Python, using multiprocess is slower than not usin

问题:

After spending a lot of time trying to wrap my head around multiprocessing I came up with this code which is a benchmark test:

Example 1:

from multiprocessing  import Process

class Alter(Process):
    def __init__(self, word):
        Process.__init__(self)
        self.word = word
        self.word2 = ''

    def run(self):
        # Alter string + test processing speed
        for i in range(80000):
            self.word2 = self.word2 + self.word

if __name__=='__main__':
    # Send a string to be altered
    thread1 = Alter('foo')
    thread2 = Alter('bar')
    thread1.start()
    thread2.start()

    # wait for both to finish

    thread1.join()
    thread2.join()

    print(thread1.word2)
    print(thread2.word2)

This completes in 2 seconds (half the time of multithreading). Out of curiosity I decided to run this next:

Example 2:

word2 = 'foo'
word3 = 'bar'

word = 'foo'
for i in range(80000):
    word2 = word2 + word

word  = 'bar'
for i in range(80000):
    word3 = word3 + word

print(word2)
print(word3)

To my horror this ran in less than half a second!

What is going on here? I expected multiprocessing to run faster - shouldn't it complete in half Example 2's time given that Example 1 is Example 2 split into two processes?

Update:

After considering Chris' feedback, I have included the 'actual' code consuming the most process time, and lead me to consider multiprocessing:

self.ListVar = [[13379+ strings],[13379+ strings],
                [13379+ strings],[13379+ strings]]

for b in range(len(self.ListVar)):
    self.list1 = []
    self.temp = []
    for n in range(len(self.ListVar[b])):
        if not self.ListVar[b][n] in self.temp:
            self.list1.insert(n, self.ListVar[b][n] + '(' + 
                              str(self.ListVar[b].count(self.ListVar[b][n])) +
                              ')')
           self.temp.insert(0, self.ListVar[b][n])

   self.ListVar[b] = list(self.list1)

回答1:

ETA: Now that you've posted your code, I can tell you there is a simple way to do what you're doing MUCH faster (>100 times faster).

I see that what you're doing is adding a frequency in parentheses to each item in a list of strings. Instead of counting all the elements each time (which, as you can confirm using cProfile, is by far the largest bottleneck in your code), you can just create a dictionary that maps from each element to its frequency. That way, you only have to go through the list twice- once to create the frequency dictionary, once to use it to add frequency.

Here I'll show my new method, time it, and compare it to the old method using a generated test case. The test case even shows the new result to be exactly identical to the old one. Note: All you really need to pay attention to below is the new_method.

import random
import time
import collections
import cProfile

LIST_LEN = 14000

def timefunc(f):
    t = time.time()
    f()
    return time.time() - t


def random_string(length=3):
    """Return a random string of given length"""
    return "".join([chr(random.randint(65, 90)) for i in range(length)])


class Profiler:
    def __init__(self):
        self.original = [[random_string() for i in range(LIST_LEN)]
                            for j in range(4)]

    def old_method(self):
        self.ListVar = self.original[:]
        for b in range(len(self.ListVar)):
            self.list1 = []
            self.temp = []
            for n in range(len(self.ListVar[b])):
                if not self.ListVar[b][n] in self.temp:
                    self.list1.insert(n, self.ListVar[b][n] + '(' +    str(self.ListVar[b].count(self.ListVar[b][n])) + ')')
                    self.temp.insert(0, self.ListVar[b][n])

            self.ListVar[b] = list(self.list1)
        return self.ListVar

    def new_method(self):
        self.ListVar = self.original[:]
        for i, inner_lst in enumerate(self.ListVar):
            freq_dict = collections.defaultdict(int)
            # create frequency dictionary
            for e in inner_lst:
                freq_dict[e] += 1
            temp = set()
            ret = []
            for e in inner_lst:
                if e not in temp:
                    ret.append(e + '(' + str(freq_dict[e]) + ')')
                    temp.add(e)
            self.ListVar[i] = ret
        return self.ListVar

    def time_and_confirm(self):
        """
        Time the old and new methods, and confirm they return the same value
        """
        time_a = time.time()
        l1 = self.old_method()
        time_b = time.time()
        l2 = self.new_method()
        time_c = time.time()

        # confirm that the two are the same
        assert l1 == l2, "The old and new methods don't return the same value"

        return time_b - time_a, time_c - time_b

p = Profiler()
print p.time_and_confirm()

