接受文件对象或路径的 Python 函数

Question

我想编写一个函数，它可以接受路径作为字符串或文件对象。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)

do_stuff 接受一个打开的文件对象。

有没有更好的方法来做到这一点？ with f as f:有影响吗？

谢谢！

Answer 1

您的代码的奇怪之处在于，如果传递一个打开的文件，它将关闭它。这不好。无论打开文件的代码应该负责关闭它。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将其抽象出来：

@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，当您声明任何对打开的文件对象执行操作的函数时：

@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：有关装饰器的更多详细信息。