Python:如何在嵌套字典中更新键值对的值？

Question

我正在尝试制作一个反向的文档索引,因此我需要从集合中的所有独特单词中了解它们发生在哪些文档中以及发生的频率.

我已经使用这个答案,以便两个创建一个嵌套字典.提供的解决方案工作正常,但有一个问题.

首先,我打开文件并列出一个独特的单词列表.这些独特的单词我想要与原始文件进行比较.当存在匹配时,应更新频率计数器并将其值存储在二维数组中.

输出最终应该如下所示:

word1, {doc1 : freq}, {doc2 : freq} <br>
word2, {doc1 : freq}, {doc2 : freq}, {doc3:freq}
etc....

问题是我无法更新字典变量.尝试这样做时,我收到错误:

  File "scriptV3.py", line 45, in main
    freq = dictionary[keyword][filename] + 1
TypeError: unsupported operand type(s) for +: 'AutoVivification' and 'int'

我想我需要以某种方式将AutoVivification的实例转换为int ....

怎么去？

提前致谢

我的代码:

#!/usr/bin/env python 
# encoding: utf-8

import sys
import os
import re
import glob
import string
import sets

class AutoVivification(dict):
    """Implementation of perl's autovivification feature."""
    def __getitem__(self, item):
        try:
            return dict.__getitem__(self, item)
        except KeyError:
            value = self[item] = type(self)()
            return value

def main():
    pad = 'temp/'
    dictionary  = AutoVivification()
    docID = 0
    for files in glob.glob( os.path.join(pad, '*.html') ):  #for all files in specified folder:
        docID = docID + 1
        filename = "doc_"+str(docID)
        text = open(files, 'r').read()                      #returns content of file as string
        text = extract(text, '<pre>', '</pre>')             #call extract function to extract text from within <pre> tags
        text = text.lower()                                 #all words to lowercase
        exclude = set(string.punctuation)                   #sets list of all punctuation characters
        text = ''.join(char for char in text if char not in exclude) # use created exclude list to remove characters from files
        text = text.split()                                 #creates list (array) from string
        uniques = set(text)                                 #make list unique (is dat handig? we moeten nog tellen)

        for keyword in uniques:                             #For every unique word do   
            for word in text:                               #for every word in doc:
                if (word == keyword and dictionary[keyword][filename] is not None): #if there is an occurence of keyword increment counter 
                    freq = dictionary[keyword][filename]    #here we fail, cannot cast object instance to integer.
                    freq = dictionary[keyword][filename] + 1
                    print(keyword,dictionary[keyword])
                else:
                    dictionary[word][filename] = 1

#extract text between substring 1 and 2 
def extract(text, sub1, sub2): 
    return text.split(sub1, 1)[-1].split(sub2, 1)[0]    

if __name__ == '__main__':
    main()

Answer 1

可以使用Python的collections.defaultdict而不是创建AutoVivification类,然后将字典实例化为该类型的对象.

import collections
dictionary = collections.defaultdict(lambda: collections.defaultdict(int))

这将创建一个字典字典,其默认值为0.如果要增加条目,请使用:

dictionary[keyword][filename] += 1