在python中是否有一种更清晰的方法来链接空列表检查?

try:
  bs = a["key"][0][0]
# Note: the syntax for catching exceptions is different in old versions
# of Python. Use whichever one of these lines is appropriate to your version.
except KeyError, IndexError, TypeError:   # Python 3
except (KeyError, IndexError, TypeError): # Python 2
  bs = []
for b in bs:

如果你不介意长线,你可以把它打包成一个函数:

def maybe_list(f):
  try:
    return f()
  except KeyError, IndexError, TypeError:
    return []

for b in maybe_list(lambda: a["key"][0][0]):

我将编写这样的自定义索引器函数:

def safe_indexer(obj, *indices):
    for idx in indices:
        if not obj: break

        if hasattr(obj, "get"):
            obj = obj.get(idx)
        else:
            obj = obj[idx]

    return obj

用途:

a = {"key": {0: {0: "foo"} } };
print safe_indexer(a, "key", 0, 0)
print safe_indexer(a, "bad", 0, 0)

输出:

foo
None

这个怎么样:

try:
    for b in a['key'][0][0]:
        # do stuff.
except KeyError, TypeError, IndexError:
    # respond appropriately.