我有一个json数据(来自mongodb),包含数千个记录(所以json对象的数组/列表),每个对象的结构如下所示:
{
"id":1,
"first_name":"Mead",
"last_name":"Lantaph",
"email":"mlantaph0@opensource.org",
"gender":"Male",
"ip_address":"231.126.209.31",
"nested_array_to_expand":[
{
"property":"Quaxo",
"json_obj":{
"prop1":"Chevrolet",
"prop2":"Mercy Streets"
}
},
{
"property":"Blogpad",
"json_obj":{
"prop1":"Hyundai",
"prop2":"Flashback"
}
},
{
"property":"Yabox",
"json_obj":{
"prop1":"Nissan",
"prop2":"Welcome Mr. Marshall (Bienvenido Mister Marshall)"
}
}
]
}
在数据框中加载" nested_array_to_expand"是一个包含json的字符串(我使用" json_normalize"在加载期间)。预期的结果是获得一个包含3行的数据帧(给出上面的示例)和嵌套对象的新列,如下所示:
index email first_name gender id ip_address last_name \
0 mlantaph0@opensource.org Mead Male 1 231.126.209.31 Lantaph
1 mlantaph0@opensource.org Mead Male 1 231.126.209.31 Lantaph
2 mlantaph0@opensource.org Mead Male 1 231.126.209.31 Lantaph
test.name test.obj.ahah test.obj.buzz
0 Quaxo Mercy Streets Chevrolet
1 Blogpad Flashback Hyundai
2 Yabox Welcome Mr. Marshall (Bienvenido Mister Marshall) Nissan
我能够通过以下函数获得该结果,但速度非常慢(1k记录大约2s),所以我想改进现有代码或者找到一种完全不同的方法来获得这个结果。
def expand_field(field, df, parent_id='id'):
all_sub = pd.DataFrame()
# we need an id per row to be able to merge back dataframes
# if no id, then we will create one based on index of rows
if parent_id not in df:
df[parent_id] = df.index
# go through all rows and create a new dataframe with values
for i, row in df.iterrows():
try:
sub = json_normalize(df[field].values[i])
sub = sub.add_prefix(field + '.')
sub['parent_id'] = row[parent_id]
all_sub = all_sub.append(sub)
except:
print('crash')
pass
df = pd.merge(df, all_sub, left_on=parent_id, right_on='parent_id', how='left')
#remove old columns
del df["parent_id"]
del df[field]
#return expanded dataframe
return df
非常感谢你的帮助。
=====编辑回答评论====
从mongodb加载的数据是一个对象数组。 我用以下代码加载它:
data = json.loads(my_json_string)
df = json_normalize(data)
输出给我一个df [" nested_array_to_expand"]的数据帧作为dtype对象(字符串)
0 [{'property': 'Quaxo', 'json_obj': {'prop1': '...
Name: nested_array_to_expand, dtype: object
答案 0 :(得分:4)
我提出一个有趣的答案,我认为使用pandas.io.json.json_normalize
查看documentation。我用它来扩展嵌套的json - 也许有更好的方法,但你明确地应该考虑使用这个功能。然后你只需要根据需要重命名列。
import io
from pandas.io.json import json_normalize
# Loading the json string into a structure
json_dict = json.load(io.StringIO(json_str))
df = pd.concat([pd.DataFrame(json_dict),
json_normalize(json_dict['nested_array_to_expand'])],
axis=1).drop('nested_array_to_expand', 1)
答案 1 :(得分:1)
下面的代码是您想要的。您可以使用python的内置列表功能展开嵌套列表,并将其作为新数据框传递。
pd.DataFrame(list(json_dict['nested_col']))
您可能必须对此进行多次迭代,具体取决于数据的嵌套方式。
from pandas.io.json import json_normalize
df= pd.concat([pd.DataFrame(json_dict), pd.DataFrame(list(json_dict['nested_array_to_expand']))], axis=1).drop('nested_array_to_expand', 1)
答案 2 :(得分:0)
import pandas as pd
import json
data = '''
[
{
"id":1,
"first_name":"Mead",
"last_name":"Lantaph",
"email":"mlantaph0@opensource.org",
"gender":"Male",
"ip_address":"231.126.209.31",
"nested_array_to_expand":[
{
"property":"Quaxo",
"json_obj":{
"prop1":"Chevrolet",
"prop2":"Mercy Streets"
}
},
{
"property":"Blogpad",
"json_obj":{
"prop1":"Hyundai",
"prop2":"Flashback"
}
},
{
"property":"Yabox",
"json_obj":{
"prop1":"Nissan",
"prop2":"Welcome Mr. Marshall (Bienvenido Mister Marshall)"
}
}
]
}
]
'''
data = json.loads(data)
result = pd.json_normalize(data, "nested_array_to_expand",
['email', 'first_name', 'gender', 'id', 'ip_address', 'last_name'])
结果
property json_obj.prop1 json_obj.prop2 \
0 Quaxo Chevrolet Mercy Streets
1 Blogpad Hyundai Flashback
2 Yabox Nissan Welcome Mr. Marshall (Bienvenido Mister Marshall)
email first_name gender id ip_address last_name
0 mlantaph0@opensource.org Mead Male 1 231.126.209.31 Lantaph
1 mlantaph0@opensource.org Mead Male 1 231.126.209.31 Lantaph
2 mlantaph0@opensource.org Mead Male 1 231.126.209.31 Lantaph
有关json_normalize
的更多信息:
https://pandas.pydata.org/docs/reference/api/pandas.json_normalize.html