更新
考虑以下dict。如何通过以下方式提取符合的4元组：
lema，original_form，tag以及当且仅当其id时。到目前为止，这是我尝试过的：
def gettuples(data, level = 0):
if isinstance(data, dict):
if 'semtheme_list' in data:
print(data['semtheme_list'][0])
yield data['semtheme_list'][0]
elif 'analysis_list' in data:
print(data['analysis_list'][0])
yield data['analysis_list'][0]
for val in data.values():
yield from gettuples(val)
elif isinstance(data, list):
for val in data:
yield from gettuples(val)
使用上述功能，我得到以下（*）：
{'lemma': '*', 'tag': 'Z-----------', 'original_form': "Robert Downey Jr has topped Forbes magazine's annual list John Deere"}
{'lemma': 'Robert Downey Jr', 'tag': 'GNUS3S--', 'original_form': 'Robert Downey Jr'}
{'sense_id_list': [{'sense_id': '__12123288058840445720'}], 'lemma': 'Robert Downey Jr', 'tag': 'NPUU-N-', 'original_form': 'Robert Downey Jr'}
{'lemma': 'top', 'tag': 'VI-S3PPA-N-N9', 'original_form': 'has topped'}
{'lemma': 'John Deere', 'tag': 'GN-S3D--', 'original_form': "Forbes magazine's annual list John Deere"}
{'lemma': 'magazine', 'tag': 'GN-S3---', 'original_form': 'Forbes magazine'}
{'sense_id_list': [{'sense_id': 'db0f9829ff'}], 'lemma': 'Forbes', 'tag': 'NP-S-N-', 'original_form': 'Forbes'}
{'type': 'Top>SocialSciences>Economy', 'id': 'ODTHEME_ECONOMY'}
这与我正在寻找的4元组非常相似（**）：
[[['Z-----------', "Robert Downey Jr has topped Forbes magazine's annual list John Deere", '*'], ['GNUS3S--', 'Robert Downey Jr', 'Robert Downey Jr'], ['NPUU-N-', 'Robert Downey Jr', 'Robert Downey Jr'], ['VI-S3PPA-N-N9', 'has topped', 'top'], ['GN-S3D--', "Forbes magazine's annual list John Deere", 'John Deere'], ['GN-S3---', 'Forbes magazine', 'magazine'], ['NP-S-N-', 'Forbes', 'Forbes'], ['NC-S-N5', 'magazine', 'magazine'], ['WN-', "'s", "'s"], ['GN-S3---', 'annual list John Deere', 'John Deere'], ['GN-S3---', 'annual list', 'list'], ['AP-N5', 'annual', 'annual'], ['NC-S-N5', 'list', 'list'], ['GN-S3Y--', 'John Deere', 'John Deere'], ['NP-S-N-', 'John Deere', 'John Deere']]]
但entity_list id：
entity_list: [{ form: "John Deere", official_form: "Deere & Company", id: "d5250a54a8", sementity: { class: "instance", fiction: "nonfiction", id: "ODENTITY_INDUSTRIAL_COMPANY", type: "Top>Organization>Company>IndustrialCompany"
}
然后，当我打印时：
result = [['lema:',obj['lemma'], 'original_form', obj['original_form'], 'tag:',obj['tag']] for obj in gettuples(json_data)]
print(result)
我收到了这个错误：
File "/Users/user/PycharmProjects/Tests/test.py", line 51, in pos_tag2
result = [['lema:',obj['lemma'], 'original_form', obj['original_form'], 'tag:',obj['tag']] for obj in gettuples(json_data)]
File "/Users/user/PycharmProjects/Tests/test.py", line 51, in <listcomp>
result = [['lema:',obj['lemma'], 'original_form', obj['original_form'], 'tag:',obj['tag']] for obj in gettuples(json_data)]
KeyError: 'lemma'
所以，我的问题是：如何从（*）？获得4元组格式，或者我应该采用哪种其他方法来提取符合{{1}的4元组}，lema，original_form，以及当前是否tag？
更新2
或者，我尝试的另一件事是使用json_normalize：
在：
id输出：
from pandas.io.json import json_normalize
df = json_normalize(request, ['token_list',['token_list']])
df = pd.DataFrame(df)
