我有一个DataFrame,其中包含一个带字典的特定列。
我想在DataFrame中为包含dicts的列中的每个元素上找到的每个键添加一个新标头,如果该元素不包含,则分配给这些新单元格的每个新值应对应f = open('data.txt', 'r')
data_txt = f.read()
data_txt = data_txt.replace('[', '')
data_txt = data_txt.replace(']', '')
np.loadtxt(StringIO.StringIO(data_txt), delimiter = '\n')
那个标题密钥和相应的密钥值。
以下是测试和可视化我所说内容的数据:
导入依赖项:
None
创建包含内部字典列表的字典:
import pandas as pd
import numpy as np
为测试创建适当的初始DataFrame:
data = {'string_info': ['User1', 'User2', 'User3'],
'dict_info': [{'elm1': 'attr5', 'elm2': 'attr9', 'elm3': 'attr33'},
{'elm5': 'attr31', 'elm7': 'attr13'},
{'elm5': 'attr28', 'elm1': 'attr23', 'elm2': 'attr33','elm6': 'attr33'}],
'int_info': [4, 24, 31],}
手动说明我想要的输出:
df = pd.DataFrame.from_dict(data)
df
所需的输出是:
data2 = {'string_info': ['User1', 'User2', 'User3'],
'elm1': ['attr5',None,'attr23'],
'elm2': ['attr9',None,'attr33'],
'elm3': ['attr33',None,None],
'elm4': [None,None,None],
'elm5': [None,'attr31',None],
'elm6': [None,None,'attr33'],
'elm7': [None,None,'attr13'],
'int_info': [4, 24, 31]}
谢谢!
答案 0 :(得分:1)
您可以将concat
与DataFrame
构造函数一起使用,以将dict
替换为列:
print (pd.DataFrame(df.dict_info.values.tolist()))
elm1 elm2 elm3 elm5 elm6 elm7
0 attr5 attr9 attr33 NaN NaN NaN
1 NaN NaN NaN attr31 NaN attr13
2 attr23 attr33 NaN attr28 attr33 NaN
print (pd.concat([pd.DataFrame(df.dict_info.values.tolist()),
df[['int_info','string_info']]], axis=1))
elm1 elm2 elm3 elm5 elm6 elm7 int_info string_info
0 attr5 attr9 attr33 NaN NaN NaN 4 User1
1 NaN NaN NaN attr31 NaN attr13 24 User2
2 attr23 attr33 NaN attr28 attr33 NaN 31 User3
如果需要None
添加replace
:
print (pd.concat([pd.DataFrame(df.dict_info.values.tolist()).replace({np.nan:None}),
df[['int_info','string_info']]], axis=1))
elm1 elm2 elm3 elm5 elm6 elm7 int_info string_info
0 attr5 attr9 attr33 None None None 4 User1
1 None None None attr31 None attr13 24 User2
2 attr23 attr33 None attr28 attr33 None 31 User3