I'm trying to optimize the performance of my python program and I think I have identified this piece of code as bottleneck:

  for i in range(len(green_list)):
    rgb_list = []

    for j in range(len(green_list[i])):
      rgb_list.append('xxx' % (red_list[i][j], green_list[i][j], blue_list[i][j]))

    write_file(str(i), rgb_list)

Where red_list, green_list and blue_list are numpy arrays with values like this:

red_list = [[1, 2, 3, 4, 5], [51, 52, 53, 54, 55]]

green_list = [[6, 7, 8, 9, 10], [56, 57, 58, 59, 60]]

blue_list = [[11, 12, 13, 14, 15], [61, 62, 63, 64, 65]]

At the end of each execution of the inner-for rgb_list is containing the hex values:

rgb_list = ['01060b', '02070c', '03080d', '04090e', '050a01']

Now, it is not clear to me how to exploit the potential of numpy arrays but I think there is a way to optimize those two nested loops. Any suggestions?

CodePudding user response：

I assume the essential traits of your code could be summarized in the following generator:

import numpy as np


def as_str_OP(r_arr, g_arr, b_arr):
    n, m = r_arr.shape
    rgbs = []
    for i in range(n):
        rgb = []
        for j in range(m):
            rgb.append('xxx' % (r_arr[i, j], g_arr[i, j], b_arr[i, j]))
        yield rgb

which can be consumed with a for loop, for example to write to disk:

for x in as_str_OP(r_arr, g_arr, b_arr):
    write_to_disk(x)

The generator itself can be written in a Numba-friendly way, resulting in massive speed up, especially as the size of the input grows (and particularly the second dimension):

import numba as nb


@nb.njit
def hex_to_ascii(x):
    ascii_num_offset = 48  # ord(b'0') == 48
    ascii_alp_offset = 87  # ord(b'a') == 97, (num of non-alpha digits) == 10
    return x   (ascii_num_offset if x < 10 else ascii_alp_offset)


@nb.njit
def _to_hex_2d(x):
    base = 16
    a, b = divmod(x, 16)
    return hex_to_ascii(a), hex_to_ascii(b)


@nb.njit
def _as_str_nb(r_arr, g_arr, b_arr):
    l = 3
    n, m = r_arr.shape
    for i in range(n):
        rgb = np.empty((m, 2 * l), dtype=np.uint32)
        for j in range(m):
            rgb[j, 0:2] = _to_hex_2d(r_arr[i, j])
            rgb[j, 2:4] = _to_hex_2d(g_arr[i, j])
            rgb[j, 4:6] = _to_hex_2d(b_arr[i, j])
        yield rgb


def as_str_nb(r_arr, g_arr, b_arr):
    l = 3
    n, m = r_arr.shape
    for x in _as_str_nb(r_arr, g_arr, b_arr):
        yield x.view(f'<U{2 * l}').reshape(m).tolist()

This essentially involves manually writing the number, correctly converted to hexadecimal ASCII chars, into a properly typed array, which can then be converted to give the desired output.

Note that the final numpy.ndarray.tolist() could be avoided if whatever will consume the generator is capable of dealing with the NumPy array itself, thus saving some potentially large and definitely appreciable time.

def as_str_nba(r_arr, g_arr, b_arr):
    l = 3
    n, m = r_arr.shape
    for x in _as_str_nb(r_arr, g_arr, b_arr):
        yield x.view(f'<U{2 * l}').reshape(m)

---

Overcoming IO-bound bottleneck

However, if you are IO-bounded you should modify your code to write in blocks, e.g using the grouper recipe from itertools recipes:

from itertools import zip_longest


def grouper(iterable, n, *, incomplete='fill', fillvalue=None):
    "Collect data into non-overlapping fixed-length chunks or blocks"
    # grouper('ABCDEFG', 3, fillvalue='x') --> ABC DEF Gxx
    # grouper('ABCDEFG', 3, incomplete='strict') --> ABC DEF ValueError
    # grouper('ABCDEFG', 3, incomplete='ignore') --> ABC DEF
    args = [iter(iterable)] * n
    if incomplete == 'fill':
        return zip_longest(*args, fillvalue=fillvalue)
    if incomplete == 'strict':
        return zip(*args, strict=True)
    if incomplete == 'ignore':
        return zip(*args)
    else:
        raise ValueError('Expected fill, strict, or ignore')

to be used like:

group_size = 3
for x in grouper(as_str_OP(r_arr, g_arr, b_arr), group_size):
    write_many_to_disk(x)

