问题
TLDR; Of the various compression algorithms available in python gzip, bz2, lzma, etc, which has the best decompression performance?
Full discussion:
Python 3 has various modules for compressing/decompressing data
including gzip, bz2 and lzma. gzip and bz2 additionally have different compression levels you can set.
If my goal is to balance file size (/compression ratio) and decompression speed (compression speed is not a concern), which is going to be the best choice? Decompression speed is more important than file size, but as the uncompressed files in question would be around 600-800MB each (32-bit RGB .png image files), and I have a dozen of them, I do want some compression.
My use case is that I am loading a dozen images from disk, doing some processing on them (as a numpy array) and then using the processed array data in my program.
- The images never change, I just have to load them each time I run my program.
- The processing takes about the same length of time as the loading (several seconds), so I'm trying to save some loading time by saving the processed data (using
pickle) rather than loading the raw, unprocessed, images every time. Initial tests were promising - loading the raw/uncompressed pickled data took less than a second, vs 3 or 4 seconds to load and process the original image - but as mentioned resulted in file sizes of around 600-800MB, while the original png images were only around 5MB. So I'm hoping I can strike a balance between loading time and file size by storing the picked data in a compressed format.
UPDATE: The situation is actually a bit more complicated than I represented above. My application uses
PySide2, so I have access to theQtlibraries.- If I read the images and convert to a numpy array using
pillow(PIL.Image), I actually don't have to do any processing, but the total time to read the image into the array is around 4 seconds. - If instead I use
QImageto read the image, I then have to do some processing on the result to make it usable for the rest of my program due to the endian-ness of howQImageloads the data - basically I have to swap the bit order and then rotate each "pixel" so that the alpha channel (which is apparently added by QImage) comes last rather than first. This whole process takes about 3.8 seconds, so marginally faster than just using PIL. - If I save the
numpyarray uncompressed, then I can load them back in in .8 seconds, so by far the fastest, but with large file size.
- If I read the images and convert to a numpy array using
┌────────────┬────────────────────────┬───────────────┬─────────────┐
│ Python Ver │ Library/Method │ Read/unpack + │ Compression │
│ │ │ Decompress (s)│ Ratio │
├────────────┼────────────────────────┼───────────────┼─────────────┤
│ 3.7.2 │ pillow (PIL.Image) │ 4.0 │ ~0.006 │
│ 3.7.2 │ Qt (QImage) │ 3.8 │ ~0.006 │
│ 3.7.2 │ numpy (uncompressed) │ 0.8 │ 1.0 │
│ 3.7.2 │ gzip (compresslevel=9) │ ? │ ? │
│ 3.7.2 │ gzip (compresslevel=?) │ ? │ ? │
│ 3.7.2 │ bz2 (compresslevel=9) │ ? │ ? │
│ 3.7.2 │ bz2 (compresslevel=?) │ ? │ ? │
│ 3.7.2 │ lzma │ ? │ ? │
├────────────┼────────────────────────┼───────────────┼─────────────┤
│ 3.7.3 │ ? │ ? │ ? │
├────────────┼────────────────────────┼───────────────┼─────────────┤
│ 3.8beta1 │ ? │ ? │ ? │
├────────────┼────────────────────────┼───────────────┼─────────────┤
│ 3.8.0final │ ? │ ? │ ? │
├────────────┼────────────────────────┼───────────────┼─────────────┤
│ 3.5.7 │ ? │ ? │ ? │
├────────────┼────────────────────────┼───────────────┼─────────────┤
│ 3.6.10 │ ? │ ? │ ? │
└────────────┴────────────────────────┴───────────────┴─────────────┘
Sample .png image: As an example, take this 5.0Mb png image, a fairly high resolution image of the coastline of Alaska.
Code for the png/PIL case (load into a numpy array):
from PIL import Image
import time
import numpy
start = time.time()
FILE = '/path/to/file/AlaskaCoast.png'
Image.MAX_IMAGE_PIXELS = None
img = Image.open(FILE)
arr = numpy.array(img)
print("Loaded in", time.time()-start)
this load takes around 4.2s on my machine with Python 3.7.2.
Alternatively, I can instead load the uncompressed pickle file generated by picking the array created above.
Code for the uncompressed pickle load case:
import pickle
import time
start = time.time()
with open('/tmp/test_file.pickle','rb') as picklefile:
arr = pickle.load(picklefile)
print("Loaded in", time.time()-start)
Loading from this uncompressed pickle file takes ~0.8s on my machine.
回答1:
You can use Python-blosc
It is very fast and for small arrays (<2GB) also quite easy to use. On easily compressable data like your example, it is often faster to compress the data for IO operations. (SATA-SSD: about 500 MB/s, PCIe- SSD: up to 3500MB/s) In the decompression step the array allocation is the most costly part. If your images are of similar shape you can avoid repeated memory allocation.
Example
A contigous array is assumed for the following example.
