Python：将列表列表转换为分层字典答案

【问题标题】：Python: Transform a list of lists into a hierarchical dictionaryPython：将列表列表转换为分层字典
【发布时间】：2020-10-07 00:21:58
【问题描述】：

我有一些基因测序数据如下：

data = [{'sequence': 'gene1__gene2__gene3', 'occurrence': 10},
        {'sequence': 'gene2__gene3', 'occurrence': 5},
        {'sequence': 'gene2', 'occurrence': 2},
        {'sequence': 'gene4', 'occurrence': 4}
       ]

我想将其转换为以下（树状）dictionary 数据结构，其中任何子路径都会告诉我该组基因的共现计数：

tree_dict = {
        'gene1': {'occurrence': 10, 'self': 0, 'children': {'gene2': {'occurrence': 10, 'self': 0, 'children': {'gene3': {'occurrence': 10, 'self': 10, 'children': {}}}},
                                                            'gene3': {'occurrence': 10, 'self': 0, 'children': {'gene2': {'occurrence': 10, 'self': 10, 'children': {}}}},
                                                           }
                 },
        'gene2': {'occurrence': 17, 'self': 2, 'children': {'gene1': {'occurrence': 10, 'self': 0, 'children': {'gene3': {'occurrence': 10, 'self': 10, 'children': {}}}},
                                                            'gene3': {'occurrence': 15, 'self': 5, 'children': {'gene1': {'occurrence': 10, 'self': 10, 'children': {}}}},
                                                           }
                 },
        'gene3': {'occurrence': 15, 'self': 0, 'children': {'gene1': {'occurrence': 10, 'self': 0, 'children': {'gene2': {'occurrence': 10, 'self': 10, 'children': {}}}},
                                                            'gene2': {'occurrence': 15, 'self': 5, 'children': {'gene1': {'occurrence': 10, 'self': 10, 'children': {}}}},
                                                           }
                 },
        'gene4': {'occurrence': 4, 'self': 4, 'children': {}}
       }

在上面的tree_dict：

self 指的是（子）路径中仅出现节点。例如：gene3 永远不会单独存在，因此 self 的值为 0；而gene2 单独存在2 次，因此self 的值为2。
occurrence 指的是（子）路径中节点作为子字符串和整体的出现。

我尝试过的代码？
当我知道解决方案必须是递归函数时，我正在尝试失败迭代方法。类似于这个问题的东西：How to transform a list into a hierarchy dict。但我无法在这个方向上取得任何进展。

【问题讨论】：

标签： python python-3.x recursion data-structures tree

【解决方案1】：

试试这个：

data = [{'sequence': 'gene1__gene2__gene3', 'occurrence': 10},
        {'sequence': 'gene2__gene3', 'occurrence': 5},
        {'sequence': 'gene2', 'occurrence': 2},
        {'sequence': 'gene4', 'occurrence': 4}]

tree_dict = {}

def generate_tree(sequence, occurrence, curr_dict):
    gene_list = sequence.split('__')
    for gene in gene_list:
        if gene in curr_dict:
            curr_dict[gene]['occurrence'] += occurrence
        else:
            curr_dict[gene] = {'occurrence': occurrence, 'self': 0, 'children': {}}
        updated_list = gene_list.copy()
        updated_list.remove(gene)
        updated_sequence = '__'.join(updated_list)
        if updated_sequence != '':
            generate_tree(updated_sequence, occurrence, curr_dict[gene]['children'])
        else:
            curr_dict[gene]['self'] += occurrence

for item in data:
    generate_tree(item['sequence'], item['occurrence'], tree_dict)

print(tree_dict)

【讨论】：