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out-of-core/external-memory combinatorics in python

I am iterating the search space of valid Python3 ASTs. With max recursion depth = 3, my laptop runs out of memory. My implementation makes heavy use of generators, specifically 'yield' and itertools.product().
Ideally, I'd replace product() and the max recursion depth with some sort of iterative deepening, but first things first:
Are there any libraries or useful SO posts for out-of-core/external-memory combinatorics?
If not... I am considering the feasibility of using either dask or joblib's delayed()... or perhaps wendelin-core's ZBigArray, though I don't like the looks of its interface:
root = dbopen('test.fs')
root['A'] = A = ZBigArray((10,), np.int)
transaction.commit()
Based on this example, I think that my solution would involve an annotation/wrapper function that eagerly converts the generators to ZBigArrays, replacing root['A'] with something like root[get_key(function_name, *function_args)] It's not pretty, since my generators are not entirely pure--the output is shuffled. In my current version, this shouldn't be a big deal, but the previous and next versions involve using various NNs and RL rather mere shuffling.

First things first- the reason you're getting the out of memory error is because itertools.product() caches intermediate values. It has no idea if the function that gave you your generator is idempotent, and even if it did, it couldn't be able to infer how to call it again given just the generator. This means itertools.product must cache values of each iterable its passed.
The solution here is to bite the small performance bullet and either write explicit for loops, or write your own cartesian product function, which takes functions that would produce each generator. For instance:
def product(*funcs, repeat=None):
if not funcs:
yield ()
return
if repeat is not None:
funcs *= repeat
func, *rest = funcs
for val in func():
for res in product(*rest):
yield (val, ) + res
from functools import partial
values = product(partial(gen1, arg1, arg2), partial(gen2, arg1))
The bonus from rolling your own here is that you can also change how it goes about traversing the A by B by C ... dimensional search space, so that you could do maybe a breadth-first search instead of an iteratively deepening DFS. Or, maybe pick some random space-filling curve, such as the Hilbert Curve which would iterate all indices/depths of each dimension in your product() in a local-centric fashion.
Apart from that, I have one more thing to point out- you can also implement BFS lazily (using generators) to avoid building a queue that could bloat memory usage as well. See this SO answer, copied below for convenience:
def breadth_first(self):
yield self
for c in self.breadth_first():
if not c.children:
return # stop the recursion as soon as we hit a leaf
yield from c.children
Overall, you will take a performance hit from using semi-coroutines, with zero caching, all in python-land (in comparison to the baked in and heavily optimized C of CPython). However, it should still be doable- algorithmic optimizations (avoiding generating semantically nonsensical ASTs, prioritizing ASTs that suit your goal, etc.) will have a larger impact than the constant-factor performance hit.

What does self do inside a Python array embedded in a Class?

def traverse(self):
print("Traversing...")
nodes_to_visit = [self]
while len(nodes_to_visit) != 0:
current_node = nodes_to_visit.pop()
print(current_node.value)
nodes_to_visit += current_node.children
I have this function inside a class(I'm learning Data Structures), and on the third line, there's a self inside an array, that is then used. What does it do, what does it return? (And while asking, are Data Structures advanced? Can I consider my self an 'advanced' programmer now ;)?)

Data Structures can be indeed advance. Most likely it involves performance for processing many data like thousands and millions. You will learn terms like run time complexity O(n), log(n).
An example benefit of having good knowledge in Data Structure is when you parse an excel file that contains a million rows. I have a colleage that made a script that took an hour to finish the job while mine only took 5 minutes.
Take note that the most basic data structure for python is the dictionary.. It is always O(n) runtime complexity

