Question

所以，我有文本文件（段落），我需要读取文件并创建一个包含文件中每个不同单词的字典作为键，每个键的相应值将是一个整数，显示频率文本文件中的单词。字典应该是什么样子的一个例子：

{'and':2, 'all':1, 'be':1, 'is':3}等

到目前为止，我有这个，

def create_word_frequency_dictionary () :
filename = 'dictionary.txt'
infile = open(filename, 'r') 
line = infile.readline()

my_dictionary = {}
frequency = 0

while line != '' :
    row = line.lower()
    word_list = row.split()
    print(word_list)
    print (word_list[0])
    words = word_list[0]
    my_dictionary[words] = frequency+1
    line = infile.readline()

infile.close()

print (my_dictionary)

create_word_frequency_dictionary()

任何帮助将不胜感激。

Answer 1

文档将collections模块定义为“高性能容器数据类型”。考虑使用collections.Counter而不是重新发明轮子。

from collections import Counter
filename = 'dictionary.txt'
infile = open(filename, 'r') 
text = str(infile.read())
print(Counter(text.split()))

<强>更新好的，我修复了你的代码，现在它可以工作，但Counter仍然是一个更好的选择：

def create_word_frequency_dictionary () :
    filename = 'dictionary.txt'
    infile = open(filename, 'r') 
    lines = infile.readlines()

    my_dictionary = {}

    for line in lines:
        row = str(line.lower())
        for word in row.split():
            if word in my_dictionary:
                 my_dictionary[word] = my_dictionary[word] + 1
            else:
                 my_dictionary[word] = 1

    infile.close()
    print (my_dictionary)

create_word_frequency_dictionary()

Answer 2

如果你没有使用具有Counter的python版本：

>>> import collections
>>> words = ["a", "b", "a", "c"]
>>> word_frequency = collections.defaultdict(int)
>>> for w in words:
...   word_frequency[w] += 1
... 
>>> print word_frequency
defaultdict(<type 'int'>, {'a': 2, 'c': 1, 'b': 1})

Answer 3

只需将my_dictionary[words] = frequency+1替换为my_dictionary[words] = my_dictionary[words]+1。

试图从文本文件创建字典但是

3 个答案: