Why does functools.lru_cache not cache __call__ while working on normal methods

Why does functools.lru_cache not cache call while working on normal methods

I have been trying to make functools.lru_cache instance specific as described in this answer, but their solution fails when used on the __call__ method.

class test:

    def __init__(self):

        self.method = lru_cache()(self.method)

        self.__call__ = lru_cache()(self.__call__)



    def method(self, x):

        print('method', end=' ')

        return x



    def __call__(self, x):

        print('__call__', end=' ')

        return x



b = test()

# b.method is cached as expected

print(b.method(1)) # method 1

print(b.method(1)) # 1



# __call__ is executed every time

print(b(1)) # __call__ 1

print(b(1)) # __call__ 1

So the results of __call__ are not getting cached when wrapped using this method. The cache on __call__ does not even register the function having been called, and unhashable values do not throw errors.

print(b.method.cache_info())

# CacheInfo(hits=1, misses=1, maxsize=128, currsize=1)

print(b.__call__.cache_info())

# CacheInfo(hits=0, misses=0, maxsize=128, currsize=0)



print(b.call({})) # __call__ {}

print(b.method({})) # ... TypeError: unhashable type: 'dict'

edited Nov 22 '18 at 18:23

asked Nov 22 '18 at 2:38

Will

149411

1

i have tried you code, and if applying @lru_cache decorator to __call__ , it works fine, but for you case, you are assigning __call__ to self when initialize, so the lru version of __call__ is added in self.__dict__, without actually changing __call__ method.

– Enix
Nov 22 '18 at 4:37

@Enix If you mean decorating __call__ normally then yes that works fine for one instance, but as soon as you have two instances then the cache is shared, and resetting it on one resets it on the other instance, so using the decorator normally does not work. (I have tried not to make this question about how to have instance specific caches, since the linked question covers that)

– Will
Nov 22 '18 at 18:22

add a comment |

I have been trying to make functools.lru_cache instance specific as described in this answer, but their solution fails when used on the __call__ method.

class test:

    def __init__(self):

        self.method = lru_cache()(self.method)

        self.__call__ = lru_cache()(self.__call__)



    def method(self, x):

        print('method', end=' ')

        return x



    def __call__(self, x):

        print('__call__', end=' ')

        return x



b = test()

# b.method is cached as expected

print(b.method(1)) # method 1

print(b.method(1)) # 1



# __call__ is executed every time

print(b(1)) # __call__ 1

print(b(1)) # __call__ 1

print(b.method.cache_info())

# CacheInfo(hits=1, misses=1, maxsize=128, currsize=1)

print(b.__call__.cache_info())

# CacheInfo(hits=0, misses=0, maxsize=128, currsize=0)



print(b.call({})) # __call__ {}

print(b.method({})) # ... TypeError: unhashable type: 'dict'

edited Nov 22 '18 at 18:23

asked Nov 22 '18 at 2:38

Will

149411

1

i have tried you code, and if applying @lru_cache decorator to __call__ , it works fine, but for you case, you are assigning __call__ to self when initialize, so the lru version of __call__ is added in self.__dict__, without actually changing __call__ method.

– Enix
Nov 22 '18 at 4:37

@Enix If you mean decorating __call__ normally then yes that works fine for one instance, but as soon as you have two instances then the cache is shared, and resetting it on one resets it on the other instance, so using the decorator normally does not work. (I have tried not to make this question about how to have instance specific caches, since the linked question covers that)

– Will
Nov 22 '18 at 18:22

add a comment |

I have been trying to make functools.lru_cache instance specific as described in this answer, but their solution fails when used on the __call__ method.

class test:

    def __init__(self):

        self.method = lru_cache()(self.method)

        self.__call__ = lru_cache()(self.__call__)



    def method(self, x):

        print('method', end=' ')

        return x



    def __call__(self, x):

        print('__call__', end=' ')

        return x



b = test()

# b.method is cached as expected

print(b.method(1)) # method 1

print(b.method(1)) # 1



# __call__ is executed every time

print(b(1)) # __call__ 1

print(b(1)) # __call__ 1

print(b.method.cache_info())

# CacheInfo(hits=1, misses=1, maxsize=128, currsize=1)

print(b.__call__.cache_info())

# CacheInfo(hits=0, misses=0, maxsize=128, currsize=0)



print(b.call({})) # __call__ {}

print(b.method({})) # ... TypeError: unhashable type: 'dict'

edited Nov 22 '18 at 18:23

asked Nov 22 '18 at 2:38

Will

149411

I have been trying to make functools.lru_cache instance specific as described in this answer, but their solution fails when used on the __call__ method.