LFU cache implementation in python

I have implemented LFU cache in python with the help of Priority Queue Implementation given at
https://docs.python.org/2/library/heapq.html#priority-queue-implementation-notes
I have given code in the end of the post.
But I feel that code has some serious problems:
1. To give a scenario, suppose there is only one page is continuously getting visited (say 50 times). But this code will always mark the already added node as "removed" and add it to heap again. So basically it will have 50 different nodes for the same page. Hence increasing heap size enormously.
2. This question is almost similar to Q1 of Telephonic Interview of
http://www.geeksforgeeks.org/flipkart-interview-set-2-sde-2/
And the person mentioned that doubly linked list can give better efficiency as compared to heap. Can anyone explain me, how?
from llist import dllist
import sys
from heapq import heappush, heappop
class LFUCache:
heap = []
cache_map = {}
REMOVED = "<removed-task>"
def __init__(self, cache_size):
self.cache_size = cache_size
def get_page_content(self, page_no):
if self.cache_map.has_key(page_no):
self.update_frequency_of_page_in_cache(page_no)
else:
self.add_page_in_cache(page_no)
return self.cache_map[page_no][2]
def add_page_in_cache(self, page_no):
if (len(self.cache_map) == self.cache_size):
self.delete_page_from_cache()
heap_node = [1, page_no, "content of page " + str(page_no)]
heappush(self.heap, heap_node)
self.cache_map[page_no] = heap_node
def delete_page_from_cache(self):
while self.heap:
count, page_no, page_content = heappop(self.heap)
if page_content is not self.REMOVED:
del self.cache_map[page_no]
return
def update_frequency_of_page_in_cache(self, page_no):
heap_node = self.cache_map[page_no]
heap_node[2] = self.REMOVED
count = heap_node[0]
heap_node = [count+1, page_no, "content of page " + str(page_no)]
heappush(self.heap, heap_node)
self.cache_map[page_no] = heap_node
def main():
cache_size = int(raw_input("Enter cache size "))
cache = LFUCache(cache_size)
while 1:
page_no = int(raw_input("Enter page no needed "))
print cache.get_page_content(page_no)
print cache.heap, cache.cache_map, "\n"
if __name__ == "__main__":
main()

Efficiency is a tricky thing. In real-world applications, it's often a good idea to use the simplest and easiest algorithm, and only start to optimize when that's measurably slow. And then you optimize by doing profiling to figure out where the code is slow.
If you are using CPython, it gets especially tricky, as even an inefficient algorithm implemented in C can beat an efficient algorithm implemented in Python due to the large constant factors; e.g. a double-linked list implemented in Python tends to be a lot slower than simply using the normal Python list, even for cases where in theory it should be faster.
Simple algorithm:
For an LFU, the simplest algorithm is to use a dictionary that maps keys to (item, frequency) objects, and update the frequency on each access. This makes access very fast (O(1)), but pruning the cache is slower as you need to sort by frequency to cut off the least-used elements. For certain usage characteristics, this is actually faster than other "smarter" solutions, though.
You can optimize for this pattern by not simply pruning your LFU cache to the maximum length, but to prune it to, say, 50% of the maximum length when it grows too large. That means your prune operation is called infrequently, so it can be inefficient compared to the read operation.
Using a heap:
In (1), you used a heap because that's an efficient way of storing a priority queue. But you are not implementing a priority queue. The resulting algorithm is optimized for pruning, but not access: You can easily find the n smallest elements, but it's not quite as obvious how to update the priority of an existing element. In theory, you'd have to rebalance the heap after every access, which is highly inefficient.
To avoid that, you added a trick by keeping elements around even if they are deleted. But this trades in space for time.
If you don't want to trade in time, you could update the frequencies in-place, and simply rebalance the heap before pruning the cache. You regain fast access times at the expense of slower pruning time, like the simple algorithm above. (I doubt there is any speed difference between the two, but I have not measured this.)
Using a double-linked list:
The double-linked list mentioned in (2) takes advantage of the nature of the possible changes here: An element is either added as the lowest priority (0 accesses), or an existing element's priority is incremented exactly by 1. You can use these attributes to your advantage if you design your data structures like this:
You have a double-linked list of elements which is ordered by the frequency of the elements. In addition, you have a dictionary that maps items to elements within that list.
Accessing an element then means:
Either it's not in the dictionary, that is, it's a new item, in which case you can simply append it to the end of the double-linked list (O(1))
or it's in the dictionary, in which case you increment the frequency in the element and move it leftwards through the double-linked list until the list is ordered again (O(n) worst-case, but usually closer to O(1)).
To prune the cache, you simply cut off n elements from the end of the list (O(n)).

