pandas 在熊猫数据框中展平嵌套的 Json
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flattening nested Json in pandas data frame
提问by Zephyr
I am trying to load the json file to pandas data frame. I found that there were some nested json. Below is the sample json:
我正在尝试将 json 文件加载到 Pandas 数据框。我发现有一些嵌套的json。下面是示例 json：
{'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'}}]}
I used the following code to load json into dataframe:
我使用以下代码将 json 加载到数据帧中：
with open('EVENTS.json') as f:
jsonstr = json.load(f)
df = pd.io.json.json_normalize(jsonstr['events'])
Below is the output of df.head()
下面是 df.head() 的输出
But I found two nested columns such as positions and tags.
但是我发现了两个嵌套的列，例如位置和标签。
I tried using the following code to flatten it:
我尝试使用以下代码将其展平：
Position_data = json_normalize(data =jsonstr['events'], record_path='positions', meta = ['x','y','x','y'] )
It showed me an error as follow:
它向我显示了如下错误：
KeyError: "Try running with errors='ignore' as key 'x' is not always present"
Can you advise me how to flatten positions and tags ( those having nested data).
你能告诉我如何展平位置和标签（那些有嵌套数据的）。
Thanks, Zep
谢谢，泽普
回答by calestini
If you are looking for a more general way to unfold multiple hierarchies from a json you can use recursionand list comprehension to reshape your data. One alternative is presented below:
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，您可以使用recursion和列表理解来重塑您的数据。下面介绍了一种替代方案：
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
Then you can apply to your data, independent of nested levels:
然后您可以应用到您的数据，独立于嵌套级别：
New sample data
新样本数据
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'}}]}
]}
Usage
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
Note that this flatten_jsoncode is not mine, I have seen it hereand herewithout much certainty of the original source.
回答by Trenton McKinney
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'}}]}]}
Create the DataFrame
创建数据框
df = pd.DataFrame.from_dict(data)
df = df['events'].apply(pd.Series)
Flatten positionswith pd.Series
拼合positions与pd.Series
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)
Rename positions[0]& positions[1]:
重命名positions[0]& positions[1]:
df_p_0.columns = ['pos_0_x', 'pos_0_y']
df_p_1.columns = ['pos_1_x', 'pos_1_y']
Flatten tagswith pd.Series:
拼合tags有pd.Series：
df_t = df.tags.apply(pd.Series)
df_t = df_t[0].apply(pd.Series)
df_t_t = df_t.tag.apply(pd.Series)
Rename id& label:
重命名id& label:
df_t = df_t.rename(columns={'id': 'tags_id'})
df_t_t.columns = ['tags_tag_label']
Combine them all with pd.concat:
将它们全部与pd.concat：
df_new = pd.concat([df, df_p_0, df_p_1, df_t.tags_id, df_t_t], axis=1)
Drop the old columns:
删除旧列：
df_new = df_new.drop(['positions', 'tags'], axis=1)
