我正在尝试将json文件加载到熊猫数据框。 我发现有一些嵌套的json。 下面是示例json:
ChildComponentTwo
我使用以下代码将json加载到数据框中:
AnotherComponent
下面是df.head()的输出
但是我发现了两个嵌套的列,例如位置和标签。
我尝试使用以下代码对其进行展平:
{'events': [{'id': 142896214,
'playerId': 37831,
'teamId': 3157,
'matchId': 2214569,
'matchPeriod': '1H',
'eventSec': 0.8935539999999946,
'eventId': 8,
'eventName': 'Pass',
'subEventId': 85,
'subEventName': 'Simple pass',
'positions': [{'x': 51, 'y': 49}, {'x': 40, 'y': 53}],
'tags': [{'id': 1801, 'tag': {'label': 'accurate'}}]}
它向我显示了以下错误:
with open('EVENTS.json') as f:
jsonstr = json.load(f)
df = pd.io.json.json_normalize(jsonstr['events'])
您能建议我如何展平位置和标签(那些具有嵌套数据的位置和标签)吗?
谢谢, 压缩
答案 0 :(得分:5)
如果您正在寻找一种更通用的方法来从json展开多个层次结构,则可以使用recursion
并列出理解来重塑数据。下面介绍了一种替代方法:
def flatten_json(nested_json, exclude=['']):
"""Flatten json object with nested keys into a single level.
Args:
nested_json: A nested json object.
exclude: Keys to exclude from output.
Returns:
The flattened json object if successful, None otherwise.
"""
out = {}
def flatten(x, name='', exclude=exclude):
if type(x) is dict:
for a in x:
if a not in exclude: flatten(x[a], name + a + '_')
elif type(x) is list:
i = 0
for a in x:
flatten(a, name + str(i) + '_')
i += 1
else:
out[name[:-1]] = x
flatten(nested_json)
return out
然后,您可以应用于数据,而与嵌套级别无关:
新示例数据
this_dict = {'events': [
{'id': 142896214,
'playerId': 37831,
'teamId': 3157,
'matchId': 2214569,
'matchPeriod': '1H',
'eventSec': 0.8935539999999946,
'eventId': 8,
'eventName': 'Pass',
'subEventId': 85,
'subEventName': 'Simple pass',
'positions': [{'x': 51, 'y': 49}, {'x': 40, 'y': 53}],
'tags': [{'id': 1801, 'tag': {'label': 'accurate'}}]},
{'id': 142896214,
'playerId': 37831,
'teamId': 3157,
'matchId': 2214569,
'matchPeriod': '1H',
'eventSec': 0.8935539999999946,
'eventId': 8,
'eventName': 'Pass',
'subEventId': 85,
'subEventName': 'Simple pass',
'positions': [{'x': 51, 'y': 49}, {'x': 40, 'y': 53},{'x': 51, 'y': 49}],
'tags': [{'id': 1801, 'tag': {'label': 'accurate'}}]}
]}
用法
pd.DataFrame([flatten_json(x) for x in this_dict['events']])
Out[1]:
id playerId teamId matchId matchPeriod eventSec eventId \
0 142896214 37831 3157 2214569 1H 0.893554 8
1 142896214 37831 3157 2214569 1H 0.893554 8
eventName subEventId subEventName positions_0_x positions_0_y \
0 Pass 85 Simple pass 51 49
1 Pass 85 Simple pass 51 49
positions_1_x positions_1_y tags_0_id tags_0_tag_label positions_2_x \
0 40 53 1801 accurate NaN
1 40 53 1801 accurate 51.0
positions_2_y
0 NaN
1 49.0
答案 1 :(得分:2)
data = {'events': [{'id': 142896214,
'playerId': 37831,
'teamId': 3157,
'matchId': 2214569,
'matchPeriod': '1H',
'eventSec': 0.8935539999999946,
'eventId': 8,
'eventName': 'Pass',
'subEventId': 85,
'subEventName': 'Simple pass',
'positions': [{'x': 51, 'y': 49}, {'x': 40, 'y': 53}],
'tags': [{'id': 1801, 'tag': {'label': 'accurate'}}]}]}
创建数据框
df = pd.DataFrame.from_dict(data)
df = df['events'].apply(pd.Series)
将positions
与pd.Series
一起
df_p = df['positions'].apply(pd.Series)
df_p_0 = df_p[0].apply(pd.Series)
df_p_1 = df_p[1].apply(pd.Series)
重命名positions[0]
和positions[1]
:
df_p_0.columns = ['pos_0_x', 'pos_0_y']
df_p_1.columns = ['pos_1_x', 'pos_1_y']
将tags
与pd.Series
对齐:
df_t = df.tags.apply(pd.Series)
df_t = df_t[0].apply(pd.Series)
df_t_t = df_t.tag.apply(pd.Series)
重命名id
和label
:
df_t = df_t.rename(columns={'id': 'tags_id'})
df_t_t.columns = ['tags_tag_label']
使用pd.concat
将它们全部合并:
df_new = pd.concat([df, df_p_0, df_p_1, df_t.tags_id, df_t_t], axis=1)
删除旧列:
df_new = df_new.drop(['positions', 'tags'], axis=1)