如何根据 Python 中的另一个集合对集合进行排序？

Question

我是 Python 的新手，我需要帮助解决以下问题。我在下面发布的这个定义将采用表示矩形 Ex 的元组集合。（宽度，高度）。

在方法中，我首先对矩形集合进行排序，使宽度最大的元素位于列表的开头，然后我稍后使用这个集合来分配 UPPER_LEFT_X 和 UPPER_LEFT_Y每个矩形的坐标。

我的问题是，如何获取这个坐标集合并按照与 original_rectangle_collection 完全相同的顺序对其进行排序？ Driver 的工作方式是确定的坐标必须按照与原始矩形集合相同的顺序给出。谢谢！让我知道是否需要更多解释。

def find_best_coordinates(rectangles):
    placement = []
    upper_left_x = 0
    upper_left_y = 0
    loop_count = 0

    original_rectangle_collection = rectangles
    #sort the rectangles according to width.
    rectangles.sort(key=operator.itemgetter(0))
    #reverse the list so the tallest rectanlgle is placed first
    rectangles.reverse()


    #set the max width to the tallest rectangle's width
    max_width = rectangles[0][0]
    #print("Max width",max_width)

    #loop through and place the rectangles
    for rectangle in rectangles:
        if loop_count == 0:
            max_width = rectangle[0]
        height = rectangle[1]
        coordinate = (upper_left_x, upper_left_y)  
        placement.insert(0, coordinate)             
        upper_left_y = upper_left_y - height - 990
        loop_count = loop_count + 1
        if loop_count == 50:
            upper_left_x = upper_left_x + max_width + 990
           # print("x = ", upper_left_x)
            loop_count = 0
            #print("y = ", upper_left_y)
            upper_left_y = 0

    #reverse the list before it gets returned
    placement.reverse()                            
    return placement

Answer 1

您可以这样做，始终将原始索引与矩形一起传递，并将其与生成的坐标保持一致。

# Store indexes
rdicts = [{'index': i, 'rectangle': r} for i, r in enumerate(rectangles)]
# Sort by width
rdicts.sort(key=lambda d: d['rectangle'][0], reverse=True)

placement = []
for rdict in rdicts:
    rectangle = rdict['rectangle']
    ... # do the rest of your algorithm
    # Add the calculated coordinate, along with the original index
    placement.append({'index': rdict['index'], 'coordinate': coordinate})

# Sort by the original index
placement.sort(key=lambda p: p['index'])
# Return the coordinates without the indexes.
ordered_placement = [p['coordinate'] for p in placement]

Answer 2

您可以通过对索引进行排序来加快速度：

processing_order = sorted(
    range(len(rectangles)), 
    key=lambda i: rectangles[i][0]
)

for rect_i in processing_order:
    r = rectangles[rect_i]
    # ... blah blah with r as current rectangle

# Done processing, output them in original order:
for r in rectangles:
    # Still in order!

如何根据 Python 中的另一个集合对集合进行排序？

How to sort a collection based on another collection in Python?

python

sorting

bin-packing