我在一个目录中有500个文本文件。我必须在所有文本文件中找到500个最常用的单词。我怎样才能实现呢?
PS:我已经搜索了很多但找不到解决方案。答案 0 :(得分:4)
import os
from collections import Counter
c, directory = Counter(), 'path_to_your_directory'
for x in os.listdir(directory):
fname = os.path.join(directory, x)
if os.path.isfile(fname):
with open(fname) as f:
c += Counter(f.read().split())
for word, _ in c.most_common(500):
print(word)
当然,它会读取该目录中的每个文件。如果这不是预期的行为,请使用glob.glob
or glob.iglob
所需的模式而不是os.listdir
(请参阅Reut对我的回答的评论)。
答案 1 :(得分:1)
这是我可以考虑使用字典进行计数的最简单的方法,使用键作为单词ad计算值:
import os
# word counts are stored in a dictionary
# for fast access and duplication prevention
count = {}
# your text files should be in this folder
DIR = "files"
# iterate over all files in the folder
for filename in os.listdir(DIR):
with open(os.path.sep.join([DIR, filename]), 'r') as f:
for line in f.readlines():
# strip line separators from end of line
line = line.strip()
# once we have a line from the file, split it to words, and
# add word to the scores (if it's new), or increase it's count
for word in line.split():
if word in count:
# existing
count[word] = count[word] + 1
else:
# new
count[word] = 1
print sorted(count.items(), key=lambda x: x[1], reverse=True)[:500]
使用collections.Counter(如Padraic建议的那样):
import os
from collections import Counter
count = Counter()
DIR = "files"
for filename in os.listdir(DIR):
with open(os.path.sep.join([DIR, filename]), 'r') as f:
for line in f.readlines():
line = line.strip()
# count all words in line
count.update(line.split())
print count.most_common(500)
答案 2 :(得分:0)
您可以为每个新单词和一个单词数组创建一个计数器。将每个新单词添加到数组中。使用"索引"将文本文件中的每个单词与数组中的单词进行比较,递增单词的计数器。或者你可以创建一个数组,填充每个NEW 从文本文件中的单词,使用数组的第二个元素作为计数器。
答案 3 :(得分:0)
我们可以使用集合模块中的Counter方法。
glob
for loop
迭代步骤1中的所有文件。with
语句和read()
文件对象方法以读取模式打开文件。split()
字符串方法拆分文件内容,并使用Counter
创建可数字典。添加两个计数器。 https://docs.python.org/2/library/collections.html most_common(3)
方法从计数器中获取最常用的字词。代码:
from glob import glob
from collections import Counter
p = "/home/vivek/Desktop/test/*.txt"
main_counter = Counter()
for i in glob(p):
with open(i, "rb") as fp:
main_counter += Counter(fp.read().split())
print "main_counter:-", main_counter
print "most common 3:-", main_counter.most_common(3)
输出: -
vivek@vivek:~/Desktop$ python 4.py
main_counter:- Counter({'This': 3, 'try': 2, 'again.': 2, 'is': 2, 'can': 2, 'file': 2, 'you': 2, 'my': 2, '1': 1, 'this': 1, '2': 1})
most common 3:- [('This', 3), ('try', 2), ('again.', 2)]