zope.bforest 1.1.1

Mappings based transparently on multiple BTrees; good for rotating caches and logs.

BForests are dictionary-like objects that use multiple BTrees for a backend and support rotation of the composite trees. This supports various implementations of timed member expirations, enabling caches and semi-persistent storage. A useful and simple subclass would be to promote a key-value pair to the first (newest) bucket whenever the key is accessed, for instance. It also is useful with disabling the rotation capability.

Like btrees, bforests come in four flavors: Integer-Integer (IIBForest), Integer-Object (IOBForest), Object-Integer (OIBForest), and Object-Object (OOBForest). The examples here will deal with them in the abstract: we will create classes from the imaginary and representative BForest class, and generate keys from KeyGenerator and values from ValueGenerator. From the examples you should be able to extrapolate usage of all four types.

First let’s instantiate a bforest and look at an empty example. By default, a new bforest creates two composite btree buckets.

>>> d = BForest()
>>> list(d.keys())
[]
>>> list(d.values())
[]
>>> len(d.buckets)
2
>>> dummy_key = KeyGenerator()
>>> d.get(dummy_key)
>>> d.get(dummy_key, 42)
42

Now we’ll populate it. We’ll first create a dictionary we’ll use to compare.

>>> original = {}
>>> for i in range(10):
...     original[KeyGenerator()] = ValueGenerator()
...
>>> d.update(original)
>>> d == original
True
>>> d_keys = list(d.keys())
>>> d_keys.sort()
>>> o_keys = original.keys()
>>> o_keys.sort()
>>> d_keys == o_keys
True
>>> d_values = list(d.values())
>>> d_values.sort()
>>> o_values = original.values()
>>> o_values.sort()
>>> o_values == d_values
True
>>> d_items = list(d.items())
>>> d_items.sort()
>>> o_items = original.items()
>>> o_items.sort()
>>> o_items == d_items
True
>>> key, value = d.popitem()
>>> value == original.pop(key)
True
>>> key, value = original.popitem()
>>> value == d.pop(key)
True
>>> len(d) == len(original)
True

Now let’s rotate the buckets.

>>> d.rotateBucket()

…and we’ll do the exact same test as above, first.

>>> d == original
True
>>> d_keys = list(d.keys())
>>> d_keys.sort()
>>> o_keys = original.keys()
>>> o_keys.sort()
>>> d_keys == o_keys
True
>>> d_values = list(d.values())
>>> d_values.sort()
>>> o_values = original.values()
>>> o_values.sort()
>>> o_values == d_values
True
>>> d_items = list(d.items())
>>> d_items.sort()
>>> o_items = original.items()
>>> o_items.sort()
>>> o_items == d_items
True
>>> key, value = d.popitem()
>>> value == original.pop(key)
True
>>> key, value = original.popitem()
>>> value == d.pop(key)
True
>>> len(d) == len(original)
True

Now we’ll make a new dictionary to represent changes made after the bucket rotation.

>>> second = {}
>>> for i in range(10):
...     key = KeyGenerator()
...     value = ValueGenerator()
...     second[key] = value
...     d[key] = value
...
>>> original.update(second)

…and we’ll do almost the exact same test as above, first.

>>> d == original
True
>>> d_keys = list(d.keys())
>>> d_keys.sort()
>>> o_keys = original.keys()
>>> o_keys.sort()
>>> d_keys == o_keys
True
>>> d_values = list(d.values())
>>> d_values.sort()
>>> o_values = original.values()
>>> o_values.sort()
>>> o_values == d_values
True
>>> d_items = list(d.items())
>>> d_items.sort()
>>> o_items = original.items()
>>> o_items.sort()
>>> o_items == d_items
True
>>> key, value = d.popitem()
>>> ignore = second.pop(key, None) # keep second up-to-date
>>> value == original.pop(key)
True
>>> key, value = original.popitem()
>>> ignore = second.pop(key, None) # keep second up-to-date
>>> value == d.pop(key)
True
>>> len(d) == len(original)
True

Now if we rotate the buckets, the first set of items will be gone, but the second will remain.

>>> d.rotateBucket()
>>> d == original
False
>>> d == second
True

Let’s set a value, check the copy behavior, and then rotate it one more time.

>>> third = {KeyGenerator(): ValueGenerator()}
>>> d.update(third)
>>> copy = d.copy()
>>> copy == d
True
>>> copy != second # because second doesn't have the values of third
True
>>> list(copy.buckets[0].items()) == list(d.buckets[0].items())
True
>>> list(copy.buckets[1].items()) == list(d.buckets[1].items())
True
>>> copy[KeyGenerator()] = ValueGenerator()
>>> copy == d
False
>>> d.rotateBucket()
>>> d == third
True
>>> d.clear()
>>> d == BForest() == {}
True

>>> d.update(second)

We’ll make a value in one bucket that we’ll override in another.

>>> d[third.keys()[0]] = ValueGenerator()
>>> d.rotateBucket()
>>> d.update(third)
>>> second.update(third)
>>> d == second
True
>>> second == d
True

The tree method converts the bforest to a btree as efficiently as I know how for a common case of more items in buckets than buckets.

>>> tree = d.tree()
>>> d_items = list(d.items())
>>> d_items.sort()
>>> t_items = list(tree.items())
>>> t_items.sort()
>>> t_items == d_items
True

Finally, comparisons work similarly to dicts but in a simpleminded way–improvements welcome! We’ve already looked at a lot of examples above, but here are some additional cases

>>> d == None
False
>>> d == [1, 2]
False
>>> d != None
True
>>> None == d
False
>>> d != None
True
>>> d >= second
True
>>> d >= dict(second)
True
>>> d <= second
True
>>> d <= dict(second)
True
>>> d > second
False
>>> d > dict(second)
False
>>> d < second
False
>>> d > dict(second)
False
>>> second.popitem()[0] in d
True
>>> d > second
True
>>> d < second
False
>>> d >= second
True
>>> d <= second
False
>>> second < d
True
>>> second > d
False
>>> second <= d
True
>>> second >= d
False