Python Numpy vectorize nested for-loops for combin

问题:

Given an nxn array A of real positive numbers, I'm trying to find the minimum of the maximum of the element-wise minimum of all combinations of three rows of the 2-d array. Using for-loops, that comes out to something like this:

import numpy as np

n = 100
np.random.seed(2)
A = np.random.rand(n,n)
global_best = np.inf

for i in range(n-2):
    for j in range(i+1, n-1):
        for k in range(j+1, n):
            # find the maximum of the element-wise minimum of the three vectors
            local_best = np.amax(np.array([A[i,:], A[j,:], A[k,:]]).min(0))
            # if local_best is lower than global_best, update global_best
            if (local_best < global_best):
                global_best = local_best
                save_rows = [i, j, k]

print global_best, save_rows

In the case for n = 100, the output should be this:

Out[]: 0.492652949593 [6, 41, 58]

I have a feeling though that I could do this much faster using Numpy vectorization, and would certainly appreciate any help on doing this. Thanks.

回答1:

Don't try to vectorize loops that are not simple to vectorize. Instead use a jit compiler like Numba or use Cython. Vectorized solutions are good if the resulting code is more readable, but in terms of performance a compiled solution is usually faster or in a worst case scenario as fast as a vectorized solution (except BLAS routines).

Single-threaded example

import numba as nb
import numpy as np

#Min and max library calls may be costly for only 3 values
@nb.njit()
def max_min_3(A,B,C):
  max_of_min=-np.inf
  for i in range(A.shape[0]):
    loc_min=A[i]
    if (B[i]<loc_min):
      loc_min=B[i]
    if (C[i]<loc_min):
      loc_min=C[i]

    if (max_of_min<loc_min):
      max_of_min=loc_min

  return max_of_min

@nb.njit()
def your_func(A):
  n=A.shape[0]
  save_rows=np.zeros(3,dtype=np.uint64)
  global_best=np.inf
  for i in range(n):
      for j in range(i+1, n):
          for k in range(j+1, n):
              # find the maximum of the element-wise minimum of the three vectors
              local_best = max_min_3(A[i,:], A[j,:], A[k,:])
              # if local_best is lower than global_best, update global_best
              if (local_best < global_best):
                  global_best = local_best
                  save_rows[0] = i
                  save_rows[1] = j
                  save_rows[2] = k

  return global_best, save_rows

Performance of single-threaded version

n=100
your_version: 1.56s
compiled_version: 0.0168s (92x speedup)

n=150
your_version: 5.41s
compiled_version: 0.08122s (66x speedup)

n=500
your_version: 283s
compiled_version: 8.86s (31x speedup)

The first call has a constant overhead of about 0.3-1s. For performance measurement of the calculation time itself, call it once and then measure performance.

With a few code changes this task can also be parallelized.

Multi-threaded example

@nb.njit(parallel=True)
def your_func(A):
  n=A.shape[0]
  all_global_best=np.inf
  rows=np.empty((3),dtype=np.uint64)

  save_rows=np.empty((n,3),dtype=np.uint64)
  global_best_Temp=np.empty((n),dtype=A.dtype)
  global_best_Temp[:]=np.inf

  for i in range(n):
      for j in nb.prange(i+1, n):
          row_1=0
          row_2=0
          row_3=0
          global_best=np.inf
          for k in range(j+1, n):
              # find the maximum of the element-wise minimum of the three vectors

              local_best = max_min_3(A[i,:], A[j,:], A[k,:])
              # if local_best is lower than global_best, update global_best
              if (local_best < global_best):
                  global_best = local_best
                  row_1 = i
                  row_2 = j
                  row_3 = k

          save_rows[j,0]=row_1
          save_rows[j,1]=row_2
          save_rows[j,2]=row_3
          global_best_Temp[j]=global_best

      ind=np.argmin(global_best_Temp)
      if (global_best_Temp[ind]<all_global_best):
          rows[0] = save_rows[ind,0]
          rows[1] = save_rows[ind,1]
          rows[2] = save_rows[ind,2]
          all_global_best=global_best_Temp[ind]

  return all_global_best, rows

Performance of multi-threaded version

n=100
your_version: 1.56s
compiled_version: 0.0078s (200x speedup)

n=150
your_version: 5.41s
compiled_version: 0.0282s (191x speedup)

n=500
your_version: 283s
compiled_version: 2.95s (96x speedup)

Edit

In a newer Numba Version (installed through the Anaconda Python Distribution) I have to manually install tbb to get a working parallelization.

