獲取python嵌套字典中子值的平均值

我有一個情況，我有一個表（從 excel 匯入）機智要求和子要求，可以像這個例子一樣看到：

 ------------- -------- 
| 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           |        |
 ------------- -------- 

所有需求最大值為'1.0'，如果存在需求有子需求的情況，則需求值= sum(subrequirments)/total subrequirments，它也適用于子需求和子需求。

在這里，我展示了如何對沒有子要求的要求進行評估，如果它不完整，則為 0，如果已完成，則為 1.0。

帶有子要求的要求將通過以下公式進行評估：要求點=子要求的總和/子要求的值。值將是這樣的：

這張表被轉換成這樣的字典：

{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: {}}
 }
}

At the end I can not get a good way to evaluate the requirments aplying the function requirment value = sum(subrequirments)/total subrequirments in the nested sub requirments.

Do you know a good way to aply it to a dictionary or at least a good way do do it to a lost or a table?

uj5u.com熱心網友回復：

使用遞回函式：

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)))

輸入：

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': {}}}}

輸出：

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': {}}}}

標籤：python list dictionary nested evaluation

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