Quick way to extend a set if we know elements are unique

I am performing multiple iterations of the type:
masterSet=masterSet.union(setA)
As the set grows the length of time taken to perform these operations is growing (as one would expect, I guess).
I expect that the time is taken up checking whether each element of setA is already in masterSet?
My question is that if i KNOW that masterSet does not already contain any of elements in setA can I do this quicker?
[UPDATE]
Given that this question is still attracting views I thought I would clear up a few of the things from the comments and answers below:
When iterating though there were many iterations where I knew setA would be distinct from masterSet because of how it was constructed (without having to process any checks) but a few iterations I needed the uniqueness check.
I wondered if there was a way to 'tell' the masterSet.union() procedure not to bother with the uniquness check this time around as I know this one is distinct from masterSet just add these elements quickly trusting the programmer's assertion they were definately distict. Perhpas through calling some different ".unionWithDistinctSet()" procedure or something.
I think the responses have suggested that this isnt possible (and that really set operations should be quick enough anyway) but to use masterSet.update(setA) instead of union as its slightly quicker still.
I have accepted the clearest reponse along those lines, resolved the issue I was having at the time and got on with my life but would still love to hear if my hypothesised .unionWithDistinctSet() could ever exist?

You can use set.update to update your master set in place. This saves allocating a new set all the time so it should be a little faster than set.union...
>>> s = set(range(3))
>>> s.update(range(4))
>>> s
set([0, 1, 2, 3])
Of course, if you're doing this in a loop:
masterSet = set()
for setA in iterable:
masterSet = masterSet.union(setA)
You might get a performance boost by doing something like:
masterSet = set().union(*iterable)
Ultimately, membership testing of a set is O(1) (in the average case), so testing if the element is already contained in the set isn't really a big performance hit.

As mgilson points out, you can use update to update a set in-place from another set. That actually works out slightly quicker:
def union():
i = set(range(10000))
j = set(range(5000, 15000))
return i.union(j)
def update():
i = set(range(10000))
j = set(range(5000, 15000))
i.update(j)
return i
timeit.Timer(union).timeit(10000) # 10.351907968521118
timeit.Timer(update).timeit(10000) # 8.83384895324707

If you know your elements are unique, a set is not necessarily the best structure.
A simple list is way faster to extend.
masterList = list(masterSet)
masterList.extend(setA)

For sure, forgoing this check could be a big saving when the __eq__(..) method is very expensive. In the CPython implementation, __eq__(..) is called with every element already in the set that hashes to the same number. (Reference: source code for set.)
However, there will never be this functionality in a million years, because it opens up another way to violate the integrity of a set. The trouble associated with that far outweighs the (typically negligible) performance gain. While if this is determined as a performance bottleneck, it's not hard to write a C++ extension, and use its STL <set>, which should be faster by one or more orders of magnitude.

How to optimize operations on large (75,000 items) sets of booleans in Python?

There's this script called svnmerge.py that I'm trying to tweak and optimize a bit. I'm completely new to Python though, so it's not easy.
The current problem seems to be related to a class called RevisionSet in the script. In essence what it does is create a large hashtable(?) of integer-keyed boolean values. In the worst case - one for each revision in our SVN repository, which is near 75,000 now.
After that it performs set operations on such huge arrays - addition, subtraction, intersection, and so forth. The implementation is the simplest O(n) implementation, which, naturally, gets pretty slow on such large sets. The whole data structure could be optimized because there are long spans of continuous values. For example, all keys from 1 to 74,000 might contain true. Also the script is written for Python 2.2, which is a pretty old version and we're using 2.6 anyway, so there could be something to gain there too.
I could try to cobble this together myself, but it would be difficult and take a lot of time - not to mention that it might be already implemented somewhere. Although I'd like the learning experience, the result is more important right now. What would you suggest I do?

