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Merge sorted lists in python

Asked 2009-07-21T09:21:00.210
13

I have a bunch of sorted lists of objects, and a comparison function

class Obj :
    def __init__(p) :
        self.points = p
def cmp(a, b) :
    return a.points < b.points

a = [Obj(1), Obj(3), Obj(8), ...]
b = [Obj(1), Obj(2), Obj(3), ...]
c = [Obj(100), Obj(300), Obj(800), ...]

result = magic(a, b, c)
assert result == [Obj(1), Obj(1), Obj(2), Obj(3), Obj(3), Obj(8), ...]

what does magic look like? My current implementation is 

def magic(*args) :
    r = []
    for a in args : r += a
    return sorted(r, cmp)

but that is quite inefficient. Better answers?

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2 Answers

13

Python standard library offers a method for it: heapq.merge.
As the documentation says, it is very similar to using itertools (but with more limitations); if you cannot live with those limitations (or if you do not use Python 2.6) you can do something like this:

sorted(itertools.chain(args), cmp)

However, I think it has the same complexity as your own solution, although using iterators should give some quite good optimization and speed increase.

answered 2009-07-21T09:38:03.077
0

Below is an example of a function that runs in O(n) comparisons.

You could make this faster by making a and b iterators and incrementing them.

I have simply called the function twice to merge 3 lists:

def zip_sorted(a, b):
    '''
    zips two iterables, assuming they are already sorted
    '''
    i = 0
    j = 0
    result = []
    while i < len(a) and j < len(b):
        if a[i] < b[j]:
            result.append(a[i])
            i += 1
        else:
            result.append(b[j])
            j += 1
    if i < len(a):
        result.extend(a[i:])
    else:
        result.extend(b[j:])
    return result

def genSortedList(num,seed):
    result = [] 
    for i in range(num):
        result.append(i*seed)
    return result

if __name__ == '__main__':
    a = genSortedList(10000,2.0)
    b = genSortedList(6666,3.0)
    c = genSortedList(5000,4.0)
    d = zip_sorted(zip_sorted(a,b),c)
    print d

However, heapq.merge uses a mix of this method and heaping the current elements of all lists, so should perform much better

answered 2013-04-09T04:51:05.343

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