paralellize loop over iter

Question

I am having performance issues with my code. step # IIII consumes hours of time. I used to materialize the the itertools.prodct before, but thanks to a user I dont do pro_data = product(array_b,array_a) anymore. This helped me with memory issues, but the still is heavily time consuming. I would like to paralellize it with multithreading or multiprocesisng, whatever you can suggest, I am grateful.

Explanation. I have two arrays that contain x and y values of particles. For each particle (defined by two coordinates) I want to calculate a function with another. For combinations I use the itertools.product method and loop over every particle. I run over 50000 particels in total, so I have N*N/2 combinations to calculate.

Thanks in advance

import numpy as np
import matplotlib.pyplot as plt
from itertools import product,combinations_with_replacement

def func(ar1,ar2,ar3,ar4): #example func that takes four arguments
  return (ar1*ar2**22+np.sin(ar3)+ar4)

def newdist(a):
  return func(a[0][0],a[0][1],a[1][0],a[1][1])    

x_edges = np.logspace(-3,1, num=25) #prepare x-axis for histogram 

x_mean = 10**((np.log10(x_edges[:-1])+np.log10(x_edges[1:]))/2)
x_width=x_edges[1:]-x_edges[:-1]

hist_data=np.zeros([len(x_edges)-1])

array1=np.random.uniform(0.,10.,100)
array2=np.random.uniform(0.,10.,100)

array_a = np.dstack((array1,array1))[0]
array_b = np.dstack((array2,array2))[0]
# IIII
for i in product(array_a,array_b):
  (result,bins) = np.histogram(newdist(i),bins=x_edges)
  hist_data+=result

hist_data = np.array(map(float, hist_data))
plt.bar(x_mean,hist_data,width=x_width,color='r')
plt.show()

-----EDIT----- I used this code now:

def mp_dist(array_a,array_b, d, bins): #d chunks AND processes
  def worker(array_ab, out_q):
      """ push result in queue """
      outdict = {}
      outdict = vec_chunk(array_ab, bins)
      out_q.put(outdict)
  out_q = mp.Queue()
  a = np.swapaxes(array_a, 0 ,1)
  b = np.swapaxes(array_b, 0 ,1)
  array_size_a=len(array_a)-(len(array_a)%d)
  array_size_b=len(array_b)-(len(array_b)%d)
  a_chunk = array_size_a / d
  b_chunk = array_size_b / d
  procs = []
  #prepare arrays for mp
  array_ab = np.empty((4, a_chunk, b_chunk))
  for j in xrange(d):
    for k in xrange(d):
      array_ab[[0, 1]] = a[:, a_chunk * j:a_chunk * (j + 1), None]
      array_ab[[2, 3]] = b[:, None, b_chunk * k:b_chunk * (k + 1)]
      p = mp.Process(target=worker, args=(array_ab, out_q))
      procs.append(p)
      p.start()
  resultarray = np.empty(len(bins)-1)
  for i in range(d):
      resultarray+=out_q.get() 
  # Wait for all worker processes to finish
  for pro in procs:
      pro.join()
  print resultarray
  return resultarray

Problem here is that I cannot control the numbers of processes. How Can I use a mp.Pool() instead? than

Answer 1

Use vectorized numpy operations. Replace the for-loop over product() with a single newdist() call by creating arguments using meshgrid().

To parallize the problem compute newdist() on slices of array_a, array_b that correspond to subblocks of meshgrid(). Here's an example using slices and multiprocessing.

Here's another example to demonstrate the steps: python loop -> vectorized numpy version -> parallel:

#!/usr/bin/env python
from __future__ import division
import math
import multiprocessing as mp
import numpy as np

try:
    from itertools import izip as zip
except ImportError:
    zip = zip # Python 3

def pi_loop(x, y, npoints):
    """Compute pi using Monte-Carlo method."""
    #  note: the method converges to pi very slowly.
    return 4 * sum(1 for xx, yy in zip(x, y) if (xx**2 + yy**2) < 1) / npoints

def pi_vectorized(x, y, npoints):
    return 4 * ((x**2 + y**2) < 1).sum() / npoints # or just .mean()

def mp_init(x_shared, y_shared):
    global mp_x, mp_y
    mp_x, mp_y = map(np.frombuffer, [x_shared, y_shared]) # no copy

def mp_pi(args):
    # perform computations on slices of mp_x, mp_y
    start, end = args
    x = mp_x[start:end] # no copy
    y = mp_y[start:end]
    return ((x**2 + y**2) < 1).sum()

def pi_parallel(x, y, npoints):
    # compute pi using multiple processes
    pool = mp.Pool(initializer=mp_init, initargs=[x, y])
    step = 100000
    slices = ((start, start + step) for start in range(0, npoints, step))
    return 4 * sum(pool.imap_unordered(mp_pi, slices)) / npoints

def main():
    npoints = 1000000

    # create shared arrays
    x_sh, y_sh = [mp.RawArray('d', npoints) for _ in range(2)]

    # initialize arrays
    x, y = map(np.frombuffer, [x_sh, y_sh])
    x[:] = np.random.uniform(size=npoints)
    y[:] = np.random.uniform(size=npoints)

    for f, a, b in [(pi_loop, x, y), 
                    (pi_vectorized, x, y), 
                    (pi_parallel, x_sh, y_sh)]:
        pi = f(a, b, npoints)
        precision = int(math.floor(math.log10(npoints)) / 2 - 1 + 0.5)
        print("%.*f %.1e" % (precision + 1, pi, abs(pi - math.pi)))

if __name__=="__main__":
    main()

Time performance for npoints = 10_000_000:

pi_loop pi_vectorized pi_parallel 
   32.6         0.159       0.069 # seconds

It shows that the main performance benefit is from converting the python loop to its vectorized numpy analog.

