将 n 个可变高度图像拟合为 3 个（相似长度）列布局

Question

我希望制作一个类似于 piccsy.com 的 3 列布局。给定许多宽度相同但高度不同的图像，有什么算法可以对它们进行排序以使列长度的差异最小？最好使用 Python 或 JavaScript...

非常感谢您提前提供的帮助！

马丁

Answer 1

有多少图像？

如果限制最大页面大小，并具有最小图片高度的值，则可以计算每页的最大图像数。在评估任何解决方案时您都需要这个。

我认为您提供的链接上有 27 张图片。

下面使用 Robin Green 之前提到的first_fit 算法，然后通过贪婪交换对此进行改进。

交换例程找到距离平均列高最远的列，然后系统地寻找其中一张图片与另一列中的第一张图片之间的交换，以最小化与平均值的最大偏差。

我使用了 30 张图片的随机样本，高度在 5 到 50 个“单位”范围内。在我的例子中，收敛速度很快，并且比first_fit算法有了显着的改进。

代码（Python 3.2：

def first_fit(items, bincount=3):
    items = sorted(items, reverse=1) # New - improves first fit.
    bins     = [[] for c in range(bincount)]
    binsizes = [0] * bincount
    for item in items:
        minbinindex = binsizes.index(min(binsizes))
        bins[minbinindex].append(item)
        binsizes[minbinindex] += item
    average = sum(binsizes) / float(bincount)
    maxdeviation = max(abs(average - bs) for bs in binsizes)

    return bins, binsizes, average, maxdeviation

def swap1(columns, colsize, average, margin=0):
    'See if you can do a swap to smooth the heights'
    colcount = len(columns)
    maxdeviation, i_a = max((abs(average - cs), i)
                              for i,cs in enumerate(colsize))
    col_a = columns[i_a]
    for pic_a in set(col_a): # use set as if same height then only do once
        for i_b, col_b in enumerate(columns):
            if i_a != i_b: # Not same column
                for pic_b in set(col_b):
                    if (abs(pic_a - pic_b) > margin): # Not same heights
                        # new heights if swapped
                        new_a = colsize[i_a] - pic_a + pic_b
                        new_b = colsize[i_b] - pic_b + pic_a
                        if all(abs(average - new) < maxdeviation
                               for new in (new_a, new_b)):
                            # Better to swap (in-place)
                            colsize[i_a] = new_a
                            colsize[i_b] = new_b
                            columns[i_a].remove(pic_a)
                            columns[i_a].append(pic_b)
                            columns[i_b].remove(pic_b)
                            columns[i_b].append(pic_a)
                            maxdeviation = max(abs(average - cs)
                                               for cs in colsize)
                            return True, maxdeviation
    return False, maxdeviation

def printit(columns, colsize, average, maxdeviation):
    print('columns')
    pp(columns)
    print('colsize:', colsize)
    print('average, maxdeviation:', average, maxdeviation)
    print('deviations:', [abs(average - cs) for cs in colsize])
    print()


if __name__ == '__main__':
    ## Some data
    #import random
    #heights = [random.randint(5, 50) for i in range(30)]
    ## Here's some from the above, but 'fixed'.
    from pprint import pprint as pp

    heights = [45, 7, 46, 34, 12, 12, 34, 19, 17, 41,
               28, 9, 37, 32, 30, 44, 17, 16, 44, 7,
               23, 30, 36, 5, 40, 20, 28, 42, 8, 38]

    columns, colsize, average, maxdeviation = first_fit(heights)
    printit(columns, colsize, average, maxdeviation)
    while 1:
        swapped, maxdeviation = swap1(columns, colsize, average, maxdeviation)
        printit(columns, colsize, average, maxdeviation)
        if not swapped:
            break
        #input('Paused: ')

输出：

columns
[[45, 12, 17, 28, 32, 17, 44, 5, 40, 8, 38],
 [7, 34, 12, 19, 41, 30, 16, 7, 23, 36, 42],
 [46, 34, 9, 37, 44, 30, 20, 28]]
colsize: [286, 267, 248]
average, maxdeviation: 267.0 19.0
deviations: [19.0, 0.0, 19.0]

columns
[[45, 12, 17, 28, 17, 44, 5, 40, 8, 38, 9],
 [7, 34, 12, 19, 41, 30, 16, 7, 23, 36, 42],
 [46, 34, 37, 44, 30, 20, 28, 32]]
colsize: [263, 267, 271]
average, maxdeviation: 267.0 4.0
deviations: [4.0, 0.0, 4.0]

columns
[[45, 12, 17, 17, 44, 5, 40, 8, 38, 9, 34],
 [7, 34, 12, 19, 41, 30, 16, 7, 23, 36, 42],
 [46, 37, 44, 30, 20, 28, 32, 28]]
colsize: [269, 267, 265]
average, maxdeviation: 267.0 2.0
deviations: [2.0, 0.0, 2.0]

columns
[[45, 12, 17, 17, 44, 5, 8, 38, 9, 34, 37],
 [7, 34, 12, 19, 41, 30, 16, 7, 23, 36, 42],
 [46, 44, 30, 20, 28, 32, 28, 40]]
colsize: [266, 267, 268]
average, maxdeviation: 267.0 1.0
deviations: [1.0, 0.0, 1.0]

columns
[[45, 12, 17, 17, 44, 5, 8, 38, 9, 34, 37],
 [7, 34, 12, 19, 41, 30, 16, 7, 23, 36, 42],
 [46, 44, 30, 20, 28, 32, 28, 40]]
colsize: [266, 267, 268]
average, maxdeviation: 267.0 1.0
deviations: [1.0, 0.0, 1.0]

