问题描述
我的示例代码如下:
import pandas as pd
dictx = {'col1':[1,'nan','nan','nan',5,'nan',7,'nan',9,'nan','nan','nan',13],\
'col2':[20,'nan','nan','nan',22,'nan',25,'nan',30,'nan','nan','nan',25],\
'col3':[15,'nan','nan','nan',10,'nan',14,'nan',13,'nan','nan','nan',9]}
df = pd.DataFrame(dictx).astype(float)
我正在尝试插入包含值nan"的各个段.
就上下文而言，我正在尝试使用城市(巴西圣保罗)提供的 GPS 数据来跟踪公交车速度，但数据很少，并且部分不提供信息，例如，但有部分我知道它们会停止，例如黎明，但信息也以nan"的形式出现.
I'm trying to interpolate various segments which contain the value 'nan'.
For context, I'm trying to track bus speeds using GPS data provided by the city (São Paulo, Brazil), but the data is scarce and with parts that do not provide the information, as the e.g., but there're segments which I know for a fact that they are stopped, such as dawn, but the information come as 'nan' as well.
我需要什么:
我一直在试验 dataframe.interpolate() 参数(limit 和 limit_diretcion)，但结果很短.如果我设置 df.interpolate(limit=2) 我不仅会插入我需要的数据，还会插入不应该插入的数据.所以我需要在由限制定义的部分之间进行插值
What I need:
I've been experimenting with dataframe.interpolate() parameters (limit and limit_diretcion) but came up short. If I set df.interpolate(limit=2) I will not only interpolate the data that I need but the data where it shouldn't. So I need to interpolate between sections defined by a limit
期望的输出:
Out[7]:
col1 col2 col3
0 1.0 20.00 15.00
1 nan nan nan
2 nan nan nan
3 nan nan nan
4 5.0 22.00 10.00
5 6.0 23.50 12.00
6 7.0 25.00 14.00
7 8.0 27.50 13.50
8 9.0 30.00 13.00
9 nan nan nan
10 nan nan nan
11 nan nan nan
12 13.0 25.00 9.00
我一直在尝试应用的逻辑基本上是试图找到 nan 并计算它们的索引之间的差异，因此创建一个新的 dataframe_temp 进行插值，然后将其添加到另一个创建新的 dataframe_final 中.但是由于 'nan'=='nan' 返回 False
The logic that I've been trying to apply is basically trying to find nan's and calculating the difference between their indexes and so createing a new dataframe_temp to interpolate and only than add it to another creating a new dataframe_final. But this has become hard to achieve due to the fact that 'nan'=='nan' return False
推荐答案
这是一个技巧，但可能仍然有用.可能 Pandas 0.23 会有更好的解决方案.
This is a hack but may still be useful. Likely Pandas 0.23 will have a better solution.
df_fw = df.interpolate(limit=1)
df_bk = df.interpolate(limit=1, limit_direction='backward')
df_fw.where(df_bk.notna())
col1 col2 col3
0 1.0 20.0 15.0
1 NaN NaN NaN
2 NaN NaN NaN
3 NaN NaN NaN
4 5.0 22.0 10.0
5 6.0 23.5 12.0
6 7.0 25.0 14.0
7 8.0 27.5 13.5
8 9.0 30.0 13.0
9 NaN NaN NaN
10 NaN NaN NaN
11 NaN NaN NaN
12 13.0 25.0 9.0
不是黑客
更合法的处理方式.
泛化处理任何限制.
Not a Hack
More legitimate way of handling it.
Generalized to handle any limit.
def interp(df, limit):
d = df.notna().rolling(limit + 1).agg(any).fillna(1)
d = pd.concat({
i: d.shift(-i).fillna(1)
for i in range(limit + 1)
}).prod(level=1)
return df.interpolate(limit=limit).where(d.astype(bool))
df.pipe(interp, 1)
col1 col2 col3
0 1.0 20.0 15.0
1 NaN NaN NaN
2 NaN NaN NaN
3 NaN NaN NaN
4 5.0 22.0 10.0
5 6.0 23.5 12.0
6 7.0 25.0 14.0
7 8.0 27.5 13.5
8 9.0 30.0 13.0
9 NaN NaN NaN
10 NaN NaN NaN
11 NaN NaN NaN
12 13.0 25.0 9.0
还可以处理列与列之间 NaN 的变化.考虑一个不同的 df
Can also handle variation in NaN from column to column. Consider a different df
dictx = {'col1':[1,'nan','nan','nan',5,'nan','nan',7,'nan',9,'nan','nan','nan',13],\
'col2':[20,'nan','nan','nan',22,'nan',25,'nan','nan',30,'nan','nan','nan',25],\
'col3':[15,'nan','nan','nan',10,'nan',14,'nan',13,'nan','nan','nan',9,'nan']}
df = pd.DataFrame(dictx).astype(float)
df
col1 col2 col3
0 1.0 20.0 15.0
1 NaN NaN NaN
2 NaN NaN NaN
3 NaN NaN NaN
4 5.0 22.0 10.0
5 NaN NaN NaN
6 NaN 25.0 14.0
7 7.0 NaN NaN
8 NaN NaN 13.0
9 9.0 30.0 NaN
10 NaN NaN NaN
11 NaN NaN NaN
12 NaN NaN 9.0
13 13.0 25.0 NaN
然后用 limit=1
df.pipe(interp, 1)
col1 col2 col3
0 1.0 20.0 15.0
1 NaN NaN NaN
2 NaN NaN NaN
3 NaN NaN NaN
4 5.0 22.0 10.0
5 NaN 23.5 12.0
6 NaN 25.0 14.0
7 7.0 NaN 13.5
8 8.0 NaN 13.0
9 9.0 30.0 NaN
10 NaN NaN NaN
11 NaN NaN NaN
12 NaN NaN 9.0
13 13.0 25.0 9.0
并使用 limit=2
df.pipe(interp, 2).round(2)
col1 col2 col3
0 1.00 20.00 15.0
1 NaN NaN NaN
2 NaN NaN NaN
3 NaN NaN NaN
4 5.00 22.00 10.0
5 5.67 23.50 12.0
6 6.33 25.00 14.0
7 7.00 26.67 13.5
8 8.00 28.33 13.0
9 9.00 30.00 NaN
10 NaN NaN NaN
11 NaN NaN NaN
12 NaN NaN 9.0
13 13.00 25.00 9.0
这篇关于在由索引分隔的部分中插入 Pandas Dataframe的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！