Question

我想比较两个文本文件f1.txt和f2.txt，从f2.txt中删除两个文件中的常用词，然后按频率降序对新的f2.txt进行排序

我的方法：

列出f1.txt和f2.txt的单词。
从文本输入中删除不需要的字符。
比较两个列表，并从f2.txt生成的列表中删除常用词
按频率对f2.txt生成的列表中的单词进行排序

with open(sys.argv[1]) as f1,open(sys.argv[2]) as f2:
    passage = f2.read()
    common = f1.read()
words = re.findall(r'\w+', passage)
common_words = re.findall(r'\w+', common)
passage_text = [words.lower() for words in words]
final = set(passage_text) - set(common_words)
word_count = Counter(final)
for word, count in word_count.items():
    print(word, ":", count)

我希望输出如下：

Foo:          12
Bar:          11
Baz:           3
Longword:      1

但是我得到每个单词的计数频率为1

Answer 1

您的值final仅包含唯一的单词（每个单词一个），这就是Counter仅显示1次出现的原因。您需要使用这组单词过滤passage_text，然后将过滤后的列表传递给Counter：

import re
from collections import Counter

passage = '''
    Foo and Bar and Baz or Longword
    Bar or Baz
    Foo foo foo
'''

common = '''and or'''

words = re.findall(r'\w+', passage)
common_words = re.findall(r'\w+', common)
passage_text = [words.lower() for words in words]
final_set = set(passage_text) - set(common_words)
word_count = Counter([w for w in passage_text if w in final_set])
for word, count in sorted(word_count.items(), key=lambda k: -k[1]): # or word_count.most_common()
    print(word, ":", count)

打印：

foo : 4
bar : 2
baz : 2
longword : 1

Answer 2

有两种方法可以计算文本文件中的单词。

from re import split


def process_line(words, word_dict):
    for word in words:
        if word in word_dict:
            word_dict[word] += 1
        else:
            word_dict[word] = 1


def process_dict(word_dict):
    temp_list = []
    for key, value in word_dict.items():
        temp_list.append((value, key))

    temp_list.sort()
    return temp_list


def format_print(input_list, reverse, word_num):
    if reverse:
        input_list.sort(reverse=True)

    print("\n", ("[Unique Words: " + str(word_num) + "]").center(35, "="))
    print("-"*35 + "\n", "%-16s %s %16s" % ("Word", "|", "Count"), "\n", "-"*35)
    for count, word in input_list:
        print("%-16s %s %16d" % (word, "|", count))


def word_count(_file, max_to_min=False):
    txt = open(_file, "rU")
    word_dict = {}
    for line in txt:
        if line.replace(" ", "") != ("\n" or None):
            process_line(filter(None, split("[^a-zA-Z']+", line.lower())), word_dict)

    txt.close()
    final_list = process_dict(word_dict)
    format_print(final_list, max_to_min, len(word_dict))


word_count("C:\\your_path_here\\Test.txt", True)


#########################################################


from collections import Counter
import re

def openfile(filename):
    fh = open(filename, "r+")
    str = fh.read()
    fh.close()
    return str

def removegarbage(str):
    # Replace one or more non-word (non-alphanumeric) chars with a space
    str = re.sub(r'\W+', ' ', str)
    str = str.lower()
    return str

def getwordbins(words):
    cnt = Counter()
    for word in words:
        cnt[word] += 1
    return cnt

def main(filename, topwords):
    txt = openfile(filename)
    txt = removegarbage(txt)
    words = txt.split(' ')
    bins = getwordbins(words)
    for key, value in bins.most_common(topwords):
        print(key,value)

main('C:\\your_path_here\\Test.txt', 500)

这里是比较两个文本文件并保留共同元素的一种方法。

with open('C:\\your_path_here\\text1.txt', 'r') as file1:
    with open('C:\\your_path_here\\text2.txt', 'r') as file2:
        same = set(file1).intersection(file2)

same.discard('\n')

with open('C:\\your_path_here\\some_output_file.txt', 'w') as file_out:
    for line in same:
        file_out.write(line)

# For differences, use the code below:
with open('C:\\your_path_here\\text1.txt', 'r') as file1:
    with open('C:\\your_path_here\\text2.txt', 'r') as file2:
        same = set(file1).symmetric_difference(file2)

same.discard('\n')

with open('C:\\your_path_here\\some_output_file.txt', 'w') as file_out:
    for line in same:
        file_out.write(line)

比较，删除和计算文本文件中的单词

2 个答案: