python - Python 的 shelve.open 可以嵌套调用吗？

Question

我正在尝试编写一个使用 shelve 的内存库持久存储返回值。如果我有内存函数调用其他内存函数，我想知道如何正确打开 shelf 文件。

import shelve
import functools


def cache(filename):
    def decorating_function(user_function):
        def wrapper(*args, **kwds):
            key = str(hash(functools._make_key(args, kwds, typed=False)))
            with shelve.open(filename, writeback=True) as cache:
                if key in cache:
                    return cache[key]
                else:
                    result = user_function(*args, **kwds)
                    cache[key] = result
                    return result

        return functools.update_wrapper(wrapper, user_function)

    return decorating_function


@cache(filename='cache')
def expensive_calculation():
    print('inside function')
    return


@cache(filename='cache')
def other_expensive_calculation():
    print('outside function')
    return expensive_calculation()

other_expensive_calculation()

除了这行不通

$ python3 shelve_test.py
outside function
Traceback (most recent call last):
  File "shelve_test.py", line 33, in <module>
    other_expensive_calculation()
  File "shelve_test.py", line 13, in wrapper
    result = user_function(*args, **kwds)
  File "shelve_test.py", line 31, in other_expensive_calculation
    return expensive_calculation()
  File "shelve_test.py", line 9, in wrapper
    with shelve.open(filename, writeback=True) as cache:
  File "/usr/local/Cellar/python3/3.4.1/Frameworks/Python.framework/Versions/3.4/lib/python3.4/shelve.py", line 239, in open
    return DbfilenameShelf(filename, flag, protocol, writeback)
  File "/usr/local/Cellar/python3/3.4.1/Frameworks/Python.framework/Versions/3.4/lib/python3.4/shelve.py", line 223, in __init__
    Shelf.__init__(self, dbm.open(filename, flag), protocol, writeback)
  File "/usr/local/Cellar/python3/3.4.1/Frameworks/Python.framework/Versions/3.4/lib/python3.4/dbm/__init__.py", line 94, in open
    return mod.open(file, flag, mode)
_gdbm.error: [Errno 35] Resource temporarily unavailable

您对解决这类问题的建议是什么。

Answer 1

不，你可能没有嵌套 shelve具有相同文件名的实例。

The shelve module does not support concurrent read/write access to shelved objects. (Multiple simultaneous read accesses are safe.) When a program has a shelf open for writing, no other program should have it open for reading or writing. Unix file locking can be used to solve this, but this differs across Unix versions and requires knowledge about the database implementation used.

https://docs.python.org/3/library/shelve.html#restrictions