How to limit the size of a dictionary?

醉酒当歌 提交于 2019-12-18 10:05:26

问题


I'd like to work with a dict in python, but limit the number of key/value pairs to X. In other words, if the dict is currently storing X key/value pairs and I perform an insertion, I would like one of the existing pairs to be dropped. It would be nice if it was the least recently inserted/accesses key but that's not completely necessary.

If this exists in the standard library please save me some time and point it out!


回答1:


Python 2.7 and 3.1 have OrderedDict and there are pure-Python implementations for earlier Pythons.

from collections import OrderedDict

class LimitedSizeDict(OrderedDict):
    def __init__(self, *args, **kwds):
        self.size_limit = kwds.pop("size_limit", None)
        OrderedDict.__init__(self, *args, **kwds)
        self._check_size_limit()

    def __setitem__(self, key, value):
        OrderedDict.__setitem__(self, key, value)
        self._check_size_limit()

    def _check_size_limit(self):
        if self.size_limit is not None:
            while len(self) > self.size_limit:
                self.popitem(last=False)

You would also have to override other methods that can insert items, such as update. The primary use of OrderedDict is so you can control what gets popped easily, otherwise a normal dict would work.




回答2:


cachetools will provide you nice implementation of Mapping Hashes that does this (and it works on python 2 and 3).

Excerpt of the documentation:

For the purpose of this module, a cache is a mutable mapping of a fixed maximum size. When the cache is full, i.e. by adding another item the cache would exceed its maximum size, the cache must choose which item(s) to discard based on a suitable cache algorithm.




回答3:


Here's a simple, no-LRU Python 2.6+ solution (in older Pythons you could do something similar with UserDict.DictMixin, but in 2.6 and better that's not recommended, and the ABCs from collections are preferable anyway...):

import collections

class MyDict(collections.MutableMapping):
    def __init__(self, maxlen, *a, **k):
        self.maxlen = maxlen
        self.d = dict(*a, **k)
        while len(self) > maxlen:
            self.popitem()
    def __iter__(self):
        return iter(self.d)
    def __len__(self):
        return len(self.d)
    def __getitem__(self, k):
        return self.d[k]
    def __delitem__(self, k):
        del self.d[k]
    def __setitem__(self, k, v):
        if k not in self and len(self) == self.maxlen:
            self.popitem()
        self.d[k] = v

d = MyDict(5)
for i in range(10):
    d[i] = i
    print(sorted(d))

As other answers mentioned, you probably don't want to subclass dict -- the explicit delegation to self.d is unfortunately boilerplatey but it does guarantee that every other method is properly supplied by collections.MutableMapping.




回答4:


Here is a simple and efficient LRU cache written with dirt simple Python code that runs on any python version 1.5.2 or later:

class LRU_Cache:

    def __init__(self, original_function, maxsize=1000):
        self.original_function = original_function
        self.maxsize = maxsize
        self.mapping = {}

        PREV, NEXT, KEY, VALUE = 0, 1, 2, 3         # link fields
        self.head = [None, None, None, None]        # oldest
        self.tail = [self.head, None, None, None]   # newest
        self.head[NEXT] = self.tail

    def __call__(self, *key):
        PREV, NEXT = 0, 1
        mapping, head, tail = self.mapping, self.head, self.tail

        link = mapping.get(key, head)
        if link is head:
            value = self.original_function(*key)
            if len(mapping) >= self.maxsize:
                old_prev, old_next, old_key, old_value = head[NEXT]
                head[NEXT] = old_next
                old_next[PREV] = head
                del mapping[old_key]
            last = tail[PREV]
            link = [last, tail, key, value]
            mapping[key] = last[NEXT] = tail[PREV] = link
        else:
            link_prev, link_next, key, value = link
            link_prev[NEXT] = link_next
            link_next[PREV] = link_prev
            last = tail[PREV]
            last[NEXT] = tail[PREV] = link
            link[PREV] = last
            link[NEXT] = tail
        return value

if __name__ == '__main__':
    p = LRU_Cache(pow, maxsize=3)
    for i in [1,2,3,4,5,3,1,5,1,1]:
        print(i, p(i, 2))



回答5:


A dict does not have this behavior. You could make your own class that does this, for example something like

class MaxSizeDict(object):
    def __init__(self, max_size):
        self.max_size = max_size
        self.dict = {}
    def __setitem__(self, key, value):
        if key in self.dict:
            self.dict[key] = value    
            return

        if len(self.dict) >= self.max_size:
      ...

A few notes about this

  • It would be tempting for some to subclass dict here. You can technically do this, but it is bug-prone because the methods do not depend on each other. You can use UserDict.DictMixin to save having to define all methods. There are few methods you would be able re-use if you subclass dict.
  • A dict does not know what the least recently added key is, since dicts are unordered.
    • 2.7 will introduce collections.OrderedDict, but for now keeping the keys in order separately should work fine (use a collections.deque as a queue).
    • If getting the oldest isn't all that imporant, you can just use the popitem method to delete one arbitrary item.
  • I interprettered oldest to mean first insertion, approximately. You would have to do something a bit different to eliminate the LRU items. The most obvious efficient strategy would involve keeping a doubly-linked list of keys with references to the nodes themselves stored as dict values (along with the real values). This gets more complicated and implementing it in pure Python carries a lot of overhead.



回答6:


You can create a custom dictionary class by subclassing dict. In your case, you would have to override __setitem__ to have check your own length and delete something if the limit is recahed. The following example would print the current lenght after every insertion:

class mydict(dict):
    def __setitem__(self, k, v):
        dict.__setitem__(self, k, v)
        print len(self)

d = mydict()
d['foo'] = 'bar'
d['bar'] = 'baz'



回答7:


There have been many good answers, but I want to point out a simple, pythonic implementation for LRU cache. It's similar to Alex Martelli's answer.

from collections import OrderedDict, MutableMapping

class Cache(MutableMapping):
    def __init__(self, maxlen, items=None):
        self._maxlen = maxlen
        self.d = OrderedDict()
        if items:
            for k, v in items:
                self[k] = v

    @property
    def maxlen(self):
        return self._maxlen

    def __getitem__(self, key):
        self.d.move_to_end(key)
        return self.d[key]

    def __setitem__(self, key, value):
        if key in self.d:
            self.d.move_to_end(key)
        elif len(self.d) == self.maxlen:
            self.d.popitem(last=False)
        self.d[key] = value

    def __delitem__(self, key):
        del self.d[key]

    def __iter__(self):
        return self.d.__iter__()

    def __len__(self):
        return len(self.d)


来源:https://stackoverflow.com/questions/2437617/how-to-limit-the-size-of-a-dictionary

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