获取python嵌套字典中子值的平均值

2024-03-28 19:20:19 发布

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我有一个表格(从excel导入),其中包含需求和子需求,可以看到如下示例:

+-------------+--------+
| Requeriment | Points |
+-------------+--------+
| 1           |        |
+-------------+--------+
| 2           |        |
+-------------+--------+
| 3           |        |
+-------------+--------+
| 3.1         |        |
+-------------+--------+
| 3.2         |        |
+-------------+--------+
| 3.3         |        |
+-------------+--------+
| 4           |        |
+-------------+--------+
| 5           |        |
+-------------+--------+
| 5.1         |        |
+-------------+--------+
| 5.2         |        |
+-------------+--------+
| 5.2.1       |        |
+-------------+--------+
| 5.2.1.1     |        |
+-------------+--------+
| 5.3         |        |
+-------------+--------+
| 6           |        |
+-------------+--------+


所有需求的最大值均为“1.0”,如果存在需求有子需求的情况,则需求值=总和(子需求)/总子需求,且该值也适用于子需求和子需求

在这里,我展示了没有子需求的需求是如何用0(如果它不完整)和1.0(如果它完成)来评估的

enter image description here

带有子要求的要求将通过以下公式进行评估: 再查询点=子查询的总和/子查询的值。该值如下所示:

enter image description here

此表在字典中进行了如下转换:

{requeriments:{
 '1':{'points':'', subrequeriments: {}},
 '2':{'points':'', subrequeriments: {}},
 '3':{'points':'', subrequeriments: {
    '3.1':{'points':'', subrequeriments: {} },
    '3.2':{'points':'', subrequeriments: {} },
    '3.3':{'points':'', subrequeriments: {} }
   }
  },
 '4':{'points':'', subrequeriments: {}},
 '5':{'points':'', subrequeriments: {
    '5.1':{'points':'', subrequeriments: {} },
    '5.2':{'points':'', subrequeriments: {
      '5.2.1':{'points':'', subrequeriments: {
         '5.2.1.1':{'points':'', subrequeriments: {}
         }

        } 
      },
    '5.3':{'points':'', subrequeriments: {} }
   }
  },
 '6':{'points':'', subrequeriments: {}}
 }
}

最后,在嵌套子需求中使用函数requirementvalue=sum(subrequirements)/total subrequirements,我无法找到一个好方法来评估需求

你知道一个好方法可以把它应用到字典里吗,或者至少有一个好方法可以把它应用到失物招领处或桌子上吗


Tags: 方法示例字典情况excelpoints表格公式
1条回答
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1楼 · 发布于 2024-03-28 19:20:19

使用递归函数:

def calcPoints(d):
    points = 0
    for k,v in d.items():
        if v['points'] == '':
            v['points'] = calcPoints(v['subrequirements'])
        points += float(v['points'])
    return str(points / max(1,len(d)))

输入:

data = {'requeriments': 
{'1': {'points': '1', 'subrequirements': {}}, 
 '2': {'points': '0', 'subrequirements': {}}, 
 '3': {'points': '', 'subrequirements': 
    {'3.1': {'points': '1', 'subrequirements': {}},
     '3.2': {'points': '0', 'subrequirements': {}},
     '3.3': {'points': '1', 'subrequirements': {}}}}, 
 '4': {'points': '1', 'subrequirements': {}}, 
 '5': {'points': '', 'subrequirements': 
    {'5.1': {'points': '1', 'subrequirements': {}},
     '5.2': {'points': '', 'subrequirements': 
          {'5.2.1': {'points': '', 'subrequirements': 
                  {'5.2.1.1': {'points': '1', 'subrequirements': {}}, 
                   '5.2.1.2': {'points': '0', 'subrequirements': {}}}}, 
           '5.2.2': {'points': '0', 'subrequirements': {}}}},
     '5.3': {'points': '1', 'subrequirements': {}}}}, 
 '6': {'points': '0', 'subrequirements': {}}}}

输出:

calcPoints(data['requeriments']) # '0.5694444444444444'

print(data)

{'requeriments': 
 {'1': {'points': '1', 'subrequirements': {}}, 
  '2': {'points': '0', 'subrequirements': {}}, 
  '3': {'points': '0.6666666666666666', 'subrequirements': 
     {'3.1': {'points': '1', 'subrequirements': {}}, 
      '3.2': {'points': '0', 'subrequirements': {}}, 
      '3.3': {'points': '1', 'subrequirements': {}}}}, 
  '4': {'points': '1', 'subrequirements': {}}, 
  '5': {'points': '0.75', 'subrequirements': 
     {'5.1': {'points': '1', 'subrequirements': {}}, 
      '5.2': {'points': '0.25', 'subrequirements': 
           {'5.2.1': {'points': '0.5', 'subrequirements': 
                   {'5.2.1.1': {'points': '1', 'subrequirements': {}}, 
                    '5.2.1.2': {'points': '0', 'subrequirements': {}}}}, 
            '5.2.2': {'points': '0', 'subrequirements': {}}}}, 
      '5.3': {'points': '1', 'subrequirements': {}}}}, 
  '6': {'points': '0', 'subrequirements': {}}}}

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