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.

Solution 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)

Solution 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.

Solution 3

A couple of answers suggest

import struct
buf = struct.pack(f'{len(floatlist)}f', *floatlist)

but the use of '*' needlessly converts floatlist to a tuple before passing it to struct.pack. It's faster to avoid that, by first creating an empty buffer, and then populating it using slice assignment:

import ctypes
buf = (ctypes.c_double * len(floatlist))()
buf[:] = floatlist

Other performance savings some people might be able to use:

• You can reuse an existing buffer by just doing the assignment again, without having to create a new buffer.
• You can modify parts of an existing buffer by assigning to the appropriate slice.

Solution 4

As with strings, using .join() will be faster than continually concatenating. Eg:

import struct
b = bytes()
floatList = [5.4, 3.5, 7.3, 6.8, 4.6]
b = b.join((struct.pack('f', val) for val in floatList))

Results in:

b'\xcd\xcc\xac@\x00\x00`@\x9a\x99\xe9@\x9a\x99\xd9@33\x93@'

Solution 5

That should work:

return struct.pack('f' * len(floatList), *floatList)

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