How to optimize PageRank calculation?

Question

In the book Collective Intelligence Programming, I found the following function to calculate PageRank:

def calculatepagerank(self,iterations=20):
    # clear out the current PageRank tables
    self.con.execute("drop table if exists pagerank")
    self.con.execute("create table pagerank(urlid primary key,score)")
    self.con.execute("create index prankidx on pagerank(urlid)")

    # initialize every url with a PageRank of 1.0
    self.con.execute("insert into pagerank select rowid,1.0 from urllist")
    self.dbcommit()

    for i in range(iterations):
        print "Iteration %d" % i
        for (urlid,) in self.con.execute("select rowid from urllist"):
            pr=0.15

            # Loop through all the pages that link to this one
            for (linker,) in self.con.execute("select distinct fromid from link where toid=%d" % urlid):
                # Get the PageRank of the linker
                linkingpr=self.con.execute("select score from pagerank where urlid=%d" % linker).fetchone()[0]

                # Get the total number of links from the linker
                linkingcount=self.con.execute("select count(*) from link where fromid=%d" % linker).fetchone()[0]

                pr+=0.85*(linkingpr/linkingcount)

            self.con.execute("update pagerank set score=%f where urlid=%d" % (pr,urlid))
        self.dbcommit()

However, this function is very slow, due to all the SQL queries at each iteration

>>> import cProfile
>>> cProfile.run("crawler.calculatepagerank()")
         2262510 function calls in 136.006 CPU seconds

   Ordered by: standard name

ncalls  tottime  percall  cumtime  percall filename:lineno(function)
     1    0.000    0.000  136.006  136.006 <string>:1(<module>)
     1   20.826   20.826  136.006  136.006 searchengine.py:179(calculatepagerank)
    21    0.000    0.000    0.528    0.025 searchengine.py:27(dbcommit)
    21    0.528    0.025    0.528    0.025 {method 'commit' of 'sqlite3.Connecti
     1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler
1339864  112.602    0.000  112.602    0.000 {method 'execute' of 'sqlite3.Connec 
922600    2.050    0.000    2.050    0.000 {method 'fetchone' of 'sqlite3.Cursor' 
     1    0.000    0.000    0.000    0.000 {range}

So, I optimized the function and came up with the following:

def calculatepagerank2(self,iterations=20):
    # clear out the current PageRank tables
    self.con.execute("drop table if exists pagerank")
    self.con.execute("create table pagerank(urlid primary key,score)")
    self.con.execute("create index prankidx on pagerank(urlid)")

    # initialize every url with a PageRank of 1.0
    self.con.execute("insert into pagerank select rowid,1.0 from urllist")
    self.dbcommit()

    inlinks={}
    numoutlinks={}
    pagerank={}

    for (urlid,) in self.con.execute("select rowid from urllist"):
        inlinks[urlid]=[]
        numoutlinks[urlid]=0
        # Initialize pagerank vector with 1.0
        pagerank[urlid]=1.0
        # Loop through all the pages that link to this one
        for (inlink,) in self.con.execute("select distinct fromid from link where toid=%d" % urlid):
            inlinks[urlid].append(inlink)
            # get number of outgoing links from a page        
            numoutlinks[urlid]=self.con.execute("select count(*) from link where fromid=%d" % urlid).fetchone()[0]            

    for i in range(iterations):
        print "Iteration %d" % i

        for urlid in pagerank:
            pr=0.15
            for link in inlinks[urlid]:
                linkpr=pagerank[link]
                linkcount=numoutlinks[link]
                pr+=0.85*(linkpr/linkcount)
            pagerank[urlid]=pr
    for urlid in pagerank:
        self.con.execute("update pagerank set score=%f where urlid=%d" % (pagerank[urlid],urlid))
    self.dbcommit()

This function is many times faster (but uses much more memory for all temporary dictionaries), because it allows you to avoid unnecessary SQL queries at each iteration:

>>> cProfile.run("crawler.calculatepagerank2()")
     90070 function calls in 3.527 CPU seconds
Ordered by: standard name

ncalls  tottime  percall  cumtime  percall filename:lineno(function)
     1    0.004    0.004    3.527    3.527 <string>:1(<module>)
     1    1.154    1.154    3.523    3.523 searchengine.py:207(calculatepagerank2
     2    0.000    0.000    0.058    0.029 searchengine.py:27(dbcommit)
 23065    0.013    0.000    0.013    0.000 {method 'append' of 'list' objects}
     2    0.058    0.029    0.058    0.029 {method 'commit' of 'sqlite3.Connectio
     1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler
 43932    2.261    0.000    2.261    0.000 {method 'execute' of 'sqlite3.Connecti
 23065    0.037    0.000    0.037    0.000 {method 'fetchone' of 'sqlite3.Cursor'
     1    0.000    0.000    0.000    0.000 {range}

But is it possible to further reduce the number of SQL queries to speed up this function? Update . Fixed indents in calculatepagerank2 ().

