我有一个names = ['id','t','metric_1','metric_2','metric_3']的数据框。我正在每个grp in groupby('id')上运行一些信号处理。我需要扭转整个数据框中的另一个进程的时间，并在引擎盖下进行处理。简单地说，给定grp，我只需要反转时间列，保留所有其他列的完整性，并且grp内的所有行都不完整。
输入数据帧：
id t metric_1 metric_2 metric_3
0 0 86 13.333 61.989 0.017444
1 0 87 13.333 61.993 0.017569
2 0 88 13.333 61.992 0.017711
3 0 89 13.333 61.998 0.017746
4 0 90 13.333 61.993 0.017871
5 1 32 13.333 61.964 0.018511
6 1 33 20.000 61.913 0.020058
7 1 34 20.000 61.864 0.022475
8 1 35 26.667 61.802 0.025995
9 1 36 33.123 61.563 0.032345
10 1 37 33.763 61.836 0.060189
11 2 2 13.333 61.964 0.018511
12 2 3 20.000 61.613 0.020058
13 2 4 20.000 61.164 0.027475
14 2 5 26.667 61.802 0.024995
15 2 6 33.333 61.736 0.030689
我想使用产生如下数据帧的操作：
id t metric_1 metric_2 metric_3
0 0 90 13.333 61.989 0.017444
1 0 89 13.333 61.993 0.017569
2 0 88 13.333 61.992 0.017711
3 0 87 13.333 61.998 0.017746
4 0 86 13.333 61.993 0.017871
5 1 37 13.333 61.964 0.018511
6 1 36 20.000 61.913 0.020058
7 1 35 20.000 61.864 0.022475
8 1 34 26.667 61.802 0.025995
9 1 33 33.333 61.736 0.030689
10 1 32 33.763 61.836 0.060189
11 2 6 13.333 61.964 0.018511
12 2 5 20.000 61.613 0.020058
13 2 4 20.000 61.164 0.027475
14 2 3 26.667 61.802 0.024995
15 2 2 33.333 61.736 0.030689
答案 0 :(得分：4)
UPDATE2：排序/替换t列中的值，但仅适用于id == 0（as described in this comment）的行：
In [373]: df
Out[373]:
id t metric_1 metric_2 metric_3
0 0 86 13.333 61.989 0.017444
1 0 87 13.333 61.993 0.017569
2 0 88 13.333 61.992 0.017711
3 0 89 13.333 61.998 0.017746
4 0 90 13.333 61.993 0.017871
5 1 86 13.333 61.964 0.018511
6 1 87 20.000 61.913 0.020058
7 1 88 20.000 61.864 0.022475
8 1 89 26.667 61.802 0.025995
9 1 90 33.333 61.736 0.030689
In [374]: df.loc[df.id == 0, 't'] = df.loc[df.id == 0, 't'].sort_values(ascending=0).values
In [375]: df
Out[375]:
id t metric_1 metric_2 metric_3
0 0 90 13.333 61.989 0.017444
1 0 89 13.333 61.993 0.017569
2 0 88 13.333 61.992 0.017711
3 0 87 13.333 61.998 0.017746
4 0 86 13.333 61.993 0.017871
5 1 86 13.333 61.964 0.018511
6 1 87 20.000 61.913 0.020058
7 1 88 20.000 61.864 0.022475
8 1 89 26.667 61.802 0.025995
9 1 90 33.333 61.736 0.030689
更新
原创DF：
In [363]: df
Out[363]:
id t metric_1 metric_2 metric_3
0 0 86 13.333 61.989 0.017444
1 0 87 13.333 61.993 0.017569
2 0 88 13.333 61.992 0.017711
3 0 89 13.333 61.998 0.017746
4 0 90 13.333 61.993 0.017871
5 1 86 13.333 61.964 0.018511
6 1 87 20.000 61.913 0.020058
7 1 88 20.000 61.864 0.022475
8 1 89 26.667 61.802 0.025995
9 1 90 33.333 61.736 0.030689
排序完整的行：
In [364]: df.sort_values(['id','t'], ascending=[1,0])
Out[364]:
id t metric_1 metric_2 metric_3
4 0 90 13.333 61.993 0.017871
3 0 89 13.333 61.998 0.017746
2 0 88 13.333 61.992 0.017711
1 0 87 13.333 61.993 0.017569
0 0 86 13.333 61.989 0.017444
9 1 90 33.333 61.736 0.030689
8 1 89 26.667 61.802 0.025995
7 1 88 20.000 61.864 0.022475
6 1 87 20.000 61.913 0.020058
5 1 86 13.333 61.964 0.018511 # <--
为两列（['id','t']）排序值，替换它们的值：
In [366]: df[['id','t']] = df[['id','t']].sort_values(['id','t'], ascending=[1,0]).values
In [367]: df
Out[367]:
id t metric_1 metric_2 metric_3
0 0 90 13.333 61.989 0.017444
1 0 89 13.333 61.993 0.017569
2 0 88 13.333 61.992 0.017711
3 0 87 13.333 61.998 0.017746
4 0 86 13.333 61.993 0.017871
5 1 90 13.333 61.964 0.018511
6 1 89 20.000 61.913 0.020058
7 1 88 20.000 61.864 0.022475
8 1 87 26.667 61.802 0.025995
9 1 86 33.333 61.736 0.030689 # <--
OLD回答：
IIUC您只需按两列对数据进行排序：
In [349]: df.sort_values(['id','t'], ascending=[1,1])
Out[349]:
id t metric_1 metric_2 metric_3
4 0 86 13.333 61.993 0.017871
3 0 87 13.333 61.998 0.017746
2 0 88 13.333 61.992 0.017711
1 0 89 13.333 61.993 0.017569
0 0 90 13.333 61.989 0.017444
9 1 86 33.333 61.736 0.030689
8 1 87 26.667 61.802 0.025995
7 1 88 20.000 61.864 0.022475
6 1 89 20.000 61.913 0.020058
5 1 90 13.333 61.964 0.018511
如果您想按照所需的数据集对其进行排序（替换t列值）：
In [357]: df[['id','t']] = df[['id','t']].sort_values(['id','t']).values
In [358]: df
Out[358]:
id t metric_1 metric_2 metric_3
0 0 86 13.333 61.989 0.017444
1 0 87 13.333 61.993 0.017569
2 0 88 13.333 61.992 0.017711
3 0 89 13.333 61.998 0.017746
4 0 90 13.333 61.993 0.017871
5 1 86 13.333 61.964 0.018511
6 1 87 20.000 61.913 0.020058
7 1 88 20.000 61.864 0.022475
8 1 89 26.667 61.802 0.025995
9 1 90 33.333 61.736 0.030689 # 1 90 33.333 61.736 0.030689 as in your desired DF
答案 1 :(得分：0)
如果你想扭转“反对”的话。列保持所有其他列不变，您可以尝试以下代码：
df.t=df['t'].sort_values(ascending=False)