python - python中多个函数的有序reduce

Question

有序列表缩减

我需要减少一些列表，根据元素类型，二元运算的速度和实现会有所不同，即通过首先减少一些具有特定功能的对可以大大降低速度。例如 foo(a[0], bar(a[1], a[2]))可能比 bar(foo(a[0], a[1]), a[2]) 慢很多，但在这种情况下给出相同的结果。

我已经有了以元组 (pair_index, binary_function) 列表形式生成最佳排序的代码。我正在努力实现一个有效的函数来执行缩减，理想情况下是返回一个新的部分函数，​​然后可以在相同类型排序但值不同的列表上重复使用该函数。

简单而缓慢(？)的解决方案

这是我天真的解决方案，涉及 for 循环、删除元素和关闭 (pair_index, binary_function) 列表以返回“预计算”函数。

def ordered_reduce(a, pair_indexes, binary_functions, precompute=False):
    """
    a: list to reduce, length n
    pair_indexes: order of pairs to reduce, length (n-1)
    binary_functions: functions to use for each reduction, length (n-1)
    """
    def ord_red_func(x):
        y = list(x)  # copy so as not to eat up
        for p, f in zip(pair_indexes, binary_functions):
            b = f(y[p], y[p+1])
            # Replace pair
            del y[p]
            y[p] = b
        return y[0]

    return ord_red_func if precompute else ord_red_func(a)

>>> foos = (lambda a, b: a - b, lambda a, b: a + b, lambda a, b: a * b)
>>> ordered_reduce([1, 2, 3, 4], (2, 1, 0), foos)
1
>>> 1 * (2 + (3-4))
1

以及预计算的工作原理:

>>> foo = ordered_reduce(None, (0, 1, 0), foos)
>>> foo([1, 2, 3, 4])
-7
>>> (1 - 2) * (3 + 4)
-7

然而，它涉及复制整个列表，而且(因此？)速度也很慢。有没有更好/标准的方法来做到这一点？

(编辑:)一些时间:

from operators import add
from functools import reduce
from itertools import repeat
from random import random

r = 100000
xs = [random() for _ in range(r)]
# slightly trivial choices of pairs and functions, to replicate reduce
ps = [0]*(r-1)
fs = repeat(add)
foo = ordered_reduce(None, ps, fs, precompute=True)

>>> %timeit reduce(add, xs)
100 loops, best of 3: 3.59 ms per loop
>>> %timeit foo(xs)
1 loop, best of 3: 1.44 s per loop

这是一种最坏的情况，并且由于 reduce 不采用可迭代的函数而略有作弊，但采用(但无顺序)的函数仍然非常快:

def multi_reduce(fs, xs):
    xs = iter(xs)
    x = next(xs)
    for f, nx in zip(fs, xs):
        x = f(x, nx)
    return x

>>> %timeit multi_reduce(fs, xs)
100 loops, best of 3: 8.71 ms per loop

(EDIT2):为了好玩，大量作弊的“编译”版本的性能，这让我们对发生的总开销有了一些了解。

from numba import jit

@jit(nopython=True)
def numba_sum(xs):
    y = 0
    for x in xs:
        y += x
    return y

>>> %timeit numba_sum(xs)
1000 loops, best of 3: 1.46 ms per loop

Answer 1

看到这个问题，我立马想到了reverse Polish notation (RPN)。虽然这可能不是最好的方法，但在这种情况下它仍然可以显着加快速度。

我的第二个想法是，如果您只是适本地重新排序序列 xs 以摆脱 del y[p]，您可能会得到相同的结果。 (可以说，如果整个 reduce 过程是用 C 编写的，则可以实现最佳性能。但这是另一回事。)

逆波兰表示法

如果您不熟悉 RPN，请阅读维基百科文章中的简短说明。基本上，所有的操作都可以不带括号写下来，比如(1-2)*(3+4)在RPN中就是1 2 - 3 4 + *，而1-(2*(3+4)) 变为 1 2 3 4 + * -。

