Fastest way to pack a list of floats into bytes in python

Question

I have a list of say 100k floats and I want to convert it into a bytes buffer.

buf = bytes() for val in floatList:    buf += struct.pack('f', val) return buf 

This is quite slow. How can I make it faster using only standard Python 3.x libraries.

Answer 1

Just tell struct how many floats you have. 100k floats takes about a 1/100th of a second on my slow laptop.

import random import struct  floatlist = [random.random() for _ in range(10**5)] buf = struct.pack('%sf' % len(floatlist), *floatlist) 

Answer 2

You can use ctypes, and have a double-array (or float array) exactly as you'd have in C , instead of keeping your data in a list. This is fair low level, but is a recommendation if you need great performance and if your list is of a fixed size.

You can create the equivalent of a C double array[100]; in Python by doing:

array = (ctypes.c_double * 100)() 

The ctypes.c_double * 100 expression yields a Python class for an array of doubles, 100 items long. To wire it to a file, you can just use buffer to get its contents:

>>> f = open("bla.dat", "wb") >>> f.write(buffer(array)) 

If your data is already in a Python list, packing it into a double array may or may not be faster than calling structas in Agf's accepted answer - I will leave measuring which is faster as homework, but all the code you need is this:

>>> import ctypes >>> array = (ctypes.c_double * len(floatlist))(*floatlist) 

To see it as a string, just do: str(buffer(array)) - the one drawback here is that you have to take care of float size (float vs double) and CPU dependent float type - the struct module can take care of this for you.

The big win is that with a float array you can still use the elements as numbers, by accessing then just as if it where a plain Python list, while having then readily available as a planar memory region with buffer.