You could try doing it with numpy instead of plain python. I found it to be very fast for operations like these.
For example:
# Create 1000000 numbers between 0 and 1000, takes 21ms
x = numpy.random.randint(0, 1000, 1000000)
# Get all items that are larger than 500, takes 2.58ms
y = x > 500
# Add 10 to those items, takes 26.1ms
x[y] += 10
Since that's with a lot more rows, I think that 75000 should not be a problem either :)

Here's a quick replacement for RevisionSet that makes it into a set. It should be much faster. I didn't fully test it, but it worked with all of the tests that I did. There are undoubtedly other ways to speed things up, but I think that this will really help because it actually harnesses the fast implementation of sets rather than doing loops in Python which the original code was doing in functions like __sub__ and __and__. The only problem with it is that the iterator isn't sorted. You might have to change a little bit of the code to account for this. I'm sure there are other ways to improve this, but hopefully it will give you a good start.
class RevisionSet(set):
"""
A set of revisions, held in dictionary form for easy manipulation. If we
were to rewrite this script for Python 2.3+, we would subclass this from
set (or UserSet). As this class does not include branch
information, it's assumed that one instance will be used per
branch.
"""
def __init__(self, parm):
"""Constructs a RevisionSet from a string in property form, or from
a dictionary whose keys are the revisions. Raises ValueError if the
input string is invalid."""
revision_range_split_re = re.compile('[-:]')
if isinstance(parm, set):
print "1"
self.update(parm.copy())
elif isinstance(parm, list):
self.update(set(parm))
else:
parm = parm.strip()
if parm:
for R in parm.split(","):
rev_or_revs = re.split(revision_range_split_re, R)
if len(rev_or_revs) == 1:
self.add(int(rev_or_revs[0]))
elif len(rev_or_revs) == 2:
self.update(set(range(int(rev_or_revs[0]),
int(rev_or_revs[1])+1)))
else:
raise ValueError, 'Ill formatted revision range: ' + R
def sorted(self):
return sorted(self)
def normalized(self):
"""Returns a normalized version of the revision set, which is an
ordered list of couples (start,end), with the minimum number of
intervals."""
revnums = sorted(self)
revnums.reverse()
ret = []
while revnums:
s = e = revnums.pop()
while revnums and revnums[-1] in (e, e+1):
e = revnums.pop()
ret.append((s, e))
return ret
def __str__(self):
"""Convert the revision set to a string, using its normalized form."""
L = []
for s,e in self.normalized():
if s == e:
L.append(str(s))
else:
L.append(str(s) + "-" + str(e))
return ",".join(L)
Addition:
By the way, I compared doing unions, intersections and subtractions of the original RevisionSet and my RevisionSet above, and the above code is from 3x to 7x faster for those operations when operating on two RevisionSets that have 75000 elements. I know that other people are saying that numpy is the way to go, but if you aren't very experienced with Python, as your comment indicates, then you might not want to go that route because it will involve a lot more changes. I'd recommend trying my code, seeing if it works and if it does, then see if it is fast enough for you. If it isn't, then I would try profiling to see what needs to be improved. Only then would I consider using numpy (which is a great package that I use quite frequently).

For example, all keys from 1 to 74,000 contain true
Why not work on a subset? Just 74001 to the end.
Pruning 74/75th of your data is far easier than trying to write an algorithm more clever than O(n).

You should rewrite RevisionSet to have a set of revisions. I think the internal representation for a revision should be an integer and revision ranges should be created as needed.
There is no compelling reason to use code that supports python 2.3 and earlier.

Just a thought. I used to do this kind of thing using run-coding in binary image manipulation. That is, store each set as a series of numbers: number of bits off, number of bits on, number of bits off, etc.
Then you can do all sorts of boolean operations on them as decorations on a simple merge algorithm.