用索引值替换列的值

Question

我有这个数据帧：

<1-t><2-t><1>

索引	id	3-t
2022-01-06	0	5	4	2
2022-01-05	1	3	5
2022-01-04	2	4	3	5
2022-01-03	3	3	0
2022-01-02	4	2	3	0
2022-01-01	5	1	2	4

for nm in df.columns:
if nm.endswith("-t"):
tmp = df.Index[df.loc[:, nm]]
tmp.index = df.index
df.loc[:, nm] = tmp
df
#         Index  id         1-t         2-t         3-t
# 0  2022-01-06   0  2022-01-01  2022-01-02  2022-01-04
# 1  2022-01-05   1  2022-01-03  2022-01-01  2022-01-02
# 2  2022-01-04   2  2022-01-02  2022-01-03  2022-01-01
# 3  2022-01-03   3  2022-01-03  2022-01-06  2022-01-05
# 4  2022-01-02   4  2022-01-04  2022-01-03  2022-01-06
# 5  2022-01-01   5  2022-01-05  2022-01-04  2022-01-02

或

def foo(i, s):
ans = i[s]
ans.index = i.index
return ans
df.apply(lambda x: foo(df.Index, x) if x.name.endswith("-t") else x)

Answer 1

以下是使用replace():的另一种方法

d = dict(zip(df['id'],df['Index']))
df.replace({i:d for i in df.columns[df.columns.str.contains('-t')]})

输出：

Index  id         1-t         2-t         3-t
0  2022-01-06   0  2022-01-01  2022-01-02  2022-01-04
1  2022-01-05   1  2022-01-03  2022-01-01  2022-01-02
2  2022-01-04   2  2022-01-02  2022-01-03  2022-01-01
3  2022-01-03   3  2022-01-03  2022-01-06  2022-01-05
4  2022-01-02   4  2022-01-04  2022-01-03  2022-01-06
5  2022-01-01   5  2022-01-05  2022-01-04  2022-01-02

Answer 2

假设您的数据如下所示：

id  1-t  2-t  3-t
Index                        
2022-01-06   0    5    4    2
2022-01-05   1    3    5    4
2022-01-04   2    4    3    5
2022-01-03   3    3    0    1
2022-01-02   4    2    3    0
2022-01-01   5    1    2    4

即，您在上表中标记为Index的是Pandas数据帧的实际索引，您所需要做的就是使用Dataframe.filter例程，如下所示：

for col in data.filter(like='-t'):
data[col] = data.index[data[col]]
print(data)
#            id         1-t         2-t         3-t
#Index                                             
#2022-01-06   0  2022-01-01  2022-01-02  2022-01-04
#2022-01-05   1  2022-01-03  2022-01-01  2022-01-02
#2022-01-04   2  2022-01-02  2022-01-03  2022-01-01
#2022-01-03   3  2022-01-03  2022-01-06  2022-01-05
#2022-01-02   4  2022-01-04  2022-01-03  2022-01-06
#2022-01-01   5  2022-01-05  2022-01-04  2022-01-02

甚至可能有一种方法可以同时替换所有列。如果Index只是列的名称，请将data.index替换为data.Index。

编辑：我忘了在列中使用索引值。现在应该可以了。

Answer 3

这对我有效：

for i in range(len(df)):
df.loc[i,"1-t"]=df.loc[df.loc[i,"1-t"],"Index"]
df.loc[i,"2-t"]=df.loc[df.loc[i,"2-t"],"Index"]
df.loc[i,"3-t"]=df.loc[df.loc[i,"3-t"],"Index"]