Thanks Ross.
The thing is, the 'kwargs' dict that callers pass in is not the same
dict as the kwargs that my wrapper function receives. My 'wrapper'
function always receives a brand-new dictionary, created from scratch as
the function is invoked. Hence, the kwargs never has any 'history' -
they are always newly created, from the keyword args passed in to the
function. This, I suspect, is why their 'items()' always appear in the
same order.
I tried calling 'wrapper' with keyword args in a different order:
wrapper(x=1, y=2, z=3)
wrapper(y=2, x=1, z=3)
...etc, for every permutation
and by passing in **kwargs dictionaries with different histories
(detailed in my original mail):
wrapper(**kwargs)
but in every case, inside wrapper...
def wrapper(*args, **kwargs):
print kwargs.items()
...the items call always returns the dictionary's content in the same
order. So I *suspect*, but cannot yet prove, that the '**' mechanism is
already sorting the keyword args used to construct the kwargs dictionary
(or is doing some equivalent, such that the dicts always end up
identical-including-order-of-items())
I agree that 'to be sure' is enough of a justification for including the
'sorted' call anyway. But it irks me that I can't write a test to show it.
Jonathan
On 11/11/2011 08:50, Ross Lawley wrote:
Hi,
From the docs: http://docs.python.org/library/stdtypes.html#dict.items
"CPython implementation detail: Keys and values are listed in an
arbitrary order which is non-random, varies across Python
implementations, and depends on the dictionary's history of insertions
and deletions."
Given that I would use sorted to guarantee order. Then no matter what
anyone has done to the kwargs passed in the memoizing will not have a bug.
Ross
On Fri, Nov 11, 2011 at 8:42 AM, Jonathan <tart...@tartley.com
<mailto:tart...@tartley.com>> wrote:
Hey,
I've been writing my own 'memoize' function a lot recently - I'm
using it as an interview question.
Here's what I've got (with a corresponding series of tests):
def memoize(wrapped):
cache = {}
@wraps(wrapped)
def wrapper(*args, **kwargs):
key = (args, tuple(kwargs.items()))
if key not in cache:
cache[key] = wrapped(*args, **kwargs)
return cache[key]
return wrapper
Yesterday a candidate pointed out that this is wrong because
.items() for two equal dictionaries might return the (key,value)
pairs in a different order, presumably dependent on the respective
dictionary's history. This would produce a different 'key' and
hence an erroneous cache miss.
This implies that the second line of 'wrapper' should become:
key = (args, tuple(sorted(kwargs.items())))
(I've added 'sorted')
Looking at lru_cache, I see that is exactly how it is implemented
there.
http://hg.python.org/cpython/file/default/Lib/functools.py
However, I'm unable to come up with a test that proves this is
necessary. I'm can create two equal dictionaries which return
their .items() in a different order:
# The intent is that 'small.items()' comes out in a
different order
# than 'large.items()'
small = {'x':1, 'y':5}
large = {hex(i): i for i in range(257)}
large.update(small)
for i in range(257):
del large[hex(i)]
>>> print small.items()
[('y', 5), ('x', 1)]
>>> print large.items()
[('x', 1), ('y', 5)]
If I could magically transfer these dictionaries directly into the
'kwargs' of wrapper, then I think I'd be done. However, I have to
pass these dictionaries in to wrapper via the '**' mechanism.
Printing 'kwargs.items()' within 'wrapper' shows that equal
dictionaries always return their items in the same order, so the
'sorted' call is apparently not necessary.
Is 'sorted' really required? If so, how can I write a test to
demonstrate it?
Best regards,
Jonathan
--
Jonathan Hartley tart...@tartley.com <mailto:tart...@tartley.com>
http://tartley.com
Made of meat. +44 7737 062 225 <tel:%2B44%207737%20062%20225>
twitter/skype: tartley
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--
Jonathan Hartley tart...@tartley.com http://tartley.com
Made of meat. +44 7737 062 225 twitter/skype: tartley
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