Answer 2

First, lets look at a straightforward vectorization of your problem. I have a feeling that you want your array_a and array_b to be the exact same, i.e. the coordinates of the particles, but I am keeping them separate here.

I have turned your code into a function, to make timing easier:

def IIII(array_a, array_b, bins) :
    hist_data=np.zeros([len(bins)-1])
    for i in product(array_a,array_b):
        (result,bins) = np.histogram(newdist(i), bins=bins)
        hist_data+=result
    hist_data = np.array(map(float, hist_data))
    return hist_data

You can, by the way, generate your sample data in a less convoluted way as follows:

n = 100
array_a = np.random.uniform(0, 10, size=(n, 2))
array_b = np.random.uniform(0, 10, size=(n, 2))

So first we need to vectorize your func. I have done it so it can take any array of shape (4, ...). To spare memory, it is doing the calculation in place, and returning the first plane, i.e. array[0].

def func_vectorized(a) :
    a[1] **= 22
    np.sin(a[2], out=a[2])
    a[0] *= a[1]
    a[0] += a[2]
    a[0] += a[3]
    return a[0]

With this function in place, we can write a vectorized version of IIII:

def IIII_vec(array_a, array_b, bins) :
    array_ab = np.empty((4, len(array_a), len(array_b)))
    a = np.swapaxes(array_a, 0 ,1)
    b = np.swapaxes(array_b, 0 ,1)
    array_ab[[0, 1]] = a[:, :, None]
    array_ab[[2, 3]] = b[:, None, :]
    newdist = func_vectorized(array_ab)
    hist, _ = np.histogram(newdist, bins=bins)
    return hist

With n = 100 points, they both return the same:

In [2]: h1 = IIII(array_a, array_b, x_edges)

In [3]: h2 = IIII_bis(array_a, array_b, x_edges)

In [4]: np.testing.assert_almost_equal(h1, h2)

But the timing differences are already very relevant:

In [5]: %timeit IIII(array_a, array_b, x_edges)
1 loops, best of 3: 654 ms per loop

In [6]: %timeit IIII_vec(array_a, array_b, x_edges)
100 loops, best of 3: 2.08 ms per loop

A 300x speedup!. If you try it again with longer sample data, n = 1000, you can see that they both scale equally bad, as n**2, so the 300x stays there:

In [10]: %timeit IIII(array_a, array_b, x_edges)
1 loops, best of 3: 68.2 s per loop

In [11]: %timeit IIII_bis(array_a, array_b, x_edges)
1 loops, best of 3: 229 ms per loop

So you are still looking at a good 10 min. of processing, which is not really that much when compared to the more than 2 days that your current solution would require.

Of course, for things to be so nice, you will need to fit a (4, 50000, 50000) array of floats into memory, something that my system cannot handle. But you can still keep things relatively fast, by processing it in chunks. The following version of IIII_vec divides each array into d chunks. As written, the length of the array should be divisible by d. It wouldn't bee too hard to overcome that limitation, but it would obfuscate the true purpose:

def IIII_vec_bis(array_a, array_b, bins, d=1) :
    a = np.swapaxes(array_a, 0 ,1)
    b = np.swapaxes(array_b, 0 ,1)
    a_chunk = len(array_a) // d
    b_chunk = len(array_b) // d
    array_ab = np.empty((4, a_chunk, b_chunk))
    hist_data = np.zeros((len(bins) - 1,))
    for j in xrange(d) :
        for k in xrange(d) :
            array_ab[[0, 1]] = a[:, a_chunk * j:a_chunk * (j + 1), None]
            array_ab[[2, 3]] = b[:, None, b_chunk * k:b_chunk * (k + 1)]
            newdist = func_vectorized(array_ab)
            hist, _ = np.histogram(newdist, bins=bins)
            hist_data += hist
    return hist_data

First, lets check that it really works:

In [4]: h1 = IIII_vec(array_a, array_b, x_edges)

In [5]: h2 = IIII_vec_bis(array_a, array_b, x_edges, d=10)

In [6]: np.testing.assert_almost_equal(h1, h2)

And now some timings. With n = 100:

In [7]: %timeit IIII_vec(array_a, array_b, x_edges)
100 loops, best of 3: 2.02 ms per loop

In [8]: %timeit IIII_vec_bis(array_a, array_b, x_edges, d=10)
100 loops, best of 3: 12 ms per loop

But as you start having to have a larger and larger array in memory, doing it in chunks starts to pay off. With n = 1000:

In [12]: %timeit IIII_vec(array_a, array_b, x_edges)
1 loops, best of 3: 223 ms per loop

In [13]: %timeit IIII_vec_bis(array_a, array_b, x_edges, d=10)
1 loops, best of 3: 208 ms per loop

With n = 10000 I can no longer call IIII_vec without an array is too big error, but the chunky version is still running:

In [18]: %timeit IIII_vec_bis(array_a, array_b, x_edges, d=10)
1 loops, best of 3: 21.8 s per loop

And just to show that it can be done, I have run it once with n = 50000:

In [23]: %timeit -n1 -r1 IIII_vec_bis(array_a, array_b, x_edges, d=50)
1 loops, best of 1: 543 s per loop

So a good 9 minutes of number crunching, which is not all that bad given it has computed 2.5 billion interactions.