好问题。

---

这是我在下面的单独评论中提到的反向排序的信息。

>>> h = sorted(heights, reverse=1)
>>> h
[46, 45, 44, 44, 42, 41, 40, 38, 37, 36, 34, 34, 32, 30, 30, 28, 28, 23, 20, 19, 17, 17, 16, 12, 12, 9, 8, 7, 7, 5]
>>> columns, colsize, average, maxdeviation = first_fit(h)
>>> printit(columns, colsize, average, maxdeviation)
columns
[[46, 41, 40, 34, 30, 28, 19, 12, 12, 5],
 [45, 42, 38, 36, 30, 28, 17, 16, 8, 7],
 [44, 44, 37, 34, 32, 23, 20, 17, 9, 7]]
colsize: [267, 267, 267]
average, maxdeviation: 267.0 0.0
deviations: [0.0, 0.0, 0.0]

---

如果您有反向排序，这个额外的代码会附加到上面代码的底部（在“if”中） name == ...)，将对随机数据进行额外的试验：

for trial in range(2,11):
    print('\n## Trial %i' % trial)
    heights = [random.randint(5, 50) for i in range(random.randint(5, 50))]
    print('Pictures:',len(heights))
    columns, colsize, average, maxdeviation = first_fit(heights)
    print('average %7.3f' % average, '\nmaxdeviation:')
    print('%5.2f%% = %6.3f' % ((maxdeviation * 100. / average), maxdeviation))
    swapcount = 0
    while maxdeviation:
        swapped, maxdeviation = swap1(columns, colsize, average, maxdeviation)
        if not swapped:
            break
        print('%5.2f%% = %6.3f' % ((maxdeviation * 100. / average), maxdeviation))
        swapcount += 1
    print('swaps:', swapcount)

额外的输出显示了交换的效果：

## Trial 2
Pictures: 11
average  72.000 
maxdeviation:
 9.72% =  7.000
swaps: 0

## Trial 3
Pictures: 14
average 118.667 
maxdeviation:
 6.46% =  7.667
 4.78% =  5.667
 3.09% =  3.667
 0.56% =  0.667
swaps: 3

## Trial 4
Pictures: 46
average 470.333 
maxdeviation:
 0.57% =  2.667
 0.35% =  1.667
 0.14% =  0.667
swaps: 2

## Trial 5
Pictures: 40
average 388.667 
maxdeviation:
 0.43% =  1.667
 0.17% =  0.667
swaps: 1

## Trial 6
Pictures: 5
average  44.000 
maxdeviation:
 4.55% =  2.000
swaps: 0

## Trial 7
Pictures: 30
average 295.000 
maxdeviation:
 0.34% =  1.000
swaps: 0

## Trial 8
Pictures: 43
average 413.000 
maxdeviation:
 0.97% =  4.000
 0.73% =  3.000
 0.48% =  2.000
swaps: 2

## Trial 9
Pictures: 33
average 342.000 
maxdeviation:
 0.29% =  1.000
swaps: 0

## Trial 10
Pictures: 26
average 233.333 
maxdeviation:
 2.29% =  5.333
 1.86% =  4.333
 1.43% =  3.333
 1.00% =  2.333
 0.57% =  1.333
swaps: 4

Answer 2

这就是离线完工时间最小化问题，我认为这相当于多处理器调度问题。您拥有图像而不是作业，并且您拥有图像高度而不是作业持续时间，但这是完全相同的问题。 （它涉及空间而不是时间这一事实并不重要。）因此任何（大约）解决它们中任何一个的算法都可以。

Answer 3

这是一个算法（称为首次拟合递减），它将以合理的方式为您提供非常紧凑的安排时间量。可能有更好的算法，但这非常简单。

1. 按从高到低的顺序对图像进行排序。
2. 拍摄第一张图像，并将其放置在最短的列中。
   （如果多列的高度相同（且最短），请选择任意一列。）
3. 重复步骤 2，直到没有图像剩余。

完成后，如果您不喜欢从高到低的外观，则可以根据自己的选择重新排列每列中的元素。

Answer 4

这是一个：

 // Create initial solution
 <run First Fit Decreasing algorithm first>
 // Calculate "error", i.e. maximum height difference
 // after running FFD
 err = (maximum_height - minimum_height)
 minerr = err

 // Run simple greedy optimization and random search
 repeat for a number of steps: // e.g. 1000 steps
    <find any two random images a and b from two different columns such that
     swapping a and b decreases the error>
    if <found>:
         swap a and b
         err = (maximum_height - minimum_height)
         if (err < minerr):
              <store as best solution so far> // X
    else:
         swap two random images from two columns
         err = (maximum_height - minimum_height)

 <output the best solution stored on line marked with X>