以螺旋运动切过网格，为每个切片返回 xMin、xMax、yMin、yMax

Question

我当前的Python代码（下图）将图像切成n个切片，并按每行/列顺序返回每个切片的坐标。

与其从 xMin/yMin 1/1 开始，一次完成一行，我希望从图像的中间开始，然后螺旋式地完成，最终完成所有切片。如何实现这一点？

import math

slices = 11
imageWidth = 1024
imageHeight = 576
totalPixels = imageWidth * imageHeight
print 'Slices: ' + str(slices)
# Re-calculate slices
slices = int(slices/2)*2
print 'Re-calculated slices: ' + str(slices)
print 'Total pixels in image: ' + str(totalPixels)
print 'Maximum slices allowed: ' + str(totalPixels/4)
factor = math.sqrt( slices )
print 'Factor: ' + str(factor)
if (slices > totalPixels/4):
    print 'You cannot use more than ' + int(totalPixels/4) + ' slices!'
else:
    regionWidth = int(math.ceil(imageWidth / factor))
    regionHeight = int(math.ceil(imageHeight / factor))
    print 'Region size: ' + str(int(regionWidth)) + 'x' + str(int(regionHeight))
    print 'Region width: '  + str(regionWidth)
    print 'Region height: ' + str(regionHeight)
    imageWidthRounded = int( math.ceil(factor) * math.ceil( imageWidth / factor ) )
    restWidth = imageWidthRounded - imageWidth
    imageHeightRounded = int( math.ceil(factor) * math.ceil( imageHeight / factor ) )
    restHeight = imageHeightRounded - imageHeight
    print 'Rest width: ' + str(restWidth)
    print 'Rest height: ' + str(restHeight)
    factorRounded = int(math.ceil(factor))
    print 'Factor rounded: ' + str(factorRounded)

    xMin = 0
    xMax = 0
    yMin = 0
    yMax = 0

    rows = factorRounded
    columns = factorRounded
    print 'Total rows: ' + str(rows)
    print 'Total columns: ' + str(columns)

    for column in range(1, columns+1):
        xMin = 0
        xMax = 0
        if column == columns:
            print 'Col '+ str(column) + ' (last column) '
            yMin = (column*regionHeight + 1) - regionHeight
            yMax += (regionHeight - restHeight)
        else:
            print 'Col '+ str(column)
            yMin = (column*regionHeight + 1) - regionHeight
            yMax += regionHeight

        for row in range(1, rows+1):
            if row == rows:
                xMin = (row*regionWidth + 1) - regionWidth
                xMax += (regionWidth-restWidth)
                print 'Row ' + str(row) + ':  xMin=' +str(xMin) + 't xMax=' + str(xMax) + 't yMin=' + str(yMin) + 't yMax=' + str(yMax) + ' (last row)'
            else:
                xMin = (row*regionWidth + 1) - regionWidth
                xMax += regionWidth
                print 'Row ' + str(row) + ':  xMin=' +str(xMin) + 't xMax=' + str(xMax) + 't yMin=' + str(yMin) + 't yMax=' + str(yMax)

Answer 1

首先尝试枚举一些情况：

案例1：

o

返回中心值。

案例2：

o o
o o

从左上角按顺时针顺序返回值。

案例3：

o o o
o o o
o o o

返回

中心，然后从左上角按顺时针顺序返回值。

。

通常，我们看到如果网格的维度是奇数，我们需要首先返回中心值。在此之后，我们可以跟踪当前切片的边界角，并迭代地将这些角扩展到下一个级别以处理所有切片。我们可以通过网格的维度来确定切片的总数（从而确定迭代次数）。

下面是作为生成器实现的函数：

def spiral(m):
    length = len(m[0])
    last = length - 1
    mid = length // 2
    spirals_remaining = mid
    if length % 2 == 1:
        yield m[mid][mid]
        box = [(mid-1, mid-1), (mid-1, mid+1), (mid+1, mid+1), (mid+1, mid-1)]
    else:
        box = [(mid-1, mid-1), (mid-1, mid), (mid, mid), (mid, mid-1)]
    while spirals_remaining > 0:
        # yield spiral values clockwise from top left corner
        top = m[box[0][0]][slice(box[0][1], box[1][1])]
        for x in top:
            yield x
        right = [m[i][box[1][1]] for i in range(box[1][0], box[2][0])]
        for x in right:
            yield x
        bottom = m[box[2][0]][slice(box[2][1], box[3][1], -1)]
        for x in bottom:
            yield x
        left = [m[i][box[3][1]] for i in range(box[3][0], box[0][0], -1)]
        for x in left:
            yield x
        # update bounding box for next spiral
        box[0] = box[0][0] - 1, box[0][1] - 1
        box[1] = box[1][0] - 1, box[1][1] + 1
        box[2] = box[2][0] + 1, box[2][1] + 1
        box[3] = box[3][0] + 1, box[3][1] - 1
        spirals_remaining -= 1

示例输出：

>>> m = [[1, 2, 3, 4], 
...      [5, 6, 7, 8], 
...      [9, 10, 11, 12], 
...      [13, 14, 15, 16]]
>>> list(spiral(m))
[6, 7, 11, 10, 1, 2, 3, 4, 8, 12, 16, 15, 14, 13, 9, 5]