class test:

    def __init__(self):

        self.method = lru_cache()(self.method)

        self.__call__ = lru_cache()(self.__call__)



    def method(self, x):

        print('method', end=' ')

        return x



    def __call__(self, x):

        print('__call__', end=' ')

        return x



b = test()

# b.method is cached as expected

print(b.method(1)) # method 1

print(b.method(1)) # 1



# __call__ is executed every time

print(b(1)) # __call__ 1

print(b(1)) # __call__ 1

print(b.method.cache_info())

# CacheInfo(hits=1, misses=1, maxsize=128, currsize=1)

print(b.__call__.cache_info())

# CacheInfo(hits=0, misses=0, maxsize=128, currsize=0)



print(b.call({})) # __call__ {}

print(b.method({})) # ... TypeError: unhashable type: 'dict'

python caching magic-methods

edited Nov 22 '18 at 18:23

asked Nov 22 '18 at 2:38

Will

149411

edited Nov 22 '18 at 18:23

asked Nov 22 '18 at 2:38

Will

149411

edited Nov 22 '18 at 18:23

asked Nov 22 '18 at 2:38

Will

149411

asked Nov 22 '18 at 2:38

Will

149411

asked Nov 22 '18 at 2:38

Will

149411

1

i have tried you code, and if applying @lru_cache decorator to __call__ , it works fine, but for you case, you are assigning __call__ to self when initialize, so the lru version of __call__ is added in self.__dict__, without actually changing __call__ method.

– Enix
Nov 22 '18 at 4:37

@Enix If you mean decorating __call__ normally then yes that works fine for one instance, but as soon as you have two instances then the cache is shared, and resetting it on one resets it on the other instance, so using the decorator normally does not work. (I have tried not to make this question about how to have instance specific caches, since the linked question covers that)

– Will
Nov 22 '18 at 18:22

add a comment |

1

i have tried you code, and if applying @lru_cache decorator to __call__ , it works fine, but for you case, you are assigning __call__ to self when initialize, so the lru version of __call__ is added in self.__dict__, without actually changing __call__ method.

– Enix
Nov 22 '18 at 4:37

@Enix If you mean decorating __call__ normally then yes that works fine for one instance, but as soon as you have two instances then the cache is shared, and resetting it on one resets it on the other instance, so using the decorator normally does not work. (I have tried not to make this question about how to have instance specific caches, since the linked question covers that)

– Will
Nov 22 '18 at 18:22

i have tried you code, and if applying @lru_cache decorator to __call__ , it works fine, but for you case, you are assigning __call__ to self when initialize, so the lru version of __call__ is added in self.__dict__, without actually changing __call__ method.

– Enix
Nov 22 '18 at 4:37

@Enix If you mean decorating __call__ normally then yes that works fine for one instance, but as soon as you have two instances then the cache is shared, and resetting it on one resets it on the other instance, so using the decorator normally does not work. (I have tried not to make this question about how to have instance specific caches, since the linked question covers that)

– Will
Nov 22 '18 at 18:22

add a comment |

1 Answer
1

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This is due to the difference between class attributes and instance attributes. When accessing an attribute (such as method) python first checks for an instance attribute. If you have not assigned to self.method it will not find one. Then class attributes are checked, which is equivalent to self.__class__.method. The value of this function is not changed by assigning to self.method, that only updates the instance attribute.

However, b(1) becomes b.__class__.__call__(b, 1) which uses the original class definition of __call__ and b.__call__(1) will be cached the same way as method since it uses the instance definition.

edited Nov 22 '18 at 19:29

answered Nov 22 '18 at 18:56

Will

149411

add a comment |

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edited Nov 22 '18 at 19:29

answered Nov 22 '18 at 18:56

Will

149411

add a comment |

edited Nov 22 '18 at 19:29

answered Nov 22 '18 at 18:56

Will

149411

add a comment |

edited Nov 22 '18 at 19:29

answered Nov 22 '18 at 18:56

Will

149411

edited Nov 22 '18 at 19:29

answered Nov 22 '18 at 18:56

Will

149411

edited Nov 22 '18 at 19:29

answered Nov 22 '18 at 18:56

Will

149411

answered Nov 22 '18 at 18:56

Will

149411

answered Nov 22 '18 at 18:56

Will

149411

add a comment |

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