Python中的就地字典倒置 [英] In-place dictionary inversion in Python

问题描述

我需要反转一个列表字典，我不知道该如何用英语确切地解释它，所以这里有一些代码可以满足我的要求.只是占用太多内存.

I need to invert a dictionary of lists, I don't know how to explain it in English exactly, so here is some code that does what I want. It just takes too much memory.

def invert(oldDict):
    invertedDict = {}
    for key,valuelist in oldDict.iteritems():
        for value in valuelist:
            try:
                entry = invertedDict[value]
                if key not in entry:
                    entry.append(key)
            except KeyError:
                invertedDict[value] = [key]
    return invertedDict

原始是列表的字典，结果是列表的字典.这会反转"它.

The original is a dict of lists, and the result is a dict of lists. This "inverts" it.

test = {}
test[1] = [1999,2000,2001]
test[2] = [440,441]
test[3] = [440,2000]

print invert(test)

这给出了:

{2000: [1, 3], 2001: [1], 440: [2, 3], 441: [2], 1999: [1]}

我需要知道这是否可以就地完成，因为我当前的策略是使用正在使用的字典超出计算机上的物理内存量.您能想到一种使用发电机的方法吗?

I need to know if this can be done in-place, because my current strategy is exceeding the amount of physical memory on my machine with the dictionary I am working with. Can you think of a way to do it with generators?

推荐答案

这没有就位，但是使用popitem()消耗了oldDict

This doesn't do it in place, but consumes oldDict by using popitem()

from collections import defaultdict
def invert(oldDict):
    invertedDict = defaultdict(list)
    while oldDict:
        key, valuelist = oldDict.popitem()
        for value in valuelist:
            invertedDict[value].append(key)
    return invertedDict

我有一种感觉，除非增加大小，否则决不会调整字典的大小，因此您可能需要定期添加/删除虚拟项目.参见收缩率

I have a feeling that dict's are never resized unless the size increases, so you may need to add+remove a dummy item periodically. See Shrinkage rate

from collections import defaultdict
def invert(oldDict):
    invertedDict = defaultdict(list)
    i=0
    while oldDict:
        key, valuelist = oldDict.popitem()
        for value in valuelist:
            invertedDict[value].append(key)
        i+=1
        if i%1000==0: # allow the dict to release memory from time to time
            oldDict[None]=None
            del oldDict[None]
    return invertedDict

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