嘗試讀取文件的Pythonic方式以及異常回退到備用文件的情況

Question

嘗試讀取文件的Pythonic方法是什麼，如果讀取此異常讀取引發異常回退以讀取替代文件？

這是我寫的示例代碼，它使用嵌套try - except塊。這是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

對於兩個文件的方法在我看來不夠好。

如果你有多個文件回退到我會用一個循環去：

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

您可以檢查file1是否先存在，然後決定打開哪個文件。它會縮短代碼並避免重複try -- catch子句。我相信這是更pythonic，但請注意，您需要在您的模塊import os這個工作。 它可以是這樣的：

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

雖然我不能完全肯定這是否是Python的或沒有，也許是這樣的：

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