---

Testing out the output

Some dummy input can be produced easily (r_arr is essentially red_list, etc.):

def gen_color(n, m):
    return np.random.randint(0, 2 ** 8, (n, m))


N, M = 10, 3
r_arr = gen_color(N, M)
g_arr = gen_color(N, M)
b_arr = gen_color(N, M)

and tested by consuming the generator to produce a list:

res_OP = list(as_str_OP(r_arr, g_arr, b_arr))
res_nb = list(as_str_nb(r_arr, g_arr, b_arr))
res_nba = list(as_str_nba(r_arr, g_arr, b_arr))
print(np.array(res_OP))
# [['1f6984' '916d98' 'f9d779']
#  ['65f895' 'ded23e' '332fdc']
#  ['b9e059' 'ce8676' 'cb75e9']
#  ['bca0fc' '3289a9' 'cc3d3a']
#  ['6bb0be' '07134a' 'c3cf05']
#  ['152d5c' 'bac081' 'c59a08']
#  ['97efcc' '4c31c0' '957693']
#  ['15247e' 'af8f0a' 'ffb89a']
#  ['161333' '8f41ce' '187b01']
#  ['d811ae' '730b17' 'd2e269']]
print(res_OP == res_nb)
# True
print(res_OP == [x.tolist() for x in res_nba])
# True

eventually passing through some grouping:

k = 3
res_OP = list(grouper(as_str_OP(r_arr, g_arr, b_arr), k))
res_nb = list(grouper(as_str_nb(r_arr, g_arr, b_arr), k))
res_nba = list(grouper(as_str_nba(r_arr, g_arr, b_arr), k))
print(np.array(res_OP))
# [[list(['1f6984', '916d98', 'f9d779'])
#   list(['65f895', 'ded23e', '332fdc'])
#   list(['b9e059', 'ce8676', 'cb75e9'])]
#  [list(['bca0fc', '3289a9', 'cc3d3a'])
#   list(['6bb0be', '07134a', 'c3cf05'])
#   list(['152d5c', 'bac081', 'c59a08'])]
#  [list(['97efcc', '4c31c0', '957693'])
#   list(['15247e', 'af8f0a', 'ffb89a'])
#   list(['161333', '8f41ce', '187b01'])]
#  [list(['d811ae', '730b17', 'd2e269']) None None]]
print(res_OP == res_nb)
# True
print(res_OP == [tuple(y.tolist() if y is not None else y for y in x) for x in res_nba])
# True

---

Benchmarks

To give you some ideas of the numbers we could be talking, let us use %timeit on much larger inputs:

N, M = 1000, 1000
r_arr = gen_color(N, M)
g_arr = gen_color(N, M)
b_arr = gen_color(N, M)


%timeit -n 1 -r 1 list(as_str_OP(r_arr, g_arr, b_arr))
# 1 loop, best of 1: 1.1 s per loop
%timeit -n 4 -r 4 list(as_str_nb(r_arr, g_arr, b_arr))
# 1 loop, best of 1: 96.5 ms per loop
%timeit -n 4 -r 4 list(as_str_nba(r_arr, g_arr, b_arr))
# 4 loops, best of 4: 10.4 ms per loop

To simulate disk writing we could use the following consumer:

import time
import math


def consumer(gen, timeout_sec=0.001, weight=1):
    result = []
    for x in gen:
        result.append(x)
        time.sleep(timeout_sec * weight)
    return result

where disk writing is simulated with a time.sleep() call with a timeout depending on the logarithm of the object size:

N, M = 1000, 1000
r_arr = gen_color(N, M)
g_arr = gen_color(N, M)
b_arr = gen_color(N, M)


%timeit -n 1 -r 1 consumer(as_str_OP(r_arr, g_arr, b_arr), weight=math.log2(2))
# 1 loop, best of 1: 2.37 s per loop
%timeit -n 1 -r 1 consumer(as_str_nb(r_arr, g_arr, b_arr), weight=math.log2(2))
# 1 loop, best of 1: 1.27 s per loop
%timeit -n 1 -r 1 consumer(as_str_nba(r_arr, g_arr, b_arr), weight=math.log2(2))
# 1 loop, best of 1: 1.13 s per loop

k = 100
%timeit -n 1 -r 1 consumer(grouper(as_str_OP(r_arr, g_arr, b_arr), k), weight=math.log2(1   k))
# 1 loop, best of 1: 1.17 s per loop
%timeit -n 1 -r 1 consumer(grouper(as_str_nb(r_arr, g_arr, b_arr), k), weight=math.log2(1   k))
# 1 loop, best of 1: 173 ms per loop
%timeit -n 1 -r 1 consumer(grouper(as_str_nba(r_arr, g_arr, b_arr), k), weight=math.log2(1   k))
# 1 loop, best of 1: 87.4 ms per loop

Aside of the disk-writing, which is there just to give some idea, the Numba-accelerated approach gets approx. 100x faster when skipping the potentially useless conversion to list() with numpy.ndarray.tolist(). Otherwise, the acceleration factor is still a respectable 10x.

