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    <p>It sounds as if you want an implementation of the following abstract data type.</p> <pre><code>insert(x, w): add item x to the collection with weight x select(p): return the item greater than a p weighted fraction of the items </code></pre> <p>For example, <code>select(0)</code> returns the minimum, <code>select(0.5)</code> returns the weighted median, and <code>select(1)</code> returns the maximum.</p> <p>I would implement this ADT in one of two ways. If selection is infrequent, I'd put the data in an array and use a linear-time selection algorithm, for O(1)-time inserts and O(n)-time selects. If selection is frequent, I'd use a binary search tree where each node stores the total weight in its subtree. For example, after</p> <pre><code>insert(2, 10) insert(1, 5) insert(3, 100) insert(4, 20) </code></pre> <p>the tree might look like</p> <pre><code> 2 (135) / \ / \ 1 (5) 4 (120) / / 3 (100) </code></pre> <p>Now, to find the weighted median, multiply <code>135</code> by <code>0.5</code> and get <code>67.5</code> as the desired "index". Starting at the root <code>2</code>, we find that <code>5</code> is less than <code>67.5</code>, so the item is not in the left subtree and we subtract <code>5</code> to obtain <code>62.5</code>, the index into the remainder of the tree. Since <code>135 - 120 = 15</code> is less than <code>62.5</code>, the median isn't <code>2</code>. We subtract <code>15</code> from <code>62.5</code> to obtain <code>47.5</code> and descend to <code>4</code>. At <code>4</code>, we find that <code>100</code> is greater than <code>47.5</code>, so <code>3</code> is the median.</p> <p>Assuming a balanced tree, the running time of both <code>insert</code> and <code>select</code> is <code>O(log n)</code>. If I were implementing from scratch, I'd probably opt for a splay tree.</p>
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    1. POHow to keep a dynamical histogram?
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    1. POHow to keep a dynamical histogram?
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    1. COThis looks neat, and it is probably the ideal option. I can get an aproximation for the cumulative distribution function right away... I'll look into it. But csgillespie answer seem more practical for the moment.
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      1. USRafael S. Calsaverini
    2. COIf I could choose two answers, I'd choose this as a second answer.
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      1. USRafael S. Calsaverini
 

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