将字典转换为Pandas DataFrame

mitsi 发表于 Dev

105

三井

我的数据如下所示：

{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"}'}

我想将其转换为熊猫数据框。但是当我尝试

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

我得到一个包含两列的DataFrame，第一列包含第一个键，第二列包含键的值：

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

我怎样才能得到一个列每个键："ad_since"，"indoor_swimming_pool"，"indoor_swimming_pool"？并保留第一列，或获取ID作为索引。

耶斯列尔

你需要的列转换type str，以dict通过.apply(literal_eval)或.apply(json.loads)再使用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

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