Answer 1

, , :

    def calculatepagerank4(self,iterations=20):
    # clear out the current PageRank tables
    self.con.execute("drop table if exists pagerank")
    self.con.execute("create table pagerank(urlid primary key,score)")
    self.con.execute("create index prankidx on pagerank(urlid)")

    # initialize every url with a PageRank of 1.0
    self.con.execute("insert into pagerank select rowid,1.0 from urllist")
    self.dbcommit()

    inlinks={}
    numoutlinks={}
    pagerank={}

    for (urlid,) in self.con.execute("select rowid from urllist"):
        inlinks[urlid]=[]
        numoutlinks[urlid]=0
        # Initialize pagerank vector with 1.0
        pagerank[urlid]=1.0

    for src,dest in self.con.execute("select distinct fromid, toid from link"):
        inlinks[dest].append(src)
        numoutlinks[src]+=1          

    for i in range(iterations):
        print "Iteration %d" % i

        for urlid in pagerank:
            pr=0.15
            for link in inlinks[urlid]:
                linkpr=pagerank[link]
                linkcount=numoutlinks[link]
                pr+=0.85*(linkpr/linkcount)
            pagerank[urlid]=pr

    args=((pagerank[urlid],urlid) for urlid in pagerank)
    self.con.executemany("update pagerank set score=? where urlid=?" , args)
    self.dbcommit() 

, , allyourcode, , , executemany(), ˜unutbu. ˜unutbu args, , . 100 , :

>>> cProfile.run("crawler.calculatepagerank4()")
     33512 function calls in 1.377 CPU seconds
Ordered by: standard name

ncalls  tottime  percall  cumtime  percall filename:lineno(function)
     1    0.004    0.004    1.377    1.377 <string>:1(<module>)
     2    0.000    0.000    0.073    0.036 searchengine.py:27(dbcommit)
     1    0.693    0.693    1.373    1.373 searchengine.py:286(calculatepagerank4
 10432    0.011    0.000    0.011    0.000 searchengine.py:321(<genexpr>)
 23065    0.009    0.000    0.009    0.000 {method 'append' of 'list' objects}
     2    0.073    0.036    0.073    0.036 {method 'commit' of 'sqlite3.Connectio
     1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler
     6    0.379    0.063    0.379    0.063 {method 'execute' of 'sqlite3.Connecti
     1    0.209    0.209    0.220    0.220 {method 'executemany' of 'sqlite3.Conn
     1    0.000    0.000    0.000    0.000 {range}

:

• %f placeholder ? SQL, (, 2.9796095721920315 ?, 2.9796100000000001 %f.
• PageRank. .
• : , . , , . , / , , Power Iteration.

Answer 2

(, # ~ # WWW) , , , , .

, () , . :

for urlid, in self.con.execute('select rowid from urllist'):
    inlinks[urlid] = []
    numoutlinks[urlid] = 0
    pagerank[urlid] = 1.0

for src, dest in self.con.execute('select fromid, toid from link'):
    inlinks[dest].append(src)
    numoutlinks[src] += 1

2 O (n ^ 2) .

Answer 3

, SQL-:

for (urlid,) in self.con.execute("select rowid from urllist"):
    ...
    for (inlink,) in self.con.execute("select distinct fromid from link where toid=%d" % urlid):
        ...
        numoutlinks[urlid]=self.con.execute("select count(*) from link where fromid=%d" % urlid).fetchone()[0]            

, , :

1. SELECT fromid,toid FROM link WHERE toid IN (SELECT rowid FROM urllist)
   
2. SELECT fromid,count(*) FROM link WHERE fromid IN (SELECT rowid FROM urllist) GROUP BY fromid

inlinks, numoutlinks pagerank.

collections.defaultdict:

import collections
import itertools
def constant_factory(value):
    return itertools.repeat(value).next

inlinks dict . , URL-

inlinks=collections.defaultdict(set)

pagerank a dict, 1.0:

pagerank=collections.defaultdict(constant_factory(1.0))

collection.defaultdict , dicts.

, , , , :

import collections
def constant_factory(value):
    return itertools.repeat(value).next
def calculatepagerank2(self,iterations=20):
    # clear out the current PageRank tables
    self.con.execute("DROP TABLE IF EXISTS pagerank")
    self.con.execute("CREATE TABLE pagerank(urlid primary key,score)")
    self.con.execute("CREATE INDEX prankidx ON pagerank(urlid)")

    # initialize every url with a PageRank of 1.0
    self.con.execute("INSERT INTO pagerank SELECT rowid,1.0 FROM urllist")
    self.dbcommit()

    inlinks=collections.defaultdict(set)

    sql='''SELECT fromid,toid FROM link WHERE toid IN (SELECT rowid FROM urllist)'''
    for f,t in self.con.execute(sql):
        inlinks[t].add(f)

    numoutlinks={}
    sql='''SELECT fromid,count(*) FROM link WHERE fromid IN (SELECT rowid FROM urllist) GROUP BY fromid'''
    for f,c in self.con.execute(sql):
        numoutlinks[f]=c

    pagerank=collections.defaultdict(constant_factory(1.0))
    for i in range(iterations):
        print "Iteration %d" % i
        for urlid in inlinks:
            pr=0.15
            for link in inlinks[urlid]:
                linkpr=pagerank[link]
                linkcount=numoutlinks[link]
                pr+=0.85*(linkpr/linkcount)
            pagerank[urlid]=pr
    sql="UPDATE pagerank SET score=? WHERE urlid=?"
    args=((pagerank[urlid],urlid) for urlid in pagerank)
    self.con.executemany(sql, args)
    self.dbcommit()

Answer 4

Do you have enough RAM to store a sparse matrix (fromid, toid)in some form? This would allow for large optimizations (with large algorithmic changes). At the very least, in-memory caching (fromid, numlinks), which you now use with select count(*)in your inner loop, should help (I would suggest that being O(N)in space, if you deal with NURLs, will most likely fit in memory).