Python 减少嵌套循环

Question

所以我正在研究UVA问题，我有4个嵌套循环来迭代多边形列表(每个多边形包含一个点列表，其中每个点包含一个整数x和y来表示它的坐标，即多边形[0 ] 是坐标为 Polygon[0].x 和 Polygon[0].y) 的点。

我正在尝试减少程序中 for 循环的数量，以提高其效率并缩短运行时间。我的代码如下:

for i in range(len(polygons)): # iterate over all the different polygons in the test case
    for j in range(i+1, len(polygons)): # iterate over all the different polygons in the test case but starting from the second, in order to make comparations between polygons i and j
        for a in range(len(polygons[i])):
            if (isInside(polygons[i][a].x, polygons[i][a].y, polygons[j])):
                union(i,j)
        for a in range(len(polygons[j])):
            if (isInside(polygons[j][a].x, polygons[j][a].y, polygons[i])):
                union(i,j)
        f = 1
        for a in range(len(polygons[i])): # iterate over all the different points in the polygon i
            for b in range(len(polygons[j])): # iterate over all the different points in the polygon j
                if (f!=0):
                    if(doIntersect(polygons[i][a], polygons[i][(a+1)%len(polygons[i])],polygons[j][b], polygons[j][(b+1)%len(polygons[j])])): # check if every single pair of line segments, each one made up of two points, intersect with each other
                        union(i,j) # the two line segments intersect so we join them by using union
                        f = 0

我尝试通过使用 itertools.product 来提高效率，如下所示:

def solve():
global polygons, p

ranges = [range(len(polygons)), range(1,len(polygons))]

for i, j in product(*ranges):
    for a in range(len(polygons[i])):
        if (isInside(polygons[i][a].x, polygons[i][a].y, polygons[j])):
            union(i,j)
    for a in range(len(polygons[j])):
        if (isInside(polygons[j][a].x, polygons[j][a].y, polygons[i])):
            union(i,j)
    f = 1
    ranges2 = [range(len(polygons[i])), range(len(polygons[j]))]
    for a,b in product(*ranges2):
        if (f!=0):
            if(doIntersect(polygons[i][a], polygons[i][(a+1)%len(polygons[i])],polygons[j][b], polygons[j][(b+1)%len(polygons[j])])): # check if every single pair of line segments, each one made up of two points, intersect with each other
                union(i,j) # the two line segments intersect so we join them by using union
                f = 0

无论如何，我的代码在两种情况下都具有相同的运行时，有没有办法减少我的算法的嵌套循环数量？

预先感谢您提供的任何帮助，非常感谢

Answer 1

您的两个外部循环正在创建列表的组合；使用 itertools.combinations() iterator对于那些。最里面的双循环产生笛卡尔积，所以使用 itertools.product() iterator .

不要使用 range(), just loop directly over the polygon lists; use 生成索引enumerate()` 添加索引而不是使索引以相反的方式工作。

为了配对各个部分， pairwise() recipe来自itertools食谱部分；这将使您可以使用所有部分。要再次绕回起点(将最后一个坐标与第一个坐标配对)，只需将包含第一个元素的列表附加到末尾即可。

摆脱嵌套循环后，您可以使用 break结束它们而不是使用标志变量。

from itertools import combinations, product

def pairwise(iterable):
    "s -> (s0,s1), (s1,s2), (s2, s3), ..."
    a, b = tee(iterable)
    next(b, None)
    return zip(a, b)

for (i, a_poly), (j, b_poly) in combinations(enumerate(polygons), 2):
    for a in a_poly:
        if isInside(a.x, a.y, b_poly):
            union(i, j)
    for b in b_poly:
        if isInside(b.x, b.y, a_poly):
            union(j, i)

    # attach the first element at the end so you go 'round'
    a_segments = pairwise(a_poly + a_poly[:1])
    b_segments = pairwise(b_poly + b_poly[:1])
    for a_seg, b_seg in product(a_segments, b_segments):
        if doIntersect(*a_seg, *b_seg):
            union(i,j)
            break

事实上，一旦确定某个东西是并集，您就不必继续其余的测试。您可以使用 any() function停止测试 isInside()和doIntersect早期功能:

for (i, a_poly), (j, b_poly) in combinations(enumerate(polygons), 2):
    if any(isInside(a.x, a.y, b_poly) for a in a_poly):
        union(i, j)
        break  # union found, no need to look further

    for any(isInside(b.x, b.y, a_poly) for b in b_poly):
        union(i, j)
        break  # union found, no need to look further

    # attach the first element at the end so you go 'round'
    a_segments = pairwise(a_poly + a_poly[:1])
    b_segments = pairwise(b_poly + b_poly[:1])
    if any(doIntersect(*a_seg, *b_seg) 
           for a_seg, b_seg in product(a_segments, b_segments)):
        union(i,j)

现在这不仅更具可读性，而且还应该更加高效!