回答2:

This solution is 5x faster for n=100:

coms = np.fromiter(itertools.combinations(np.arange(n), 3), 'i,i,i').view(('i', 3))
best = A[coms].min(1).max(1)
at = best.argmin()
global_best = best[at]
save_rows = coms[at]

The first line is a bit convoluted but turns the result of itertools.combinations into a NumPy array which contains all possible [i,j,k] index combinations.

From there, it's a simple matter of indexing into A using all the possible index combinations, then reducing along the appropriate axes.

This solution consumes a lot more memory as it builds the concrete array of all possible combinations A[coms]. It saves time for smallish n, say under 250, but for large n the memory traffic will be very high and it may be slower than the original code.

回答3:

Working by chunks allows to combine the speed of vectorized calculus while avoiding to run into Memory Errors. Below there is an example of converting the nested loops to vectorization by chunks.

Starting from the same variables as the question, a chunk length is defined, in order to vectorize calculations inside the chunk and loop only over chunks instead of over combinations.

chunk = 2000 # define chunk length, if to small, the code won't take advantage 
             # of vectorization, if it is too large, excessive memory usage will 
             # slow down execution, or Memory Error will be risen 
combinations = itertools.combinations(range(n),3) # generate iterator containing 
                                        # all possible combinations of 3 columns
N = n*(n-1)*(n-2)//6 # number of combinations (length of combinations cannot be 
                     # retrieved because it is an iterator)
# generate a list containing how many elements of combinations will be retrieved 
# per iteration
n_chunks, remainder = divmod(N,chunk)
counts_list = [chunk for _ in range(n_chunks)]
if remainder:
    counts_list.append(remainder)

# Iterate one chunk at a time, using vectorized code to treat the chunk
for counts in counts_list:
    # retrieve combinations in current chunk
    current_comb = np.fromiter(combinations,dtype='i,i,i',count=counts)\
                     .view(('i',3)) 
    # maximum of element-wise minimum in current chunk
    chunk_best = np.minimum(np.minimum(A[current_comb[:,0],:],A[current_comb[:,1],:]),
                            A[current_comb[:,2],:]).max(axis=1) 
    ravel_save_row = chunk_best.argmin() # minimum of maximums in current chunk
    # check if current chunk contains global minimum
    if chunk_best[ravel_save_row] < global_best: 
        global_best = chunk_best[ravel_save_row]
        save_rows = current_comb[ravel_save_row]
print(global_best,save_rows)

I ran some performance comparisons with the nested loops, obtaining the following results (chunk_length = 1000):

• n=100
  • Nested loops: 1.13 s ± 16.6 ms
  • Work by chunks: 108 ms ± 565 µs
• n=150
  • Nested loops: 4.16 s ± 39.3 ms
  • Work by chunks: 523 ms ± 4.75 ms
• n=500
  • Nested loops: 3min 18s ± 3.21 s
  • Work by chunks: 1min 12s ± 1.6 s

Note

After profiling the code, I found that the np.min was what took longest by calling np.maximum.reduce. I converted it directly to np.maximum which improved performance a bit.

回答4:

You can use combinations from itertools, that it's a python standard library, and it will help you to to remove all those nested loops.

from itertools import combinations
import numpy as np

n = 100
np.random.seed(2)
A = np.random.rand(n,n)
global_best = 1000000000000000.0

for i, j, k in combinations(range(n), 3):
    local_best = np.amax(np.array([A[i,:], A[j,:], A[k,:]]).min(0))
    if local_best < global_best:
        global_best = local_best
        save_rows = [i, j, k]

print global_best, save_rows