Question

这是数据帧。其中包含一些带有字典的单元格。我想将字典项目转换为列

dfx={'name':['Alex','Jin',np.nan,'Peter'],
     'age':[np.nan,10,12,13],
     'other':[{'school':'abc','subject':'xyz'},
     np.nan,
     {'school':'abc','subject':'xyz'},
     np.nan,]
     }

dfx=pd.DataFrame(dfx)

输出

name    age        other
Alex             {'school': 'abc', 'subject': 'xyz'}
Jin     10.0    
        12.0     {'school': 'abc', 'subject': 'xyz'}
Peter   13.0

这是所需的输出

name    age      school    subject
Alex             abc         xyz
Jin     10.0        
        12.0     abc         xyz
Peter   13.0

Answer 1

您可以使用.str.get访问器来实际索引列中的字典。每当单元格值为nan而不是字典时，它也会返回nan：

clean_df = (dfx
            .assign(
               school=lambda df: df["other"].str.get("school"),
               subject=lambda df: df["other"].str.get("subject"))
            .drop("other", axis=1))

print(clean_df)
    name   age school subject
0   Alex   NaN    abc     xyz
1    Jin  10.0    NaN     NaN
2    NaN  12.0    abc     xyz
3  Peter  13.0    NaN     NaN

Answer 2

尝试一下

df_final = dfx[['name','age']].assign(**pd.DataFrame(dfx.other.to_dict()).T)

Out[41]:
    name   age school subject
0   Alex   NaN    abc     xyz
1    Jin  10.0    NaN     NaN
2    NaN  12.0    abc     xyz
3  Peter  13.0    NaN     NaN

Answer 3

从dictionary的{{1}}和dfx创建一个index。 other字典和pd.DataFrame。这将为您提供一个新的transpose。将产生的dataframe连接到dfx的前两列。

dataframe

Answer 4

您可以将Series应用于带有字典的列：

df.drop('other', 1).join(df['other'].apply(pd.Series).drop(0, 1))

输出：

    name   age school subject
0   Alex   NaN    abc     xyz
1    Jin  10.0    NaN     NaN
2    NaN  12.0    abc     xyz
3  Peter  13.0    NaN     NaN

使用新列pandas将包含字典的数据框单元格转换为数据框

4 个答案: