How to fuse multiple Python functions together

user_9577975 0
Viewed 150

I have a number of operations I want to "fuse" together. Let's say there are 3 possible operations:

sq = lambda x: x**2
add = lambda x: x+3
mul = lambda x: x*5

I also have an array of operations:

ops = [add, sq, mul, sq]

I can then create a function from these operations:

def generateF(ops):
    def inner(x):
        for op in ops:
            x = op(x)
        return x
    return inner
f = generateF(ops)
f(3) # returns 32400

fastF = lambda x: (5*(x+3)**2)**2

f and fastF does the same thing, but fastF is around 1.7-2 times faster than f on my benchmark, which makes sense. My question is, how can I write generateF function that returns a function that is as fast as fastF? The operations are restricted to basic operations like __add__, __mul__, __matmul__, __rrshift__, etc (essentially most numeric operations). generateF can take as long as you'd like, because it will be done before reaching hot code.

The context is that this is a part of my library, so I can define every legal operation, and thus know exactly what they are. The operation definitions are not given to us by the end user randomly (the user can only pick the order of the operations), so we can utilize every outside knowledge about them.

This might seem like premature optimization, but it is not, as f is hot code. Dropping to C is not an option, as the operations can be complex (think, PyTorch tensor multiply), and x can be of any type. Currently, I'm thinking about modifying python's bytecode, but that is very unpleasant, as bytecode specifications changes for every Python version, so I wanted to ask here first before diving into that solution.

python
3 Answers

Here is a very hacky version of synthesizing a new function from the bytecode of the given functions. The basic technique is to keep the LOAD_FAST opcode only at the beginning of the first function, and strip off the RETURN_VALUE opcode except at the end of the last function. This leaves the value being manipulated on the stack in between (what were originally) your functions. When you're done, you don't have any function calls.

import dis, inspect

sq = lambda x: x**2
add = lambda x: x+3
mul = lambda x: x*5

ops = [add, sq, mul, sq]

def synthF(ops):
    bytecode = bytearray()
    constants = []
    stacksize = 0
    for i, op in enumerate(ops):
        code = op.__code__
        # works only with functions having one argument and no other vars
        assert code.co_argcount == code.co_nlocals == 1
        assert not code.co_freevars
        stacksize = max(stacksize, code.co_stacksize)
        opcodes = bytearray(code.co_code)
        # starts with LOAD_FAST argument 0 (i.e. we're doing something with our arg)
        assert opcodes[0] == dis.opmap["LOAD_FAST"] and opcodes[1] == 0
        # ends with RETURN_VALUE
        assert opcodes[-2] == dis.opmap["RETURN_VALUE"] and opcodes[-1] == 0
        if bytecode:        # if this isn't our first function, our variable is already on the stock
            opcodes = opcodes[2:]
        # adjust LOAD_CONSTANT opcodes. each function can have constants,
        # but their indexes start at 0 in each function.  since we're
        # putting these into a single function we need to accumulate the
        # constants used in each function and adjust the indexes used in
        # the function's bytecode to access the value by its index in the
        # accumulated list.
        offset = 0
        if bytecode:
            while True:
                none = code.co_consts[0] is None
                offset = opcodes.find(dis.opmap["LOAD_CONST"], offset)
                if offset < 0:
                    break
                if not offset % 2 and (not none or opcodes[offset+1]):
                    opcodes[offset+1] += len(constants) - none
                offset += 2
            # first constant is always None. don't include multiple copies
            # (to be safe, we actually check that)
            constants.extend(code.co_consts[none:])
        else:
            assert code.co_consts[0] is None
            constants.extend(code.co_consts)
        # add our adjusted bytecode, cutting off the RETURN_VALUE opcode
        bytecode.extend(opcodes[:-2])
    bytecode.extend([dis.opmap["RETURN_VALUE"], 0])

    func = type(ops[0])(type(code)(1, 1, 0, 1, stacksize, inspect.CO_OPTIMIZED, bytes(bytecode),
                tuple(constants), (), ("x",), "<generated>", "<generated>", 0, b''),
                globals())

    return func

f = synthF(ops)
assert f(3) == 32400

Gross, and lots of caveats (called out in comments) but it works, and should be about as fast as your expression, since it compiles to virtually the same bytecode. It would need a bit of work to support concatenating more complex functions.

user_4164670

Here's an alternative using chaining. This way, there's only function calls in your generated function calls, no iteration.

def makeF(ops):
    f = ops[0]
    for op in ops[1:]:
        f = lambda x, op=op, f=f: op(f(x))
    return f

Bad news: it replaces each function call with two, so it's actually slower than your iterative version. :/

user_4164670

As there seems to be no solution, this is what I've settled on. Knowing that most operations will be short (<4 operations total), I just hard code in to get rid of the for loop.

def generateF(ops):
    l = len(ops)
    if l == 1:
        return ops[0]
    if l == 2:
        a, b = ops
        return lambda x: b(a(x))
    if l == 3:
        a, b, c = ops
        return lambda x: c(b(a(x)))
    if l == 4:
        a, b, c, d = ops
        return lambda x: d(c(b(a(x))))
    def inner(x):
        for op in ops:
            x = op(x)
        return x
    return inner
fastF = generateF(ops)

This is only 1.4x slower than fastF (originally 1.7-2x slower). If you have any other ideas, I will consider it.

user_95779750
Related