Improving performance of very large dictionary in Python
Question:
I find that if I initialize an empty dictionary at the beginning, and then adding elements to the dictionary in a for loop (about 110,000 keys, the value for each key is a list, also increasing in the loop), the speed goes down as for loop goes.
I suspect that the problem is, the dictionary does not know the number of keys at init time and it is not doing something very smart, so perhaps the storage collision becomes quite often and it slows down.
If I know the number of keys and exactly what are those keys, is there any way in python to make a dict (or a hashtable) work more efficiently? I vaguely remember that if you know the keys, you can design the hash function smartly (perfect hash?) and allocate the space beforehand.
Answers:
If I know the number of keys and exactly what are those keys, is there
any way in python to make a dict (or a hashtable) work more
efficiently? I vaguely remember that if you know the keys, you can
design the hash function smartly (perfect hash?) and allocate the
space beforehand.
Python doesn’t expose a pre-sizing option to speed-up the “growth phase” of a dictionary, nor does it provide any direct controls over “placement” in the dictionary.
That said, if the keys are always known in advance, you can store them in a set and build your dictionaries from the set using dict.fromkeys(). That classmethod is optimized to pre-size the dictionary based on the set size and it can populate the dictionary without any new calls to __hash__():
>>> keys = {'red', 'green', 'blue', 'yellow', 'orange', 'pink', 'black'}
>>> d = dict.fromkeys(keys) # dict is pre-sized to 32 empty slots
If reducing collisions is your goal, you can run experiments on the insertion order in the dictionary to minimize pile-ups. (Take a look at Brent’s variation on Algorithm D in Knuth’s TAOCP to get an idea of how this is done).
By instrumenting a pure Python model for dictionaries (such as this one), it is possible to count the weighted-average number of probes for an alternative insertion order. For example, inserting dict.fromkeys([11100, 22200, 44400, 33300]) averages 1.75 probes per lookup. That beats the 2.25 average probes per lookup for dict.fromkeys([33300, 22200, 11100, 44400]).
Another “trick” is to increase spareness in a fully populated dictionary by fooling it into increasing its size without adding new keys:
d = dict.fromkeys(['red', 'green', 'blue', 'yellow', 'orange'])
d.update(dict(d)) # This makes room for additional keys
# and makes the set collision-free.
Lastly, you can introduce your own custom __hash__() for your keys with the goal of eliminating all collisions (perhaps using a perfect hash generator such as gperf).
I find that if I initialize an empty dictionary at the beginning, and then adding elements to the dictionary in a for loop (about 110,000 keys, the value for each key is a list, also increasing in the loop), the speed goes down as for loop goes.
I suspect that the problem is, the dictionary does not know the number of keys at init time and it is not doing something very smart, so perhaps the storage collision becomes quite often and it slows down.
If I know the number of keys and exactly what are those keys, is there any way in python to make a dict (or a hashtable) work more efficiently? I vaguely remember that if you know the keys, you can design the hash function smartly (perfect hash?) and allocate the space beforehand.
If I know the number of keys and exactly what are those keys, is there
any way in python to make a dict (or a hashtable) work more
efficiently? I vaguely remember that if you know the keys, you can
design the hash function smartly (perfect hash?) and allocate the
space beforehand.
Python doesn’t expose a pre-sizing option to speed-up the “growth phase” of a dictionary, nor does it provide any direct controls over “placement” in the dictionary.
That said, if the keys are always known in advance, you can store them in a set and build your dictionaries from the set using dict.fromkeys(). That classmethod is optimized to pre-size the dictionary based on the set size and it can populate the dictionary without any new calls to __hash__():
>>> keys = {'red', 'green', 'blue', 'yellow', 'orange', 'pink', 'black'}
>>> d = dict.fromkeys(keys) # dict is pre-sized to 32 empty slots
If reducing collisions is your goal, you can run experiments on the insertion order in the dictionary to minimize pile-ups. (Take a look at Brent’s variation on Algorithm D in Knuth’s TAOCP to get an idea of how this is done).
By instrumenting a pure Python model for dictionaries (such as this one), it is possible to count the weighted-average number of probes for an alternative insertion order. For example, inserting dict.fromkeys([11100, 22200, 44400, 33300]) averages 1.75 probes per lookup. That beats the 2.25 average probes per lookup for dict.fromkeys([33300, 22200, 11100, 44400]).
Another “trick” is to increase spareness in a fully populated dictionary by fooling it into increasing its size without adding new keys:
d = dict.fromkeys(['red', 'green', 'blue', 'yellow', 'orange'])
d.update(dict(d)) # This makes room for additional keys
# and makes the set collision-free.
Lastly, you can introduce your own custom __hash__() for your keys with the goal of eliminating all collisions (perhaps using a perfect hash generator such as gperf).