朱利安(Julian)做python(' s)产量的方法 [英] julian way to do python's yield (and yield from)

问题描述

儒略(julian)产生(和产生)的方法是什么(与python一样?

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

我将尝试在python中添加一个小示例.

I will try to add small example in python.

思考一下4x​​4的国际象棋棋盘.找出国际象棋国王可以做的每N步长途移动.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.

如果我们在每个职位上都用数字签名:

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具有3个邻居(1、4、5).我们可以找到每个点的每个邻居的表:

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)，它从点列表或(...的生成器)点的生成器扩展给定路径:

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)

函数(生成器)，它给出具有给定长度的每条路径

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条路径，长度为10:

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

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

在ipython中，我们可以进行计时:

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实现非常丑陋，而且复杂得多.而且它似乎要慢一些.

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

推荐答案

查看 ResumableFunctions.jl

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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