代码未输出到正确的文件夹Python

Question

所以我有一些代码打开一个文本文件，其中包含文件路径列表，如下所示：

C：/用户/用户/桌面/ mini_mouse / 1980

C：/用户/用户/桌面/ mini_mouse / 1982

C：/用户/用户/桌面/ mini_mouse / 1984

然后，它逐行单独打开这些文件，并对文件进行一些过滤。然后我希望它将结果输出到一个完全不同的文件夹：

output_location = 'C:/Users/User/Desktop/test2/'

目前，我的代码当前将结果输出到打开原始文件的位置，即如果它打开文件C：/ Users / User / Desktop / mini_mouse / 1980，则输出将位于名称下的同一文件夹中1980_filtered”。但是，我希望输出进入output_location。谁能看到我目前在哪里出错？任何帮助将不胜感激！这是我的代码：

import os

def main():
stop_words_path = 'C:/Users/User/Desktop/NLTK-stop-word-list.txt'
stopwords = get_stop_words_list(stop_words_path)
output_location = 'C:/Users/User/Desktop/test2/'

list_file = 'C:/Users/User/Desktop/list_of_files.txt'

with open(list_file, 'r') as f:
    for file_name in f:
        #print(file_name)
        if file_name.endswith('\n'):
            file_name = file_name[:-1]
        #print(file_name)
        file_path = os.path.join(file_name)  # joins the new path of the file to the current file in order to access the file

        filestring = ''  # file string which will take all the lines in the file and add them to itself
        with open(file_path, 'r') as f2:  # open the file
            print('just opened ' + file_name)
            print('\n')
            for line in f2:  # read file line by line

                x = remove_stop_words(line, stopwords)  # remove stop words from line
                filestring += x  # add newly filtered line to the file string
                filestring += '\n'  # Create new line

        new_file_path = os.path.join(output_location, file_name) + '_filtered'  # creates a new file of the file that is currenlty being filtered of stopwords
        with open(new_file_path, 'a') as output_file:  # opens output file
            output_file.write(filestring)


if __name__ == "__main__":
    main()

Answer 1

假设您正在使用Windows（因为您有一个普通的Windows文件系统），您必须在路径名中使用反斜杠。请注意，这仅适用于Windows。我知道这很烦人，所以我为你换了它（欢迎你:)）。您还必须使用两个反斜杠，因为它会尝试将其用作转义字符。

import os

def main():
stop_words_path = 'C:\\Users\\User\\Desktop\\NLTK-stop-word-list.txt'
stopwords = get_stop_words_list(stop_words_path)
output_location = 'C:\\Users\\User\\Desktop\\test2\\'

list_file = 'C:\\Users\\User\\Desktop\\list_of_files.txt'

with open(list_file, 'r') as f:
    for file_name in f:
        #print(file_name)
        if file_name.endswith('\n'):
            file_name = file_name[:-1]
        #print(file_name)
        file_path = os.path.join(file_name)  # joins the new path of the file to the current file in order to access the file

        filestring = ''  # file string which will take all the lines in the file and add them to itself
        with open(file_path, 'r') as f2:  # open the file
            print('just opened ' + file_name)
            print('\n')
            for line in f2:  # read file line by line

                x = remove_stop_words(line, stopwords)  # remove stop words from line
                filestring += x  # add newly filtered line to the file string
                filestring += '\n'  # Create new line

        new_file_path = os.path.join(output_location, file_name) + '_filtered'  # creates a new file of the file that is currenlty being filtered of stopwords
        with open(new_file_path, 'a') as output_file:  # opens output file
            output_file.write(filestring)


if __name__ == "__main__":
    main()

Answer 2

根据您的代码，它看起来像行中的问题：

new_file_path = os.path.join(output_location, file_name) + '_filtered'

在Python的os.path.join（）中，输入中的任何绝对路径（或Windows中的驱动器号）都会丢弃它之前的所有内容，并从新的绝对路径（或驱动器号）重新启动连接。由于您直接从list_of_files.txt调用file_name并且每个路径都相对于C：驱动器格式化，因此每次调用os.path.join（）都会丢弃output_location并重置为原始文件路径。

有关此行为的更好解释，请参阅Why doesn't os.path.join() work in this case?。

构建输出路径时，您可以从路径“C：/ Users / User / Desktop / mini_mouse / 1980”中删除文件名“1980”，并根据output_location变量和隔离文件名连接。