python之字数缓存的嵌套字典

如果之前已经解决过这个问题，我们深表歉意。我找不到任何以前的答案来解决我的具体问题，所以就在这里。

练习要求用户输入 .txt 文件名。该代码获取该文件，并对其中的单词进行计数，创建单词字典:计数对。如果文件已经被输入，并且它的字数被计算在内，那么程序不会重新计算它，而是引用缓存，它存储了以前的计数。

我的问题是创建一个嵌套的字典字典 - 缓存。以下是我到目前为止所拥有的。目前，每个新的 .txt 文件都会重写字典，并防止它被用作缓存。

def main(): 
 
file = input("Enter the file name: ")       #Takes a file input to count the words 
 
d = {}    #open dictionary of dictionaries: a cache of word counts] 
 
with open(file) as f: 
 
    if f in d:      #check if this file is in cache. 
 
        for word in sorted(d[f]):       #print the result of the word count of an old document. 
            print("That file has already been assessed:\n%-12s:%5d" % (word, d[f][word])) 
 
    else:       #count the words in this file and add the count to the cache as a nested list. 
 
        d[f] = {}       #create a nested dictionary within 'd'.    
 
        for line in f:              #counts the unique words within the document. 
            words = line.split() 
 
            for word in words: 
                word = word.rstrip("!'?.,")     #clean up punctuation here 
                word = word.upper()             #all words to uppercase here 
 
                if word not in d[f]: 
                    d[f][word] = 1 
                else: 
                    d[f][word] = d[f][word] + 1 
 
    for word in sorted(d[f]):       #print the result of the word count of a new document. 
        print("%-12s:%5d" % (word, d[f][word])) 
 
 
    main()      #Run code again to try new file. 
 
main()  

请您参考如下方法:

简单修复:

d[file] = {} 
.... 
d[file][word] = 1  # and so on 

因为当你 cahnge f d[f] 仍然引用 d 中的同一个条目

另外，你可以重用defaultdict:

from collections import defaultdict 
 
d = defaultdict(lambda x: defaultdict(int)) 
 
def count(file): 
    with (open(file)) as f: 
        if file not in d: 
            # this is just list comprehension 
            [d[file][word.rstrip("!'?.,").upper()] += 1  
                 for word in line.split() 
                     for line  in f] 
    return d[file] 
 
def main(): 
    file = input("Enter the file name: ") 
    count(file) 
    if file in d: 
        print("That file has already been assessed, blah blah") 
    for word in sorted(d[file]):       #print the result of the word count of a new document. 
        print("%-12s:%5d" % (word, d[f][word])) 
 
if __name__ == "__main__": 
    main()