df
然后：
affected_by_negation analysis_list endp form id inip quote_level separation style token_list type
0 no [{'lemma': '*', 'tag': 'Z-----------', 'origin... 4 Deere 6 0 0 _ {'isTitle': 'no', 'isItalics': 'no', 'isUnderl... [{'form': 'Deere', 'analysis_list': [{'lemma':... phrase
输出：
df_clean = df.drop(df.columns[[0, 2,4, 5, 6, 7, 8, 10]], axis=1)
df_clean
list(df_clean.itertuples(index=False))
然而，我在访问列表的特定值时遇到问题。另一个可能的解决方案可能是熊猫......如何做到这一点的任何想法？。
答案 0 :(得分：1)
以下代码应该可以满足您的需求。这不是最优雅的方法，但希望很清楚。
import yaml
from pprint import pprint
with open('json_dict.json', 'rU') as f:
data = yaml.load(f)
results = []
sementity_map = {}
def extract_analysis(l):
for d in l:
out = {
'lemma': d['lemma'],
'original_form': d['original_form'],
'tag': d['tag']
}
if 'sense_id_list' in d:
out['id'] = d['sense_id_list'][0]['sense_id']
results.append( out )
def extract_entities(l):
for d in l:
if 'sementity' in d and 'id' in d['sementity']:
sementity_map[ d['id'] ] = d['sementity']['id']
def find_analysis_and_entities(d):
if type(d) != dict: # Added for non-dict values
return # Fail
for k, v in d.items():
if type(v) == list:
if k == 'analysis_list':
extract_analysis(v)
elif k == 'entity_list':
extract_entities(v)
else:
for do in v:
find_analysis_and_entities(do)
else:
find_analysis_and_entities(v)
def apply_entities(e, m):
for d in e:
if 'id' in d:
if d['id'] in sementity_map:
d['id'] = sementity_map[ d['id'] ]
else:
del d['id']
find_analysis_and_entities(data)
apply_entities(results, sementity_map)
pprint(results)
对于语义ID，我们保留一个单独的地图字典，并在初始查找运行后应用它。第一个查找用于使用裸ID和语义实体映射构建结果。
部分问题（我认为）源于这样一个事实，即在找到必须应用的位置之前，您无法确定是否找到/传递了匹配的语义实体ID（使用dicts无效）。
这里我们只在找到它们时应用id映射，否则我们删除那个id字段。例如，a0a1a5401f和__12123288058840445720都未列在entity_list块中，因此会从results中删除。
上面输出的示例输入文件是：
[{'lemma': 'Robert Downey Jr',
'original_form': 'Robert Downey Jr',
'tag': 'NPUU-N-'},
{'lemma': 'Robert Downey Jr',
'original_form': 'Robert Downey Jr',
'tag': 'GNUS3S--'},
{'lemma': 'top', 'original_form': 'has topped', 'tag': 'VI-S3PPA-N-N9'},
{'id': 'ODENTITY_MAGAZINE',
'lemma': 'Forbes',
'original_form': 'Forbes',
'tag': 'NP-S-N-'},
{'lemma': 'magazine', 'original_form': 'magazine', 'tag': 'NC-S-N5'},
{'lemma': 'magazine', 'original_form': 'Forbes magazine', 'tag': 'GN-S3---'},
{'lemma': "'s", 'original_form': "'s", 'tag': 'WN-'},
{'lemma': 'annual', 'original_form': 'annual', 'tag': 'AP-N5'},
{'lemma': 'list', 'original_form': 'list', 'tag': 'NC-S-N5'},
{'lemma': 'list', 'original_form': 'annual list', 'tag': 'GN-S3---'},
{'id': 'ODENTITY_INDUSTRIAL_COMPANY',
'lemma': 'John Deere',
'original_form': 'John Deere',
'tag': 'NP-S-N-'},
{'lemma': 'John Deere', 'original_form': 'John Deere', 'tag': 'GN-S3Y--'},
{'lemma': 'John Deere',
'original_form': 'annual list John Deere',
'tag': 'GN-S3---'},
{'lemma': 'John Deere',
'original_form': "Forbes magazine's annual list John Deere",
'tag': 'GN-S3D--'},
{'lemma': '*',
'original_form': "Robert Downey Jr has topped Forbes magazine's annual list "
'John Deere',
'tag': 'Z-----------'}]