朱利安方式来做python的产量(和产量) [英] julian way to do python's yield (and yield from)

问题描述

什么是 julian 的方式来做 yield(和 yield from)as python do?

What is julian way to do yield (and yield from) as python do?

我会尝试在 python 中添加小例子.

I will try to add small example in python.

想想 4x4 棋盘.找出长路棋王能做的每 N 步棋.不要浪费内存 -> 生成每条路径的生成器.

Think 4x4 chess board. Find every N moves long path chess king could do. Don't waste memory -> make generator of every path.

如果我们用数字在每个位置上签名:

if we sign every position with numbers:

0  1  2  3
4  5  6  7
8  9  10 11
12 13 14 16

点 0 有 3 个邻居 (1, 4, 5).我们可以为每个点找到每个邻居的表:

point 0 has 3 neighbors (1, 4, 5). We could find table for every neighbors for every point:

NEIG = [[1, 4, 5], [0, 2, 4, 5, 6], [1, 3, 5, 6, 7], [2, 6, 7], [0, 1, 5, 8, 9], [0, 1, 2, 4, 6, 8, 9, 10], [1, 2, 3, 5, 7, 9, 10, 11], [2, 3, 6, 10, 11], [4, 5, 9, 12, 13], [4, 5, 6, 8, 10, 12, 13, 14], [5, 6, 7, 9, 11, 13, 14, 15], [6, 7, 10, 14, 15], [8, 9, 13], [8, 9, 10, 12, 14], [9, 10, 11, 13, 15], [10, 11, 14]]

递归函数(生成器)从点列表或(...的生成器)点的生成器放大给定路径:

Recursive function (generator) which enlarge given path from list of points or from generator of (generator of ...) points:

def enlarge(path):
    if isinstance(path, list):
        for i in NEIG[path[-1]]:
            if i not in path:
                yield path[:] + [i]
    else:
        for i in path:
            yield from enlarge(i)

给定长度的每条路径的函数(生成器)

Function (generator) which give every path with given length

def paths(length):
    steps = ([i] for i in range(16))  # first steps on every point on board
    for _ in range(length-1):
        nsteps = enlarge(steps)
        steps = nsteps
    yield from steps

我们可以看到有 905776 条长度为 10 的路径:

We could see that there is 905776 paths with length 10:

sum(1 for i in paths(10))
Out[89]: 905776

在 ipython 中我们可以计时:

In ipython we could timeit:

%timeit sum(1 for i in paths(10))
1.21 s ± 15.1 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

我的 julia 实现很丑陋，而且要复杂得多.而且它似乎更慢.

My julia implementation is ugly and much more complicated. And it seems to be slower.

推荐答案

查看 ResumableFunctions.jl

来自自述文件

using ResumableFunctions

@resumable function fibonnaci(n::Int) :: Int
  a = 0
  b = 1
  for i in 1:n-1
    @yield a
    a, b = b, a+b
  end
  a
end

for fib in fibonnaci(10)
  println(fib)
end

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