在Python中的文本文件中計算單詞的頻率

我想弄清楚如何製作一個程序，該文件需要用戶選擇的文件（通過輸入文件名）並計算每個單詞的頻率投入。在Python中的文本文件中計算單詞的頻率

我有大部分，但是當我在多個詞的程序，找出輸入，僅第一字顯示正確的頻率，其餘顯示爲「0次出現」

file_name = input("What file would you like to open? ") 
f = open(file_name, "r") 
the_full_text = f.read() 
words = the_full_text.split() 
search_word = input("What words do you want to find? ").split(",") 
len_list = len(search_word) 

word_number = 0 
print() 
print ('... analyzing ... hold on ...') 
print() 
print ('Frequency of word usage within', file_name+":") 
for i in range(len_list): 

    frequency = 0 
    for word in words: 
     word = word.strip(",.") 
     if search_word[word_number].lower() == word.lower(): 
      frequency += 1 
    print (" ",format(search_word[word_number].strip(),'<20s'),"/", frequency, "occurrences") 
    word_number = word_number + 1

等的例子輸出將是：

What file would you like to open? assignment_8.txt 
What words do you want to find? wey, rights, dem 

... analyzing ... hold on ... 

Frequency of word usage within assignment_8.txt: 
    wey    /96 occurrences 
    rights    /0 occurrences 
    dem    /0 occurrences

我的程序出了什麼問題？請幫忙：o

來源

2016-12-07 Hanako Ohashi

如果你在''分裂，''，你的輸入不應該是''wey，rights，dem''，沒有空白嗎？ –

您需要去掉搜索詞中的空格。

但是，您當前的算法效率非常低，因爲它必須重新掃描每個搜索詞的整個文本。這是一個更有效的方法。首先，我們清理搜索詞並將其放入列表中。然後，我們在該列表中建立一個字典，以便在文本文件中找到它們時存儲每個這些字詞的計數。

file_name = input("What file would you like to open? ") 
with open(file_name, "r") as f: 
    words = f.read().split() 

search_words = input("What words do you want to find? ").split(',') 
search_words = [word.strip().lower() for word in search_words] 
#print(search_words) 
search_counts = dict.fromkeys(search_words, 0) 

print ('\n... analyzing ... hold on ...') 
for word in words: 
    word = word.rstrip(",.").lower() 
    if word in search_counts: 
     search_counts[word] += 1 

print ('\nFrequency of word usage within', file_name + ":") 
for word in search_words: 
    print(" {:<20s}/{} occurrences".format(word, search_counts[word]))

來源

2016-12-07 08:02:48

在Python中的文本文件中計算單詞的頻率

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