问题描述
我正在尝试编写代码，以访问名为Mini-CORE的语料库中的文件。我打印FileNotFoundError: [Errno 2] No such file or directory: '1+IN+EN+IN-IN-IN-IN+EN-EN-EN-EN+WIKI+9990014.txt'并从中提取体裁代码没有问题。但是，当我尝试访问文件本身以提取文本时，它给了我import os
import re
import spacy
from spacy import displacy
from collections import Counter
nlp = spacy.load('en')
entries = os.listdir('Mini-CORE')
entry_list = []
# this returns the genre codes for each file
def genre_code(filename):
for entry in entries:
regex1 = r'((?<=1\+)\w*)' # This captures the genre code
genre = re.findall(regex1,entry)
entry_list.append(genre)
genre_code(entries)
print(entry_list)
# FileNotFoundError???
# This captures the text after after the <h> or <p> tags
def relevant_text(filename):
for filename in entries:
with open(filename) as current_file:
text = current_file.read()
regex2 = r'((?<=<h>|<p>).*)'
text2 = re.findall(regex2,text)
print(text2)
print(relevant_text(entries))
，这是文件夹中的第一个文件名。所以我很困惑为什么它告诉我文件名是否声称它不存在？我刚刚在某个地方犯了语法错误吗？
{{ (today | date: 'MMM d,y,h:mm') + (today | date: 'a' | lowercase) }}
解决方法
os.listdir返回不带路径的文件名。打开文件时，需要文件的父目录。 pathlib是一个面向对象的路径库，它使传递路径变得更加容易，而不必担心目录和路径名。
使用Path.glob列出目录，返回的路径将同时具有文件名及其路径供您的程序使用。经过一些清理，您的代码可能会被
from pathlib import Path
import re
import spacy
from spacy import displacy
from collections import Counter
nlp = spacy.load('en')
entries = Path('Mini-CORE').glob("*")
# this returns the genre codes for each file
def genre_code(entries):
entry_list = []
for entry in entries:
regex1 = r'((?<=1\+)\w*)' # This captures the genre code
genre = re.findall(regex1,entry.name)
entry_list.append(genre)
return entry_list
entry_list = genre_code(entries)
print(entry_list)
# This captures the text after after the <h> or <p> tags
def relevant_text(entries):
for filename in entries:
with open(filename) as current_file:
text = current_file.read()
regex2 = r'((?<=<h>|<p>).*)'
text2 = re.findall(regex2,text)
print(text2)
print(relevant_text(entries))