[英]Create a column in pandas dataframe
我有一个 dataframe 如下:
df = pd.DataFrame({'ORDER':["A", "A", "A", "B", "B","B"], 'GROUP': ["A1C", "A1", "B1", "B1C", "M1", "M1C"]})
df['_A1_XYZ'] = 1
df['_A1C_XYZ'] = 2
df['_B1_XYZ'] = 3
df['_B1C_XYZ'] = 4
df['_M1_XYZ'] = 5
df
ORDER GROUP _A1_XYZ _A1C_XYZ _B1_XYZ _B1C_XYZ _M1_XYZ
0 A A1C 1 2 3 4 5
1 A A1 1 2 3 4 5
2 A B1 1 2 3 4 5
3 B B1C 1 2 3 4 5
4 B M1 1 2 3 4 5
5 B M1C 1 2 3 4 5
我想根据列“GROUP”和所有以XYZ 结尾的列创建一个列“NEW”,如下所示:基于每行 df["NEW"] = df["_XYZ"] 的 GROUP 值。
例如,对于第一行,GROUP = A1C,所以“NEW”= 2 (_A1C_XYZ),类似地对于第二行“NEW” = 1 (_A1_XYZ)
我的预期 output
ORDER GROUP _A1_XYZ _A1C_XYZ _B1_XYZ _B1C_XYZ _M1_XYZ NEW
0 A A1C 1 2 3 4 5 2
1 A A1 1 2 3 4 5 1
2 A B1 1 2 3 4 5 3
3 B B1C 1 2 3 4 5 4
4 B M1 1 2 3 4 5 5
5 B M1C 1 2 3 4 5
使用pd.DataFrame.lookup
:
df['NEW'] = df.lookup(df.index, '_'+df['GROUP']+'_XYZ')
df
Output:
ORDER GROUP _A1_XYZ _A1C_XYZ _B1_XYZ _B1C_XYZ _M1_XYZ _M1C_XYZ NEW
0 A A1C 1 2 3 4 5 6 2
1 A A1 1 2 3 4 5 6 1
2 A B1 1 2 3 4 5 6 3
3 B B1C 1 2 3 4 5 6 4
4 B M1 1 2 3 4 5 6 5
5 B M1C 1 2 3 4 5 6 6
或者使用堆栈和重新索引,
(df['New'] = df.stack().reindex(zip(df.index, '_'+dfl['GROUP']+'_XYZ'))
.rename('NEW').reset_index(level=1, drop=True))
df
Output:
ORDER GROUP _A1_XYZ _A1C_XYZ _B1_XYZ _B1C_XYZ _M1_XYZ New
0 A A1C 1 2 3 4 5 2
1 A A1 1 2 3 4 5 1
2 A B1 1 2 3 4 5 3
3 B B1C 1 2 3 4 5 4
4 B M1 1 2 3 4 5 5
5 B M1C 1 2 3 4 5 NaN
如果行中的所有值也是列,@ScottBoston 的答案会更好,但我想我会分享我的,本质上,我用相关列创建一个新的 dataframe,删除重复项,更改列名。 转置 dataframe 并将列合并回...
a = df.iloc[:,2:].drop_duplicates()
a.columns = [col.split('_')[1] for col in df.columns if '_' in col]
a = a.T.rename({0:'NEW'}, axis=1)
df = pd.merge(df, a, how='left', left_on='GROUP', right_index=True)
df
output:
ORDER GROUP _A1_XYZ _A1C_XYZ _B1_XYZ _B1C_XYZ _M1_XYZ NEW
0 A A1C 1 2 3 4 5 2.0
1 A A1 1 2 3 4 5 1.0
2 A B1 1 2 3 4 5 3.0
3 B B1C 1 2 3 4 5 4.0
4 B M1 1 2 3 4 5 5.0
5 B M1C 1 2 3 4 5 NaN
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