字典列出并比较列表python

Question

我做了一个函数，我计算一个文件中每个单词的使用次数，即单词的频率。 现在，该函数可以计算所有单词的总和，并向我显示七个最常见的单词以及使用了多少次。 现在，我想比较我的第一个文件，如果我已经分析了另一个文件的词频，我是否拥有英语中最常用的词，我想将这些词与我的第一个文件中的词进行比较，看看是否任何单词都匹配。

我要做的是列出两个文件，然后将它们相互比较。 但是我为此编写的代码没有给我任何输出，对如何解决这个问题有任何想法吗？

def CountWords():
filename = input('What is the name of the textfile you want to open?: ')
if filename == "alice" or "alice-ch1.txt" or " ":
    file = open("alice-ch1.txt","r")
    print('You want to open alice-ch1.txt')
    wordcount = {}
    for word in file.read().split():
        if word not in wordcount:
            wordcount[word] = 1
        else:
            wordcount[word] += 1                                         
    wordcount = {k.lower(): v for k, v in wordcount.items() }
    print (wordcount)

    sum = 0
    for val in wordcount.values():
        sum += val
    print ('The total amount of words in Alice adventures in wonderland: ' + str(sum))
    sortList = sorted(wordcount.values(), reverse = True)
    most_freq_7 = sortList[0:7]
    #print (most_freq_7)
    print ('Totoro says: The 7 most common words in Alice Adventures in Wonderland:')
    print(list(wordcount.keys())[list(wordcount.values()).index(most_freq_7[0])] + " " + str(most_freq_7[0]))
    print(list(wordcount.keys())[list(wordcount.values()).index(most_freq_7[1])] + " " + str(most_freq_7[1]))
    print(list(wordcount.keys())[list(wordcount.values()).index(most_freq_7[2])] + " " + str(most_freq_7[2]))
    print(list(wordcount.keys())[list(wordcount.values()).index(most_freq_7[3])] + " " + str(most_freq_7[3]))
    print(list(wordcount.keys())[list(wordcount.values()).index(most_freq_7[4])] + " " + str(most_freq_7[4]))
    print(list(wordcount.keys())[list(wordcount.values()).index(most_freq_7[5])] + " " + str(most_freq_7[5]))
    print(list(wordcount.keys())[list(wordcount.values()).index(most_freq_7[6])] + " " + str(most_freq_7[6]))

    file_common = open("common-words.txt", "r")
    commonwords = []
    contents = file_common.readlines()

    for i in range(len(contents)):
        commonwords.append(contents[i].strip('\n'))
    print(commonwords) 

#From here's the code were I need to find out how to compare the lists:
    alice_keys = wordcount.keys()
    result = set(filter(set(alice_keys).__contains__, commonwords))
    newlist = list()


    for elm in alice_keys:
        if elm not in result:
            newlist.append(elm)
    print('Here are the similar words: ' + str(newlist)) #Why doesn't show?


else:
    print ('I am sorry, that filename does not exist. Please try again.')            

Answer 1

我不在解释器的前面，所以我的代码可能会稍微有些偏离。 但是尝试更多类似这样的方法。

from collections import Counter
with open("some_file_with_words") as f_file
  counter = Counter(f_file.read())
  top_seven = counter.most_common(7)
  with open("commonwords") as f_common:
    common_words = f_common.read().split()
    for word, count in top_seven:
      if word in common_words:
        print "your word " + word + " is in the most common words!  It appeared " + str(count) + " times!"