轉換一個字典的熊貓數據幀

我的數據是這樣的：轉換一個字典的熊貓數據幀

{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"}'}

我想將其轉換爲大熊貓數據幀。但是，當我嘗試

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

我與兩列的數據幀時，先用第一個關鍵，第二個與鍵的值：

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

我怎樣才能得到一個列對於每個鍵："ad_since","indoor_swimming_pool","indoor_swimming_pool"？並保留第一列，或者將id作爲索引。

來源

2016-09-20 mitsi

嘗試read_json http://pandas.pydata.org/pandas-docs/version/0.17.0/generated/pandas.read_json.html –

您是否嘗試使用'pd.DataFrame（response.items（））' ？對我來說，它不工作。 – jezrael

@jezrael感謝您的評論，我編輯我的帖子 – mitsi

您需要通過.apply(literal_eval)或.apply(json.loads)轉換的typestr列dict然後用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

來源

2016-09-20 13:01:07 jezrael

第一種方法（使用'literal_eval'）和整個數據集，我得到錯誤'ValueError：格式不正確的字符串'它可能是由於特殊字符。但它與'json.loads'的第二種方法完美結合，謝謝 – mitsi

很高興能爲您提供幫助。 – jezrael

由於值是字符串，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

來源

2016-09-20 13:07:58

轉換一個字典的熊貓數據幀

回答

相關問題