在所有特定名称子目录中运行Python脚本

Question

我有一个Python脚本，它需要一些参数，我想在这个脚本所在的文件夹及其所有子文件夹中运行这个脚本。

我的想法是，我希望脚本的输出结果能保存到一个文件里。

这是我目前做的：

for x in `find . -type d | grep data`;
do
python /full/path/to/file/script.py -f "%a
%t" $x/*.txt -o $x/res.txt
done

但是这个方法没有成功，我不知道为什么。在循环中的grep是为了只获取那些包含.txt文件的文件夹，并对它们运行脚本。

在%a和%t之间加了一行空行，是因为我想自定义Python脚本的输出，让每两个变量之间有一个空行。

我哪里做错了呢？

Answer 1

如果你想要从当前脚本所在的文件夹开始，运行这个脚本到所有的子文件夹，那么可以试试这样做：

import os

for path, directories, files in os.walk(os.path.dirname(os.path.realpath(__file__))):
    print path, directories, files
    txt_files = [arbitrary_file for arbitrary_file in files if arbitrary_file[-4:].lower() == ".txt"]

    #run your python here
    txt_files = [txt_file for arbitrary_file in files if arbitrary_file[]

如果你原来的代码是这样的：

import sys

text_files_to_process = #Do Something with sys.argv - or whatever you're using to parse your arguments.

with open("res.txt", "w") as f:
    #do something with all the text files, and write the output to res.txt.
    for text_file in text_files_to_process:
        with open(text_file) as tf:
            for line in tf:
                #whatever your text processing is
            tf.write("something")

那么你只需要把它改成这样：

import os

for path, directories, files in os.walk(os.path.dirname(os.path.realpath(__file__))):
    print path, directories, files
    txt_files = [arbitrary_file for arbitrary_file in files if arbitrary_file[-4:].lower() == ".txt"]

    txt_files = [txt_file for arbitrary_file in files if arbitrary_file[]

    with open("res.txt", "w") as f:
        #do something with all the text files, and write the output to res.txt.
        for text_file in txt_files:
            with open(text_file) as tf:
                for line in tf:
                    #whatever your text processing is
                tf.write("something")