Merge sorted lists in python

Question

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.p         


        

Answer 1

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

Answer 2

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.

Answer 3

I like Roberto Liffredo's answer. I didn't know about heapq.merge(). Hmmmph.

Here's what the complete solution looks like using Roberto's lead:

class Obj(object):
    def __init__(self, p) :
        self.points = p
    def __cmp__(self, b) :
        return cmp(self.points, b.points)
    def __str__(self):
        return "%d" % self.points

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

import heapq

sorted = [item for item in heapq.merge(a,b,c)]
for item in sorted:
    print item

Or:

for item in heapq.merge(a,b,c):
    print item