如何按键名计算Python字典中所有值的总和和平均值？

Question

我有一个

dict

对象，我想从中计算特定名称 values 的 sum 和

mean

。

设置：

tree_str = """{
    'trees': [
        {
            'tree_idx': 0,
            'dimensions': (1120, 640),
            'branches': [
                'leaves': [
                    {'geometry': [[0.190673828125, 0.0859375], [0.74609375, 0.1181640625]]},
                    {'geometry': [[0.1171875, 0.1162109375], [0.8076171875, 0.15625]]}
                ],
                'leaves': [
                    {'geometry': [[0.2197265625, 0.1552734375], [0.7119140625, 0.1943359375]]},
                    {'geometry': [[0.2060546875, 0.1923828125], [0.730712890625, 0.23046875]]}
                ]
            ]
        }
    ]
}"""

tree_dict = yaml.load(tree_str, Loader=yaml.Loader)

地点：

# assume for the sake of coding
{'geometry': ((xmin, ymin), (xmax, ymax))}
# where dimensions are relative to an image of a tree

现在我有了

dict

对象，我该怎么办：

1. 得到所有叶子的

   count

   ？
2. 得到

   average width

   和

   average height

   所有叶子吗？

我可以使用以下方法访问值并遍历树：

tree_dict['trees'][0]['branches'][0]['leaves'][0]['geometry'][1][1]

所以我可以使用嵌套 for 循环来做到这一点：

leafcount = 0
leafwidth = 0
leafheight = 0
sumleafwidth = 0
sumleafheight = 0
avgleafwidth = 0
avgleafheight = 0

for tree in tree_dict['trees']:
    print("TREE")
    for branch in  tree['branches']:
        print("\tBRANCH")
        for leaf in branch['leaves']:
            leafcount += 1
            (lxmin, lymin), (lxmax, lymax) = leaf['geometry']
            leafwidth = lxmax - lxmin
            leafheight = lymax - lymin
            print("\t\tLEAF: x1({}), y1({}), x2({}), y2({})\n\t\t\tWIDTH: {}\n\t\t\tHEIGHT: {}".format(lxmin, lymin, lxmax, lymax, leafwidth, leafheight))
            sumleafwidth += lxmax - lxmin
            sumleafheight += lymax - lymin

avgleafwidth = sumleafwidth / leafcount
avgleafheight = sumleafheight / leafcount

print("LEAVES\n\tCOUNT: {}\n\tAVERAGE WIDTH: {}\n\tAVERAGE HEIGHT: {}".format(leafcount, avgleafwidth, avgleafheight))

但是还有更好的办法吗？

# psuedo code
leafcount = count(tree_dict['trees'][*]['branches'][*]['leaves'][*])
leaves = (tree_dict['trees'][*]['branches'][*]['leaves'][*])
sumleafwidth = sum(leaves[*]['geometry'][1][*]-leaves[*]['geometry'][0][*])
sumleafheight = sum(leaves[*]['geometry'][*][1]-leaves[*]['geometry'][*][0])
avgleafwidth = sumleafwidth / leafcount
avgleafheight = sumleafheight / leafcount

Answer 1

我认为，尽管 python dict 在大多数情况下可以用作树表示，但如果您想处理更高级的与树相关的任务，如上所述，它并不是最好的数据结构。 python中有很多树状结构的实现，例如treelib。 您可以从字典移动到树，例如像这样：

def dict_to_tree(data, parent_node=None, tree=None):
    if tree is None:
        tree = Tree()
    
    for key, value in data.items():
        if isinstance(value, dict):
            # Create a node for the key
            tree.create_node(tag=key, identifier=key, parent=parent_node)
            # Recursively call the function to process the sub-dictionary
            dict_to_tree(value, parent_node=key, tree=tree)
        else:
            # Create a node for the key and value
            tree.create_node(tag=f"{key}: {value}", identifier=key, parent=parent_node)

    return tree 

您应该能够在正确的数据结构上以更简单、更优雅的方式解决您的问题。