import blosc
import pickle
def compress(arr,Path):
#c = blosc.compress_ptr(arr.__array_interface__['data'][0], arr.size, arr.dtype.itemsize, clevel=3,cname='lz4',shuffle=blosc.SHUFFLE)
c = blosc.compress_ptr(arr.__array_interface__['data'][0], arr.size, arr.dtype.itemsize, clevel=3,cname='zstd',shuffle=blosc.SHUFFLE)
f=open(Path,"wb")
pickle.dump((arr.shape, arr.dtype),f)
f.write(c)
f.close()
return c,arr.shape, arr.dtype
def decompress(Path):
f=open(Path,"rb")
shape,dtype=pickle.load(f)
c=f.read()
#array allocation takes most of the time
arr=np.empty(shape,dtype)
blosc.decompress_ptr(c, arr.__array_interface__['data'][0])
return arr
#Pass a preallocated array if you have many similar images
def decompress_pre(Path,arr):
f=open(Path,"rb")
shape,dtype=pickle.load(f)
c=f.read()
#array allocation takes most of the time
blosc.decompress_ptr(c, arr.__array_interface__['data'][0])
return arr
Benchmarks
#blosc.SHUFFLE, cname='zstd' -> 4728KB,
%timeit compress(arr,"Test.dat")
1.03 s ± 12.5 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
#611 MB/s
%timeit decompress("Test.dat")
146 ms ± 481 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
#4310 MB/s
%timeit decompress_pre("Test.dat",arr)
50.9 ms ± 438 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
#12362 MB/s
#blosc.SHUFFLE, cname='lz4' -> 9118KB,
%timeit compress(arr,"Test.dat")
32.1 ms ± 437 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
#19602 MB/s
%timeit decompress("Test.dat")
146 ms ± 332 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
#4310 MB/s
%timeit decompress_pre("Test.dat",arr)
53.6 ms ± 82.9 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
#11740 MB/s
Timings
回答2:
The low-hanging fruit
numpy.savez_compressed('AlaskaCoast.npz', arr)
arr = numpy.load('AlaskaCoast.npz')['arr_0']
Loading is 2.3x faster than your PIL-based code.
It uses zipfile.ZIP_DEFLATED, see savez_compressed docu.
Your PIL code also has an unneeded copy: array(img) should be asarray(img). It only costs 5% of the slow loading time. But after optimization this will be significant and you have to keep in mind which numpy operators create a copy.
Fast decompression
According to the zstd benchmarks, when optimizing for decompression lz4 is a good choice. Just plugging this into pickle gives another 2.4x gain and is only 30% slower than uncompressed pickling.
import pickle
import lz4.frame
# with lz4.frame.open('AlaskaCoast.lz4', 'wb') as f:
# pickle.dump(arr, f)
with lz4.frame.open('AlaskaCoast.lz4', 'rb') as f:
arr = pickle.load(f)
Benchmarks
method size load time
------ ---- ---------
original (PNG+PIL) 5.1M 7.1
np.load (compressed) 6.7M 3.1
pickle + lz4 7.1M 1.3
pickle (uncompressed) 601M 1.0 (baseline)
The load time was measured inside Python (3.7.3), using the minimum wall-clock time over 20 runs on my desktop. According to occasional glances at top it always seemed to be running on a single core.
For the curious: profiling
I'm not sure if the Python version matters, most work is supposed to happen inside of C libraries. To validate this I've profiled the pickle + lz4 variant:
perf record ./test.py && perf report -s dso
Overhead Shared Object
60.16% [kernel.kallsyms] # mostly page_fault and alloc_pages_vma
27.53% libc-2.28.so # mainly memmove
9.75% liblz4.so.1.8.3 # only LZ4_decompress_*
2.33% python3.7
...
Most time is spent inside of the Linux kernel, doing page_fault and stuff associated with (re-)allocating memory, probably including disk I/O. The high amount of memmove looks suspicious. Probably Python is re-allocating (resizing) the final array every time a new decompressed chunk arrives. If anyone likes to have a closer look: python and perf profiles.
回答3:
You can continue to use your existing PNGs and enjoy the space saving, but gain some speed by using libvips. Here is a comparison, but rather than test the speed of my laptop versus yours, I have shown 3 different methods so you can see the relative speed. I used:
- PIL
- OpenCV
- pyvips
#!/usr/bin/env python3
import numpy as np
import pyvips
import cv2
from PIL import Image
def usingPIL(f):
im = Image.open(f)
return np.asarray(im)
def usingOpenCV(f):
arr = cv2.imread(f,cv2.IMREAD_UNCHANGED)
return arr
def usingVIPS(f):
image = pyvips.Image.new_from_file(f)
mem_img = image.write_to_memory()
imgnp=np.frombuffer(mem_img, dtype=np.uint8).reshape(image.height, image.width, 3)
return imgnp
Then I checked the performance in IPython because it has nice timing functions. As you can see, pyvips is 13 times faster than PIL even with PIL 2x faster than the original version because of avoiding array copy:
In [49]: %timeit usingPIL('Alaska1.png')
3.66 s ± 31.8 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
In [50]: %timeit usingOpenCV('Alaska1.png')
6.82 s ± 23.3 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
In [51]: %timeit usingVIPS('Alaska1.png')
276 ms ± 4.24 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
# Quick test results match
np.sum(usingVIPS('Alaska1.png') - usingPIL('Alaska1.png'))
0
回答4:
Something I think should be fast is
- use gzip (or other) for compression
- directly store the compressed data in a python module as literal bytes
- load decompressed form into numpy array directly
i.e. write a program that generates a source code like
import gzip, numpy
data = b'\x00\x01\x02\x03'
unpacked = numpy.frombuffer(gzip.uncompress(data), numpy.uint8)
the packed data ends up encoded directly into the .pyc file
For low-entropy data gzip decompression should be quite fast (edit: not really surprisingly lzma is even faster, and it's still a predefined python module)
With your "alaska" data this approach gives the following performance on my machine
compression source module size bytecode size import time
----------- ------------------ ------------- -----------
gzip -9 26,133,461 9,458,176 1.79
lzma 11,534,009 2,883,695 1.08
You can even distribute just the .pyc provided you can control the python version used; the code to load a .pyc in Python 2 was a one liner but is now more convoluted (apparently it was decided that loading .pyc isn't supposed to be convenient).
Note that the compilation of the module is reasonably fast (e.g. the lzma version compiles on my machine in just 0.1 seconds) but it's a pity to waste on disk 11Mb more for no real reason.
来源：https://stackoverflow.com/questions/56708673/python-decompression-relative-performance