这是 RPN 解析器的简单实现。我将一个对象列表从一个 RPN 序列中分离出来，以便相同的序列可以直接用于不同的列表。

def rpn(arr, seq):
    '''
    Reverse Polish Notation algorithm
    (this version works only for binary operators)
    arr: array of objects 
    seq: rpn sequence containing indices of objects from arr and functions
    '''
    stack = []
    for x in seq:
        if isinstance(x, int):
        # it's an object: push it to stack
            stack.append(arr[x])
        else:
        # it's a function: pop two objects, apply the function, push the result to stack 
            b = stack.pop()
            #a = stack.pop()
            #stack.append(x(a,b))
            ## shortcut:
            stack[-1] = x(stack[-1], b)
    return stack.pop()

使用示例:

# Say we have an array 
arr = [100, 210, 42, 13] 
# and want to calculate 
(100 - 210) * (42 + 13) 
# It translates to RPN: 
100 210 - 42 13 + * 
# or 
arr[0] arr[1] - arr[2] arr[3] + * 
# So we apply `
rpn(arr,[0, 1, subtract, 2, 3, add, multiply])

要将 RPN 应用于您的案例，您需要从头开始生成 rpn 序列或将您的 (pair_indexes, binary_functions) 转换为它们。我还没有考虑过转换器，但肯定可以做到。

测试

您的原始测试排在第一位:

r = 100000
xs = [random() for _ in range(r)]
ps = [0]*(r-1)
fs = repeat(add)
foo = ordered_reduce(None, ps, fs, precompute=True)
rpn_seq = [0] + [x for i, f in zip(range(1,r), repeat(add)) for x in (i,f)]
rpn_seq2 = list(range(r)) + list(repeat(add,r-1))
# Here rpn_seq denotes (_ + (_ + (_ +( ... )...))))
# and rpn_seq2 denotes ((...( ... _)+ _) + _).
# Obviously, they are not equivalent but with 'add' they yield the same result. 

%timeit reduce(add, xs)
100 loops, best of 3: 7.37 ms per loop
%timeit foo(xs)
1 loops, best of 3: 1.71 s per loop
%timeit rpn(xs, rpn_seq)
10 loops, best of 3: 79.5 ms per loop
%timeit rpn(xs, rpn_seq2)
10 loops, best of 3: 73 ms per loop

# Pure numpy just out of curiosity:
%timeit np.sum(np.asarray(xs))
100 loops, best of 3: 3.84 ms per loop
xs_np = np.asarray(xs)
%timeit np.sum(xs_np)
The slowest run took 4.52 times longer than the fastest. This could mean that an intermediate result is being cached 
10000 loops, best of 3: 48.5 µs per loop

因此，rpn 比 reduce 慢 10 倍，但比 ordered_reduce 快 20 倍。

现在，让我们尝试一些更复杂的事情:交替地对矩阵进行加法和乘法运算。我需要一个特殊的函数来测试 reduce。

add_or_dot_b = 1
def add_or_dot(x,y):
    '''calls 'add' and 'np.dot' alternately'''
    global add_or_dot_b
    if add_or_dot_b:
        out = x+y
    else:
        out = np.dot(x,y)
    add_or_dot_b = 1 - add_or_dot_b
    # normalizing out to avoid `inf` in results
    return out/np.max(out)

r = 100001      # +1 for convenience
                # (we apply an even number of functions) 
xs = [np.random.rand(2,2) for _ in range(r)]
ps = [0]*(r-1)
fs = repeat(add_or_dot)
foo = ordered_reduce(None, ps, fs, precompute=True)
rpn_seq = [0] + [x for i, f in zip(range(1,r), repeat(add_or_dot)) for x in (i,f)]

%timeit reduce(add_or_dot, xs)
1 loops, best of 3: 894 ms per loop
%timeit foo(xs)
1 loops, best of 3: 2.72 s per loop
%timeit rpn(xs, rpn_seq)
1 loops, best of 3: 1.17 s per loop

在这里，rpn 比 reduce 慢大约 25%，比 ordered_reduce 快 2 倍多。