可以使用递归生成器函数逐行获得完全展平的结果，然后使用 collections.defaultdict 将帐户id上的行分组。从那里，您可以抓取所需的键来构建最终帐户 pd.DataFrame :

import pandas as pd, collections
data = {'@type': 'g:Vertex', '@value': {'id': 'account-2199023262994', 'label': 'Account', 'properties': {'account_number': [{'@type': 'g:VertexProperty', '@value': {'id': {'@type': 'g:Int32', '@value': 544016139}, 'value': '0002-1990-2326-2994', 'label': 'account_number'}}], 'last_name': [{'@type': 'g:VertexProperty', '@value': {'id': {'@type': 'g:Int32', '@value': -1616372909}, 'value': 'Law-Yone', 'label': 'last_name'}}], 'first_name': [{'@type': 'g:VertexProperty', '@value': {'id': {'@type': 'g:Int32', '@value': -451458550}, 'value': 'Eric', 'label': 'first_name'}}]}}}
def flatten(d, c = [], p = []):
   t, f = [], []
   for a, b in d.items():
      (t if not isinstance(b, (dict, list)) else f).append((p+[a], b))
   if not f:
      yield {'.'.join(a):b for a, b in (c+t)}
   else:
      for a, b in f:
         if isinstance(b, dict):
             yield from flatten(b, c=c+t, p = a)
         else:
             for i in b:
                yield from flatten(i, c=c+t, p = a)

d = collections.defaultdict(dict)
for i in flatten(data):
   d[i['@value.id']].update(i)

kv = [['@type', '@type'], ['@value.id', '@value.id'], ['@value.label', '@value.label'], ['@value.properties.account_number.@value.id.@value', 'account_number.id'], ['@value.properties.account_number.@value.value', 'account_number.value'], ['@value.properties.last_name.@value.id.@value', 'last_name.id'], ['@value.properties.last_name.@value.value', 'last_name.value'], ['@value.properties.first_name.@value.id.@value', 'first_name.id'], ['@value.properties.first_name.@value.value', 'first_name.value']]
df = pd.DataFrame([{j:b[k] for k, j in kv} for b in d.values()])

输出：

@type              @value.id @value.label  ...  last_name.value first_name.id  first_name.value
0  g:Vertex  account-2199023262994      Account  ...         Law-Yone    -451458550              Eric

[1 rows x 9 columns]