Question

我的数据如下：

{u'"57e01311817bc367c030b390"': u'{"ad_since": 2016, "indoor_swimming_pool": "No", "seaside": "No", "handicapped_access": "Yes"}', u'"57e01311817bc367c030b3a8"': u'{"ad_since": 2012, "indoor_swimming_pool": "No", "seaside": "No", "handicapped_access": "Yes"}'}

我想将其转换为pandas Dataframe。但是当我尝试

时

df = pd.DataFrame(response.items())

我得到一个包含两列的DataFrame，第一列带有第一个键，第二列带有键值：

                            0                       1 
0  "57e01311817bc367c030b390"   {"ad_since": 2016, "indoor_swimming_pool": "No...
1  "57e01311817bc367c030b3a8"   {"ad_since": 2012, "indoor_swimming_pool": "No...

如何为每个密钥获取一列："ad_since"，"indoor_swimming_pool"，"indoor_swimming_pool"？并保留第一列，或将id作为索引。

Answer 1

您需要按type或str将dict .apply(literal_eval)列转换为.apply(json.loads)，然后使用DataFrame.from_records：

import pandas as pd
from ast import literal_eval

response = {u'"57e01311817bc367c030b390"': u'{"ad_since": 2016, "indoor_swimming_pool": "No", "seaside": "No", "handicapped_access": "Yes"}', 
           u'"57e01311817bc367c030b3a8"': u'{"ad_since": 2012, "indoor_swimming_pool": "No", "seaside": "No", "handicapped_access": "Yes"}'}

df = pd.DataFrame.from_dict(response, orient='index')

print (type(df.iloc[0,0]))
<class 'str'>

df.iloc[:,0] = df.iloc[:,0].apply(literal_eval)

print (pd.DataFrame.from_records(df.iloc[:,0].values.tolist(), index=df.index))
                            ad_since handicapped_access indoor_swimming_pool  \
"57e01311817bc367c030b3a8"      2012                Yes                   No   
"57e01311817bc367c030b390"      2016                Yes                   No   

                           seaside  
"57e01311817bc367c030b3a8"      No  
"57e01311817bc367c030b390"      No

import pandas as pd
import json

response = {u'"57e01311817bc367c030b390"': u'{"ad_since": 2016, "indoor_swimming_pool": "No", "seaside": "No", "handicapped_access": "Yes"}', 
           u'"57e01311817bc367c030b3a8"': u'{"ad_since": 2012, "indoor_swimming_pool": "No", "seaside": "No", "handicapped_access": "Yes"}'}


df = pd.DataFrame.from_dict(response, orient='index')
df.iloc[:,0] = df.iloc[:,0].apply(json.loads)


print (pd.DataFrame.from_records(df.iloc[:,0].values.tolist(), index=df.index))
                            ad_since handicapped_access indoor_swimming_pool  \
"57e01311817bc367c030b3a8"      2012                Yes                   No   
"57e01311817bc367c030b390"      2016                Yes                   No   

                           seaside  
"57e01311817bc367c030b3a8"      No  
"57e01311817bc367c030b390"      No

Answer 2

由于值是字符串，您可以使用json module和列表理解：

In [20]: d =     {u'"57e01311817bc367c030b390"': u'{"ad_since": 2016, "indoor_swimming_pool": "No", "seaside": "No", "handicapped_access": "Yes"}', u'"57e01311817bc367c030b3a8"': u'{"ad_since": 2012, "indoor_swimming_pool": "No", "seaside": "No", "handicapped_access": "Yes"}'}

In [21]: import json

In [22]: pd.DataFrame(dict([(k, [json.loads(e)[k] for e in d.values()]) for k in json.loads(d.values()[0])]), index=d.keys())Out[22]: 
                            ad_since handicapped_access indoor_swimming_pool  \
"57e01311817bc367c030b390"      2016                Yes                   No   
"57e01311817bc367c030b3a8"      2012                Yes                   No   

                       seaside  
"57e01311817bc367c030b390"      No  
"57e01311817bc367c030b3a8"      No

将dict转换为pandas DataFrame

2 个答案: