></pre> <p>My implementation may be faster for longer lists (not sure, though). However, writing the code for that is tedious (for a long list using repeated <code>if comp[n] == m</code>).</p>
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    1. PODetermining index of maximum value in a list - Optimalization
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    1. USRushy Panchal
    1. USRushy Panchal
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    1. COThis presumes `some_action` is the same thing each time, which doesn't really make sense. If that was the case, then there is no need for this, instead you would simply do `for _ in range(comp.count(max(comp))):`
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      1. USGareth Latty
    2. CO@Lattyware It assumes that they are similar things. However, you can manipulate it using the value of `comp[index]`. For example: `return comp[index] * 5`.
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    3. COThis is true, I presumed that wouldn't be the case given the original implementation, but the question is ambiguous. OP will have to explain further.
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      1. USGareth Latty
 

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