接受文件对象或路径的Python函数

Question

我想编写一个接受路径作为字符串或文件对象的函数.到目前为止,我有:

def awesome_parse(path_or_file):
    if isinstance(path_or_file, basestring):
        f = open(path_or_file, 'rb')
    else:
        f = path_or_file
    with f as f:
        return do_stuff(f)

这里do_stuff需要一个打开的文件对象.

有一个更好的方法吗？请问with f as f:有什么反响？

谢谢!

Answer 1

你的代码奇怪的是,如果它传递一个打开的文件,它将关闭它.这不好.无论打开什么代码,文件都应该负责关闭它.这使得函数有点复杂:

def awesome_parse(path_or_file):
    if isinstance(path_or_file, basestring):
        f = file_to_close = open(path_or_file, 'rb')
    else:
        f = path_or_file
        file_to_close = None
    try:
        return do_stuff(f)
    finally:
        if file_to_close:
            file_to_close.close()

您可以通过编写自己的上下文管理器来抽象它:

@contextlib.contextmanager
def awesome_open(path_or_file):
    if isinstance(path_or_file, basestring):
        f = file_to_close = open(path_or_file, 'rb')
    else:
        f = path_or_file
        file_to_close = None

    try:
        yield f
    finally:
        if file_to_close:
            file_to_close.close()

def awesome_parse(path_or_file):
    with awesome_open(path_or_file) as f:
        return do_stuff(f)

Answer 2

你可以这样做：

def awesome_parse(do_stuff):
    """Decorator to open a filename passed to a function
       that requires an open file object"""
    def parse_path_or_file(path_or_file):
        """Use a ternary expression to either open the file from the filename
           or just pass the extant file object on through"""
        with (open(path_or_file, 'rb') 
               if isinstance(path_or_file, basestring) 
                else path_or_file) as f:
            return do_stuff(f)
    return parse_path_or_file

然后当你声明任何在打开的文件对象上执行操作的函数时：

@awesome_parse
def do_things(open_file_object):
    """This will always get an open file object even if passed a string"""
    pass

@awesome_parse
def do_stuff(open_file_object):
    """So will this"""
    pass

编辑 2：关于装饰器的更多详细信息。