我有一个数据框，如下所示
ID Name Address
1 Kohli Country: India; State: Delhi; Sector: SE25
2 Sachin Country: India; State: Mumbai; Sector: SE39
3 Ponting Country: Australia; State: Tasmania
4 Ponting State: Tasmania; Sector: SE27
从上面我想准备下面的数据框
ID Name Country State Sector
1 Kohli India Delhi SE25
2 Sachin India Mumbai SE39
3 Ponting Australia Tasmania None
4 Ponting None Tasmania SE27
我尝试了下面的代码
df[['Country', 'State', 'Sector']] = pd.DataFrame(df['ADDRESS'].str.split(';',2).tolist(),
columns = ['Country', 'State', 'Sector'])
但从上面看来，我必须通过对列进行切片来清理数据。我想知道有没有比这更简单的方法。
使用列表理解和字典理解来获取字典列表并传递给DataFrame构造函数：
L = [{k:v for y in x.split('; ') for k, v in dict([y.split(': ')]).items()}
for x in df.pop('Address')]
df = df.join(pd.DataFrame(L, index=df.index))
print (df)
ID Name Country State Sector
0 1 Kohli India Delhi SE25
1 2 Sachin India Mumbai SE39
2 3 Ponting Australia Tasmania NaN
Or use split重塑stack:
df1 = (df.pop('Address')
.str.split('; ', expand=True)
.stack()
.reset_index(level=1, drop=True)
.str.split(': ', expand=True)
.set_index(0, append=True)[1]
.unstack()
)
print (df1)
0 Country Sector State
0 India SE25 Delhi
1 India SE39 Mumbai
2 Australia NaN Tasmania
df = df.join(df1)
print (df)
ID Name Country Sector State
0 1 Kohli India SE25 Delhi
1 2 Sachin India SE39 Mumbai
2 3 Ponting Australia NaN Tasmania
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