When I run this, it gets times of (15.963812112808228, 0.05961179733276367), meaning it's about 250 times faster, though this advantage depends on both how long the lists are and the frequency distribution within each list. I'm sure you'll agree that with this speed advantage, you probably won't need to use multiprocessing :)

(My original answer is left in below for posterity)

ETA: By the way, it is worth noting that this algorithm is roughly linear in the length of the lists, while the code you used is quadratic. This means it performs with even more of an advantage the larger the number of elements. For example, if you increase the length of each list to 1000000, it takes only 5 seconds to run. Based on extrapolation, the old code would take over a day :)

---

It depends on the operation you are performing. For example:

import time
NUM_RANGE = 100000000

from multiprocessing  import Process

def timefunc(f):
    t = time.time()
    f()
    return time.time() - t

def multi():
    class MultiProcess(Process):
        def __init__(self):
            Process.__init__(self)

        def run(self):
            # Alter string + test processing speed
            for i in xrange(NUM_RANGE):
                a = 20 * 20

    thread1 = MultiProcess()
    thread2 = MultiProcess()
    thread1.start()
    thread2.start()
    thread1.join()
    thread2.join()

def single():
    for i in xrange(NUM_RANGE):
        a = 20 * 20

    for i in xrange(NUM_RANGE):
        a = 20 * 20

print timefunc(multi) / timefunc(single)

On my machine, the multiprocessed operation takes up only ~60% the time of the singlethreaded one.

回答2:

This example is too small to benefit from multiprocessing.

There's a LOT of overhead when starting a new process. If there were heavy processing involved, it would be negligable. But your example really isn't all that intensive, and so you're bound to notice the overhead.

You'd probably notice a bigger difference with real threads, too bad python (well, CPython) has issues with CPU-bound threading.

回答3:

Multiprocessing could be useful for what you're doing, but not in the way you're thinking about using it. As you're basically doing some computation on every member of a list, you could do it using the multiprocessing.Pool.map method, to do the computation on the list members in parallel.

Here is an example that shows your code's performance using a single process and using multiprocessing.Pool.map:

from multiprocessing import Pool
from random import choice
from string import printable
from time import time

def build_test_list():
    # Builds a test list consisting of 5 sublists of 10000 strings each.
    # each string is 20 characters long
    testlist = [[], [], [], [], []]
    for sublist in testlist:
        for _ in xrange(10000):
            sublist.append(''.join(choice(printable) for _ in xrange(20)))
    return testlist

def process_list(l):
    # the time-consuming code
    result = []
    tmp = []
    for n in range(len(l)):
        if l[n] not in tmp:
            result.insert(n, l[n]+' ('+str(l.count(l[n]))+')')
            tmp.insert(0, l[n])
    return result

def single(l):
    # process the test list elements using a single process
    results = []
    for sublist in l:
        results.append(process_list(sublist))
    return results

def multi(l):
    # process the test list elements in parallel
    pool = Pool()
    results = pool.map(process_list, l)
    return results

print "Building the test list..."
testlist = build_test_list()

print "Processing the test list using a single process..."
starttime = time()
singleresults = single(testlist)
singletime = time() - starttime

print "Processing the test list using multiple processes..."
starttime = time()
multiresults = multi(testlist)
multitime = time() - starttime

# make sure they both return the same thing
assert singleresults == multiresults

print "Single process: {0:.2f}sec".format(singletime)
print "Multiple processes: {0:.2f}sec".format(multitime)

Output:

Building the test list...
Processing the test list using a single process...
Processing the test list using multiple processes...
Single process: 34.73sec
Multiple processes: 24.97sec

回答4:

This thread has been very useful!

Just a quick observation over the good second code provided by David Robinson above (answered Jan 8 '12 at 5:34), which was the code more suitable to my current needs.

In my case I had previous records of the running times of a target function without multiprocessing. When using his code to implement a multiprocessing function his timefunc(multi) didn't reflect the actual time of multi, and it rather appeared to reflect the time expended in the parent.

What i did was to externalise the timing function and the time that I got looked more like expected:

 start = timefunc()
 multi()/single()
 elapsed = (timefunc()-start)/(--number of workers--)
 print(elapsed)

In my case with a double core the total time carried out by 'x' workers using the target function was twice faster than running a simple for-loop over the target function with 'x' iterations.

I am new to multiprocessing so please be cautious with this observation though.