Question

I’m doing an iteration through 3 words, each about 5 million characters long, and I want to find sequences of 20 characters that identifies each word. That is, I want to find all sequences of length 20 in one word that is unique for that word. My problem is that the code I’ve written takes an extremely long time to run. I’ve never even completed one word running my program over night.

The function below takes a list containing dictionaries where each dictionary contains each possible word of 20 and its location from one of the 5 million long words.

If anybody has an idea how to optimize this I would be really thankful, I don’t have a clue how to continue…

here’s a sample of my code:

def findUnique(list):
    # Takes a list with dictionaries and compairs each element in the dictionaries
    # with the others and puts all unique element in new dictionaries and finally
    # puts the new dictionaries in a list.
    # The result is a list with (in this case) 3 dictionaries containing all unique
    # sequences and their locations from each string.
    dicList=[]
    listlength=len(list)
    s=0
    valuelist=[]
    for i in list:
        j=i.values()
        valuelist.append(j)
    while s<listlength:
        currdic=list[s]
        dic={}
        for key in currdic:
            currval=currdic[key]
            test=True
            n=0
            while n<listlength:
                if n!=s:
                    if currval in valuelist[n]: #this is where it takes to much time
                        n=listlength
                        test=False
                    else:
                        n+=1
                else:
                    n+=1
            if test:
                dic[key]=currval
        dicList.append(dic)
        s+=1
    return dicList

Answer 1

def slices(seq, length, prefer_last=False):
  unique = {}
  if prefer_last: # this doesn't have to be a parameter, just choose one
    for start in xrange(len(seq) - length + 1):
      unique[seq[start:start+length]] = start
  else: # prefer first
    for start in xrange(len(seq) - length, -1, -1):
      unique[seq[start:start+length]] = start
  return unique

# or find all locations for each slice:
import collections
def slices(seq, length):
  unique = collections.defaultdict(list)
  for start in xrange(len(seq) - length + 1):
    unique[seq[start:start+length]].append(start)
  return unique

This function (currently in my iter_util module) is O(n) (n being the length of each word) and you would use set(slices(..)) (with set operations such as difference) to get slices unique across all words (example below). You could also write the function to return a set, if you don’t want to track locations. Memory usage will be high (though still O(n), just a large factor), possibly mitigated (though not by much if length is only 20) with a special “lazy slice” class that stores the base sequence (the string) plus start and stop (or start and length).

Printing unique slices:

a = set(slices("aab", 2)) # {"aa", "ab"}
b = set(slices("abb", 2)) # {"ab", "bb"}
c = set(slices("abc", 2)) # {"ab", "bc"}
all = [a, b, c]
import operator
a_unique = reduce(operator.sub, (x for x in all if x is not a), a)
print a_unique # {"aa"}

Including locations:

a = slices("aab", 2)
b = slices("abb", 2)
c = slices("abc", 2)
all = [a, b, c]
import operator
a_unique = reduce(operator.sub, (set(x) for x in all if x is not a), set(a))
# a_unique is only the keys so far
a_unique = dict((k, a[k]) for k in a_unique)
# now it's a dict of slice -> location(s)
print a_unique # {"aa": 0} or {"aa": [0]}
               # (depending on which slices function used)

---

In a test script closer to your conditions, using randomly generated words of 5m characters and a slice length of 20, memory usage was so high that my test script quickly hit my 1G main memory limit and started thrashing virtual memory. At that point Python spent very little time on the CPU and I killed it. Reducing either the slice length or word length (since I used completely random words that reduces duplicates and increases memory use) to fit within main memory and it ran under a minute. This situation plus O(n**2) in your original code will take forever, and is why algorithmic time and space complexity are both important.

import operator
import random
import string

def slices(seq, length):
  unique = {}
  for start in xrange(len(seq) - length, -1, -1):
    unique[seq[start:start+length]] = start
  return unique

def sample_with_repeat(population, length, choice=random.choice):
  return "".join(choice(population) for _ in xrange(length))

word_length = 5*1000*1000
words = [sample_with_repeat(string.lowercase, word_length) for _ in xrange(3)]
slice_length = 20
words_slices_sets = [set(slices(x, slice_length)) for x in words]
unique_words_slices = [reduce(operator.sub,
                              (x for x in words_slices_sets if x is not n),
                              n)
                       for n in words_slices_sets]
print [len(x) for x in unique_words_slices]