CodePudding user response：

you could use reduce for the inner loop, making it possible for your computer to divide the computations between different threads behind the scenes

for i in range(len(green_list)):
    rgb_list = reduce(lambda ls, j: ls   ['xxx' % (red_list[i][j], green_list[i][j], blue_list[i][j])],range(len(green_list[i])),list())
    print(rgb_list)

or you could try to achive the same goal with a one-liner,

for i in range(len(green_list)):
    rgb_list = ['xxx' % (red_list[i][j], green_list[i][j], blue_list[i][j]) for j in range(len(green_list[i]))]
    print(rgb_list)

hope it will do the trick for you

CodePudding user response：

In the code you show, the slow bit is the string formatting. That we can improve somewhat.

A hex colour consists of eight bits for the red field, eight for the green, and eight for the blue (since your data does not seem to have an alpha channel, I am going to ignore that option). So we need at least twenty four bits to store the rgb colours.

You can create hex values using numpy's bitwise operators. The advantage is that this is completely vectorised. You then only have one value to format into a hex string for each (i, j), instead of three:

for i in range(len(green_list)):
    hx = red_list[i] << 16 | green_list[i] << 8 | blue_list[i]
    hex_list = ['x' % val for val in hx]

When the numpy arrays have dimensions (10, 1_000_000), this is about 5.5x faster than your original method (on my machine).

CodePudding user response：

1. for-loop

Code modifications for rgb_list.append() does not affect much to the performance.

import timeit

n = 1000000
red_list   = [list(range(1, n 0)), list(range(1, n 2))]
green_list = [list(range(2, n 1)), list(range(2, n 3))]
blue_list  = [list(range(3, n 2)), list(range(3, n 4))]

def test_1():
    for i in range(len(green_list)):
        rgb_list = ['xxx' % (red_list[i][j], green_list[i][j], blue_list[i][j]) for j in range(len(green_list[i]))]

def test_2():
    for i in range(len(green_list)):
        rgb_list = [None] * len(green_list[i])
        
        for j in range(len(green_list[i])):
            rgb_list[j] = 'xxx' % (red_list[i][j], green_list[i][j], blue_list[i][j])

def test_3():
    for i in range(len(green_list)):
        rgb_list = []
        
        for j in range(len(green_list[i])):
            rgb_list.append('xxx' % (red_list[i][j], green_list[i][j], blue_list[i][j]))

%timeit -n 1 -r 7 test_1(): 1.31 s ± 8.14 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%timeit -n 1 -r 7 test_2(): 1.33 s ± 11.5 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%timeit -n 1 -r 7 test_3(): 1.39 s ± 10.1 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

2. disk IO

Code modifications for disk IO also does not affect much to the performance.

n = 20000000

def test_write_each():
    for i in range(len(green_list)):
        rgb_list = ['xxx' % (red_list[i][j], green_list[i][j], blue_list[i][j]) for j in range(len(green_list[i]))]
        
        with open("test_%d" % i, "wb") as f:
            pickle.dump(rgb_list, f)

def test_write_once():
    
    rgb_list_list = [None] * len(green_list)
    
    for i in range(len(green_list)):
        rgb_list_list[i] = ['xxx' % (red_list[i][j], green_list[i][j], blue_list[i][j]) for j in range(len(green_list[i]))] 
        
    with open("test_all", "wb") as f:
        pickle.dump(rgb_list_list, f)

%timeit -n 1 -r 3 test_write_each(): 35.2 s ± 74.6 ms per loop (mean ± std. dev. of 3 runs, 1 loop each)

%timeit -n 1 -r 3 test_write_once(): 35.4 s ± 54.4 ms per loop (mean ± std. dev. of 3 runs, 1 loop each)

Conclusion

From the benchmark result, there seems like no bottleneck to be avoided in the question code.

If the disk IO itself is the problem, I would like to suggest to run the disk-IO code only once after every other job (including the ones that are not mentioned in this question) is finished.