Question

我有349个文本文件。我使用以下代码来读取和标记所有这些代码。

import glob
path = "C:\\texts\\*.txt"
for file in files:
   with open (file) as in_file, open ("C:\\texts\\file_tokens.txt", 'w') as out_file:
       for line in in_file:
           words = line.split()
           for word in words:
               out_file.write(word)
               out_file.write("\n")

此代码将结果（所有标记）保存在一个文件（file_tokens.txt）中。如何在新的.txt文件中保存每个文件的标记？我的意思是我想要输出349个文件，因为每个文件都包含每个文件的标记。

Answer 1

from os import path
base_path = "C:\\texts\\*.txt"  #RENAMED
for file in files:
    with open (file) as in_file:
        with open(path.join(base_path,"%s_tokenized.txt" % file)) as out_file:  #ATTENTION
            for line in in_file:
                words = line.split()
                for word in words:
                out_file.write(word)
                out_file.write("\n")

您创建一个新文件，其名称特定于您正在处理的当前文件。在这个例子中它是($file_name)_tokenized.txt。

path.join用于将文件输出到正确的目录。即

>>> path.join("~/Documents","out.txt")
'~/Documents/out.txt'

Answer 2

为每个输出文件指定一个不同的名称。

如何在新的单独文本文件中保存每个文件的标记化结果？

2 个答案: