python - 如何计算Python中的单词数达到限制？

Question

我正在编写一个代码，计算包含大约 20,000 个文件的文档中单词出现的频率，我能够获取文档中单词的总体频率，并且 到目前为止我的代码是:

import os
import re
import sys
sys.stdout=open('f2.txt','w')
from collections import Counter
from glob import iglob

def removegarbage(text):
    text=re.sub(r'\W+',' ',text)
    text=text.lower()
    return text

folderpath='d:/articles-words'
counter=Counter()
d=0

for filepath in iglob(os.path.join(folderpath,'*.txt')):
    with open(filepath,'r') as filehandle:
        d+=1

r=round(d*0.1)
for filepath in iglob(os.path.join(folderpath,'*.txt')):
    with open(filepath,'r') as filehandle:
        words=set(removegarbage(filehandle.read()).split())
        if r > counter:
             counter.update()

for word,count in counter.most_common():
    print('{}  {}'.format(word,count))

但是，我想修改我的计数器，并且仅当计数大于 r=0.1*(文件数)时才更新它 简而言之，我想读取在整个文档中出现频率大于文档数量10%的单词。 错误是:

Traceback (most recent call last):
  File "<stdin>", line 4, in <module>
TypeError: unorderable types: int() > Counter()

如何修改？

Answer 1

像这样的东西怎么样？

from glob import glob # (instead of iglob)

...

filepaths = glob(os.path.join(folderpath,'*.txt'))

num_files = len(filepaths)

# Add all words to counter
for filepath in filepaths):
    with open(filepath,'r') as filehandle:
        lines = filehandle.read()
        words = removegarbage(lines).split()
        counter.update(words)

# Display most common
for word, count in counter.most_common():

    # Break out if the frequency is less than 0.1 * the number of files
    if count < 0.1*num_files:
        break

    print('{}  {}'.format(word,count))

---

使用Counter.iteritems():

>>> from collections import Counter
>>> c = Counter()
>>> c.update(['test', 'test', 'test2'])
>>> c.iteritems()
<dictionary-itemiterator object at 0x012F4750>
>>> for word, count in c.iteritems():
...     print word, count
...     
test 2
test2 1