尝试读取文件的Pythonic方式以及异常回退到备用文件的情况

Question

尝试读取文件的Pythonic方法是什么，如果读取此异常读取引发异常回退以读取替代文件？

这是我写的示例代码，它使用嵌套try - except块。这是pythonic：

try: 
    with open(file1, "r") as f: 
     params = json.load(f) 
except IOError: 
    try: 
     with open(file2, "r") as f: 
      params = json.load(f) 
    except Exception as exc: 
     print("Error reading config file {}: {}".format(file2, str(exc))) 
     params = {} 
except Exception as exc: 
    print("Error reading config file {}: {}".format(file1, str(exc))) 
    params = {} 

Answer 1

对于两个文件的方法在我看来不够好。

如果你有多个文件回退到我会用一个循环去：

for filename in (file1, file2): 
    try: 
     with open(filename, "r") as fin: 
      params = json.load(f) 
     break 
    except IOError: 
     pass 
    except Exception as exc: 
     print("Error reading config file {}: {}".format(filename, str(exc))) 
     params = {} 
     break 
else: # else is executed if the loop wasn't terminated by break 
    print("Couldn't open any file") 
    params = {} 

Answer 2

您可以检查file1是否先存在，然后决定打开哪个文件。它会缩短代码并避免重复try -- catch子句。我相信这是更pythonic，但请注意，您需要在您的模块import os这个工作。 它可以是这样的：

fp = file1 if os.path.isfile(file1) else file2 
if os.path.isfile(fp): 
    try: 
     with open(fp, "r") as f: 
      params = json.load(f) 
    except Exception as exc: 
     print("Error reading config file {}: {}".format(fp, str(exc))) 
      params = {} 
else: 
    print 'no config file' 

Answer 3

虽然我不能完全肯定这是否是Python的或没有，也许是这样的：

file_to_open = file1 if os.path.isfile(file1) else file2