Python减少嵌套循环

Question

因此，我正在研究UVA问题，并且我有4个嵌套循环来遍历多边形列表（每个多边形都包含一个点列表，其中每个点都包含一个整数x和y来表示其坐标，即，polygon
[0]是一个点，其坐标为面[0] .x和面[0] .y）。

我试图减少程序中for循环的数量，以使其更高效并降低运行时间。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使其变得更加高效，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，是否有办法减少算法的嵌套循环数？

在此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
。为这些使用itertools.combinations()迭代器。您最里面的双循环会产生
笛卡尔积
，因此请使用itertools.product()迭代器。

不要使用range(), just loop directly over the polygon lists; useenumerate（）来生成索引以添加索引，而不要使索引相反。

到配对部分，该pairwise()配方从itertools食谱部;
这样您就可以使用所有细分。要再次从头开始绕圈（将最后一个坐标与第一个坐标配对），只需将列表的第一个元素附加到末尾即可。

一旦摆脱了嵌套循环，就可以使用break结束循环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

实际上，一旦确定某个东西是一个并集，就不必继续进行其余的测试。您可以使用any()功能停止测试isInside()和doIntersect早期的功能：

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)

这不仅现在可读性强，而且还应该更有效！