Question

因此，以下问题中的想法给出了带有一些值的字典列表，我需要的是嵌套的从属节点字典。

 Nodes = [
{
    "number": "10",
    "dependsOn": "13",
    "priority": "1649"
    "version": "1.7",
    "health": "ok"
},
{
    "number": "12",
    "dependsOn": "14",
    "priority": "2100"
    "version": "1.8",
    "health": "ok"
},
{
    "number": "3",
    "dependsOn": "2",
    "priority": "1234"
    "version": "1.1",
    "health": "ok"
}, {
    "number": "2",
    "dependsOn": "1",
    "priority": "1456",
    "version": "1.3",
    "health": "ok"
}, {
    "number": "1",
    "dependsOn": "10",
    "priority": "1890,
    "version": "1.2",
    "health": "ok"
},
{
    "number": "5",
    "dependsOn": "4",
    "priority": "1260",
    "version": "1.2",
    "health": "ok"
},
{
    "number": "4",
    "dependsOn": "12",
    "priority": "1150",
    "version": "1.4",
    "health": "ok"
}
}]

预期输出：

dependent_nodes =  { 3 : { 'dependsOn': [2,1,10,13],
                       'priority': 1234 # Lowest priority of the three
                      }, 
                  5 : { 'dependsOn': [4,12,14],
                       'priority': 1150 # Lowest priority of the three
                      }
                }

实现此输出的最佳方法是什么？

Answer 1

类似的事情应该可以帮助您入门。这不是最佳选择，但是您可以根据适当的上下文来调整它的显示方式。

mapping = {}
for item in Nodes:
    mapping[item['number']] = (item['dependsOn'], item['priority'])

ans = {}
added = set()

for key, value in mapping.items():
    if key in added:
        continue
    if key not in ans:
        ans[key] = {'dependsOn': [], 'priority': float('Inf')}
    curr = value[0]
    priority = float(value[1])
    while True:
        ans[key]['dependsOn'].append(curr)
        added.add(curr)
        ans[key]['priority'] = min(ans[key]['priority'], priority)
        
        if curr in ans:
            ans[key]['dependsOn'] += ans[curr]['dependsOn']
            ans[key]['priority'] = min(ans[key]['priority'], ans[curr]['priority'])
            del ans[curr]
            break
        elif curr in mapping and mapping[curr][0] not in added:
            priority = float(mapping[curr][1])
            curr = mapping[curr][0]
        else:
            break

这将产生：

{'3': {'dependsOn': ['2', '1', '10', '13'], 'priority': 1234.0},
 '5': {'dependsOn': ['4', '12', '14'], 'priority': 1150.0}}

Python-遍历字典列表

1 个答案: