接受文件對象或路徑的Python函數

Question

我想編寫一個接受路徑作爲字符串或文件對象的函數。到目前爲止，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) 

其中do_stuff需要一個打開的文件對象。

有沒有更好的方法來做到這一點？ with f as f:有什麼影響？

謝謝！

Answer 1

關於你的代碼奇怪的是，如果它傳遞一個打開的文件，它會關閉它。這不好。無論打開哪個代碼，該文件都應負責關閉它。這使得功能稍微複雜一些，但：

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() 

您可以抽象送人通過編寫自己的上下文管理器：

@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

你可以這樣做：

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 

然後當你宣佈不打開的文件對象上的東西，任何功能：

@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 

編輯2：在裝飾更詳細的信息。