Question

我有一个简单的记事本，我用来节省昂贵的网络电话。粗略地说，我的代码看起来像这样：

# mem.py
import functools
import time


def memoize(fn):
    """
    Decorate a function so that it results are cached in memory.

    >>> import random
    >>> random.seed(0)
    >>> f = lambda x: random.randint(0, 10)
    >>> [f(1) for _ in range(10)]
    [9, 8, 4, 2, 5, 4, 8, 3, 5, 6]
    >>> [f(2) for _ in range(10)]
    [9, 5, 3, 8, 6, 2, 10, 10, 8, 9]
    >>> g = memoize(f)
    >>> [g(1) for _ in range(10)]
    [3, 3, 3, 3, 3, 3, 3, 3, 3, 3]
    >>> [g(2) for _ in range(10)]
    [8, 8, 8, 8, 8, 8, 8, 8, 8, 8]
    """
    cache = {}

    @functools.wraps(fn)
    def wrapped(*args, **kwargs):
        key = args, tuple(sorted(kwargs))
        try:
            return cache[key]
        except KeyError:
            cache[key] = fn(*args, **kwargs)
            return cache[key]
    return wrapped


def network_call(user_id):
    time.sleep(1)
    return 1


@memoize
def search(user_id):
    response = network_call(user_id)
    # do stuff to response
    return response

我对此代码进行了测试，我在其中模拟了network_call()的不同返回值，以确保我在search()中执行的某些修改按预期工作。

import mock

import mem


@mock.patch('mem.network_call')
def test_search(mock_network_call):
    mock_network_call.return_value = 2
    assert mem.search(1) == 2


@mock.patch('mem.network_call')
def test_search_2(mock_network_call):
    mock_network_call.return_value = 3
    assert mem.search(1) == 3

但是，当我运行这些测试时，我会失败，因为search()会返回一个缓存的结果。

CAESAR-BAUTISTA:~ caesarbautista$ py.test test_mem.py
============================= test session starts ==============================
platform darwin -- Python 2.7.8 -- py-1.4.26 -- pytest-2.6.4
collected 2 items

test_mem.py .F

=================================== FAILURES ===================================
________________________________ test_search_2 _________________________________

args = (<MagicMock name='network_call' id='4438999312'>,), keywargs = {}
extra_args = [<MagicMock name='network_call' id='4438999312'>]
entered_patchers = [<mock._patch object at 0x108913dd0>]
exc_info = (<class '_pytest.assertion.reinterpret.AssertionError'>, AssertionError(u'assert 2 == 3\n +  where 2 = <function search at 0x10893f848>(1)\n +    where <function search at 0x10893f848> = mem.search',), <traceback object at 0x1089502d8>)
patching = <mock._patch object at 0x108913dd0>
arg = <MagicMock name='network_call' id='4438999312'>

    @wraps(func)
    def patched(*args, **keywargs):
        # don't use a with here (backwards compatability with Python 2.4)
        extra_args = []
        entered_patchers = []

        # can't use try...except...finally because of Python 2.4
        # compatibility
        exc_info = tuple()
        try:
            try:
                for patching in patched.patchings:
                    arg = patching.__enter__()
                    entered_patchers.append(patching)
                    if patching.attribute_name is not None:
                        keywargs.update(arg)
                    elif patching.new is DEFAULT:
                        extra_args.append(arg)

                args += tuple(extra_args)
>               return func(*args, **keywargs)

/opt/boxen/homebrew/lib/python2.7/site-packages/mock.py:1201:
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

mock_network_call = <MagicMock name='network_call' id='4438999312'>

    @mock.patch('mem.network_call')
    def test_search_2(mock_network_call):
        mock_network_call.return_value = 3
>       assert mem.search(1) == 3
E       assert 2 == 3
E        +  where 2 = <function search at 0x10893f848>(1)
E        +    where <function search at 0x10893f848> = mem.search

test_mem.py:15: AssertionError
====================== 1 failed, 1 passed in 0.03 seconds ======================

有没有办法测试记忆功能？我考虑了一些替代方案，但它们都有缺点。

一种解决方案是模拟memoize()。我不愿意这样做，因为它泄漏了测试的实现细节。从理论上讲，我应该能够在没有系统其他部分的情况下记忆和取消默认功能，包括测试，从功能角度注意。

另一个解决方案是重写代码以公开修饰函数。也就是说，我可以这样做：

def _search(user_id):
    return network_call(user_id)
search = memoize(_search)

然而，这遇到了与上述相同的问题，尽管它可能更糟，因为它不适用于递归函数。

如何测试记忆功能？

0 个答案: