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bforests are dictionary-like objects that use multiple BTrees for a backend and |
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support rotation of the composite trees. This supports various implementations |
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of timed member expirations, enabling caches and semi-persistent storage. A |
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useful and simple subclass would be to promote a key-value pair to the |
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first (newest) bucket whenever the key is accessed, for instance. |
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Like btrees, bforests come in four flavors: Integer-Integer (IIBForest), |
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Integer-Object (IOBForest), Object-Integer (OIBForest), and Object-Object |
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(OOBForest). The examples here will deal with them in the abstract: we will |
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create classes from the imaginary and representative BForest class, and |
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generate keys from KeyGenerator and values from ValueGenerator. From the |
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examples you should be able to extrapolate usage of all four types. |
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First let's instantiate a bforest and look at an empty example. By default, |
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a new bforest creates two composite btree buckets. |
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>>> d = BForest() |
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>>> list(d.keys()) |
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[] |
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>>> list(d.values()) |
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[] |
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>>> len(d.buckets) |
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2 |
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>>> dummy_key = KeyGenerator() |
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>>> d.get(dummy_key) |
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>>> d.get(dummy_key, 42) |
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42 |
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Now we'll populate it. We'll first create a dictionary we'll use to compare. |
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>>> original = {} |
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>>> for i in range(10): |
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... original[KeyGenerator()] = ValueGenerator() |
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... |
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>>> d.update(original) |
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>>> d == original |
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True |
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>>> d_keys = list(d.keys()) |
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>>> d_keys.sort() |
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>>> o_keys = original.keys() |
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>>> o_keys.sort() |
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>>> d_keys == o_keys |
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True |
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>>> d_values = list(d.values()) |
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>>> d_values.sort() |
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>>> o_values = original.values() |
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>>> o_values.sort() |
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>>> o_values == d_values |
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True |
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>>> d_items = list(d.items()) |
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>>> d_items.sort() |
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>>> o_items = original.items() |
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>>> o_items.sort() |
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>>> o_items == d_items |
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True |
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>>> key, value = d.popitem() |
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>>> value == original.pop(key) |
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True |
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>>> key, value = original.popitem() |
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>>> value == d.pop(key) |
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True |
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>>> len(d) == len(original) |
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True |
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Now let's rotate the buckets. |
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>>> d.rotateBucket() |
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...and we'll do the exact same test as above, first. |
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>>> d == original |
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True |
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>>> d_keys = list(d.keys()) |
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>>> d_keys.sort() |
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>>> o_keys = original.keys() |
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>>> o_keys.sort() |
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>>> d_keys == o_keys |
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True |
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>>> d_values = list(d.values()) |
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>>> d_values.sort() |
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>>> o_values = original.values() |
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>>> o_values.sort() |
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>>> o_values == d_values |
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True |
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>>> d_items = list(d.items()) |
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>>> d_items.sort() |
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>>> o_items = original.items() |
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>>> o_items.sort() |
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>>> o_items == d_items |
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True |
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>>> key, value = d.popitem() |
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>>> value == original.pop(key) |
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True |
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>>> key, value = original.popitem() |
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>>> value == d.pop(key) |
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True |
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>>> len(d) == len(original) |
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True |
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Now we'll make a new dictionary to represent changes made after the bucket |
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rotation. |
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>>> second = {} |
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>>> for i in range(10): |
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... key = KeyGenerator() |
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... value = ValueGenerator() |
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... second[key] = value |
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... d[key] = value |
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... |
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>>> original.update(second) |
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|
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...and we'll do almost the exact same test as above, first. |
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>>> d == original |
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True |
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>>> d_keys = list(d.keys()) |
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>>> d_keys.sort() |
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>>> o_keys = original.keys() |
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>>> o_keys.sort() |
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>>> d_keys == o_keys |
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True |
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>>> d_values = list(d.values()) |
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>>> d_values.sort() |
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>>> o_values = original.values() |
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>>> o_values.sort() |
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>>> o_values == d_values |
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True |
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>>> d_items = list(d.items()) |
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>>> d_items.sort() |
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>>> o_items = original.items() |
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>>> o_items.sort() |
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>>> o_items == d_items |
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True |
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>>> key, value = d.popitem() |
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>>> ignore = second.pop(key, None) # keep second up-to-date |
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>>> value == original.pop(key) |
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True |
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>>> key, value = original.popitem() |
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>>> ignore = second.pop(key, None) # keep second up-to-date |
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>>> value == d.pop(key) |
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True |
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>>> len(d) == len(original) |
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True |
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Now if we rotate the buckets, the first set of items will be gone, but the |
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second will remain. |
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>>> d.rotateBucket() |
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>>> d == original |
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False |
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>>> d == second |
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True |
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Let's set a value, check the copy behavior, and then rotate it one more time. |
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>>> third = {KeyGenerator(): ValueGenerator()} |
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>>> d.update(third) |
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>>> copy = d.copy() |
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>>> copy == d |
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True |
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>>> copy != second # because second doesn't have the values of third |
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True |
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>>> list(copy.buckets[0].items()) == list(d.buckets[0].items()) |
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True |
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>>> list(copy.buckets[1].items()) == list(d.buckets[1].items()) |
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True |
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>>> copy[KeyGenerator()] = ValueGenerator() |
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>>> copy == d |
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False |
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>>> d.rotateBucket() |
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>>> d == third |
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True |
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>>> d.clear() |
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>>> d == BForest() == {} |
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True |
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>>> d.update(second) |
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We'll make a value in one bucket that we'll override in another. |
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>>> d[third.keys()[0]] = ValueGenerator() |
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>>> d.rotateBucket() |
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>>> d.update(third) |
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>>> second.update(third) |
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>>> d == second |
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True |
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Notice this unpleasant bit: a bforest and dict with the same contents will |
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return True if the comparison is bforest == dict (above) but false if the |
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comparison is dict == bforest. This might be fixable if we could use rich |
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comparisons, but we can't do that with the old-style ExtensionClass used in |
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pre-Zope 2.8. |
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>>> second == d |
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False |
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The tree method converts the bforest to a btree as efficiently as I know how |
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for a common case of more items in buckets than buckets. |
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>>> tree = d.tree() |
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>>> d_items = list(d.items()) |
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>>> d_items.sort() |
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>>> t_items = list(tree.items()) |
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>>> t_items.sort() |
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>>> t_items == d_items |
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True |
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