递归方法在迭代期间更改字典

Question

尝试在 class 中使用递归方法来展平嵌套的 OrderedDicts。结果是 RuntimeError：字典在迭代期间改变了大小

我得到了一个 OrderedDicts 列表。大多数 OrderedDicts 都是简单的 Key:string-value 属性，但有些值反而包含另一个 OrderedDict。嵌套可以下降几层。这是一个非常简化的示例：

records = [
    OrderedDict([
        ('rec-1_field-1', 'r1f1_value'),
        ('rec-1_field-2', 'r1f2_value'),
        ('rec-1_nest-1', OrderedDict([
            ('n1_field-1', 'n1f1_value'),
            ('n1_field-2', 'n1f2_value')
            ])
         )
        ]),
    OrderedDict([
        ...
        ])
]

我的目标是解除这些 OrderedDicts 的嵌套，以便上面的内容开始转换成这个（注意我试图获得的 "higherKey.lowerKey" 命名法）：

flatRecords = [
    {'rec-1_field-1':'r1f1_value',
    'rec-1_field-2':'r1f2_value',
    'rec-1_nest-1.n1_field-1':'n1f1_value',
    'rec-1_nest-1.n1_field-2':'n1f2_value'},
    ...
    ]

这是我的代码的简化版本。我将每个 OrderedDict 提供给一个方法，该方法在找到嵌套的 OrderedDict 时递归。我想我正在递归中覆盖我的 flatRecord Dict 但无法确定如何更正。

class unNested():
    def __init__(self):
        pass
    def flatResults(self, OD):
        self.OD = OD
        self.flattenedRecords = []
        for eachRecord in self.OD:
            self.flattenedRecords.append(self.flatten(eachRecord))
        return self.flattenedRecords
    def flatten(self, record):
        self.record = record
        self.flatRecord = {}
        for eachKey in self.record:
            if isinstance(self.record[eachKey], dict):
                self.subRecord = self.flatten(self.record[eachKey])
                for eachSub in self.subRecord:
                    self.key = eachKey + '.' + eachSub
                    self.flatRecord[self.key] = self.record[eachSub]
            else:
                self.flatRecord[eachKey] = self.record[eachKey]
        return self.flatRecord

所以下面的代码片段导致 "RuntimeError: dictionary changed size during iteration"

records = [
    OrderedDict([
        ('rec-1_field-1', 'r1f1_value'),
        ('rec-1_field-2', 'r1f2_value'),
        ('rec-1_nest-1', OrderedDict([
            ('rec-1_nest-1_field-1', 'r1n1f1_value'),
            ('rec-1_nest-1_field-2', 'r1n1f2_value')
            ])
         )
        ]),
    OrderedDict([
        ('rec-2_field-1', 'r2f1_value'),
        ('rec-2_field-2', 'r2f2_value'),
        ('rec-2_nest-1', OrderedDict([
            ('rec-2_nest-1_field-1', 'r2n1f1_value'),
            ('rec-2_nest-1_field-2', 'r2n1f2_value')
            ])
         )
        ])
    ]
crush = unNested()
crush.flatResults(records)

我确定这是一个业余错误，但我很想听听任何想法或指导。谢谢！

Answer 1

可以使用扁平化的方法：

from collections import OrderedDict
records = [OrderedDict([('rec-1_field-1', 'r1f1_value'), ('rec-1_field-2', 'r1f2_value'), ('rec-1_nest-1', OrderedDict([('n1_field-1', 'n1f1_value'), ('n1_field-2', 'n1f2_value')]))])]
def flatten(d, last=''):
   for a, b in d.items():
      if not isinstance(b, OrderedDict):
         yield (f'{last}.{a}' if last else a, b)
      else:
         yield from flatten(b, last = a)

final_result = dict(flatten(records[0]))

输出：

{'rec-1_field-1': 'r1f1_value', 'rec-1_field-2': 'r1f2_value', 'rec-1_nest-1.n1_field-1': 'n1f1_value', 'rec-1_nest-1.n1_field-2': 'n1f2_value'}

为列表中的每个元素创建扁平化结构：

final_result = [dict(flatten(i)) for i in records]

递归方法在迭代期间更改字典

Recursive method changing dictionary during iteration

python

python-3.x

recursion

ordereddictionary