将多列转换为 interpolate/copy 个缺失值
transform on multiple columns to interpolate/copy missing values
我正在尝试通过插入或复制组内的最后一个已知值(由 trip
标识)来填充 pandas 数据框中的缺失值。我的数据如下所示:
brake speed trip
0 0.0 NaN 1
1 1.0 NaN 1
2 NaN 1.264 1
3 NaN 0.000 1
4 0.0 NaN 1
5 NaN 1.264 1
6 NaN 6.704 1
7 1.0 NaN 1
8 0.0 NaN 1
9 NaN 11.746 2
10 1.0 NaN 2
11 0.0 NaN 2
12 NaN 16.961 3
13 1.0 NaN 3
14 NaN 11.832 3
15 0.0 NaN 3
16 NaN 17.082 3
17 NaN 22.435 3
18 NaN 28.707 3
19 NaN 34.216 3
我找到了 但我需要 brake
简单地从最后已知的复制,但 speed
进行插值(我的实际数据集有 12 列,每个列都需要这样的待遇)
您可以对每一列应用不同的方法。例如:
# interpolate speed
df['speed'] = df.groupby('trip').speed.transform(lambda x: x.interpolate())
# fill brake with last known value
df['brake'] = df.groupby('trip').brake.transform(lambda x: x.fillna(method='ffill'))
>>> df
brake speed trip
0 0.0 NaN 1
1 1.0 NaN 1
2 1.0 1.2640 1
3 1.0 0.0000 1
4 0.0 0.6320 1
5 0.0 1.2640 1
6 0.0 6.7040 1
7 1.0 6.7040 1
8 0.0 6.7040 1
9 NaN 11.7460 2
10 1.0 11.7460 2
11 0.0 11.7460 2
12 NaN 16.9610 3
13 1.0 14.3965 3
14 1.0 11.8320 3
15 0.0 14.4570 3
16 0.0 17.0820 3
17 0.0 22.4350 3
18 0.0 28.7070 3
19 0.0 34.2160 3
请注意,这意味着您仍然保持 NaN
的刹车,因为行程的第一排没有 "last known value",而第一排的速度为 NaNs
几行是 NaN
。您可以将它们替换为您认为合适的 fillna()
我正在尝试通过插入或复制组内的最后一个已知值(由 trip
标识)来填充 pandas 数据框中的缺失值。我的数据如下所示:
brake speed trip
0 0.0 NaN 1
1 1.0 NaN 1
2 NaN 1.264 1
3 NaN 0.000 1
4 0.0 NaN 1
5 NaN 1.264 1
6 NaN 6.704 1
7 1.0 NaN 1
8 0.0 NaN 1
9 NaN 11.746 2
10 1.0 NaN 2
11 0.0 NaN 2
12 NaN 16.961 3
13 1.0 NaN 3
14 NaN 11.832 3
15 0.0 NaN 3
16 NaN 17.082 3
17 NaN 22.435 3
18 NaN 28.707 3
19 NaN 34.216 3
我找到了 brake
简单地从最后已知的复制,但 speed
进行插值(我的实际数据集有 12 列,每个列都需要这样的待遇)
您可以对每一列应用不同的方法。例如:
# interpolate speed
df['speed'] = df.groupby('trip').speed.transform(lambda x: x.interpolate())
# fill brake with last known value
df['brake'] = df.groupby('trip').brake.transform(lambda x: x.fillna(method='ffill'))
>>> df
brake speed trip
0 0.0 NaN 1
1 1.0 NaN 1
2 1.0 1.2640 1
3 1.0 0.0000 1
4 0.0 0.6320 1
5 0.0 1.2640 1
6 0.0 6.7040 1
7 1.0 6.7040 1
8 0.0 6.7040 1
9 NaN 11.7460 2
10 1.0 11.7460 2
11 0.0 11.7460 2
12 NaN 16.9610 3
13 1.0 14.3965 3
14 1.0 11.8320 3
15 0.0 14.4570 3
16 0.0 17.0820 3
17 0.0 22.4350 3
18 0.0 28.7070 3
19 0.0 34.2160 3
请注意,这意味着您仍然保持 NaN
的刹车,因为行程的第一排没有 "last known value",而第一排的速度为 NaNs
几行是 NaN
。您可以将它们替换为您认为合适的 fillna()