如何使用与 scipy 插值相同的时间序列特征来插值 pandas DataFrame？

Question

我无法使用 pandas 时间序列正确插入数据。

我构建了一个小型 dataframe 进行插值：

import pandas as pd

# Build a dataframe with 3 distances
df = pd.DataFrame([0., 5., 10.], columns=['distance'])
# Add a column representing the time to reach these distances
# at a constant speed of 2.572 meters per second
speed_mps = 2.572
df['time'] = df.apply(lambda r: r.distance/speed_mps, axis=1)
df

distance    time

0 0.0 0.000000
1 5.0 1.944012
2 10.0 3.888025

我可以用 scipy.interpolate 插入我想要的方式：

# Interpolate distance at every second with scipy
from scipy.interpolate import interp1d

# Get the interpolation function
f = interp1d(df.time, df.distance)
seconds = [0., 1., 2., 3.]
# Interpolate distances and build the expected dataframe
expected_df = pd.DataFrame(list(zip(f(seconds),seconds)),
                           columns=['distance', 'time'])
expected_df

distance    time

0 0.000 0.0
1 2.572 1.0
2 5.144 2.0
3 7.716 3.0

如果生成的 dataframe 包含原始数据和插值数据，则可以接受：

acceptable_df = pd.concat([expected_df, df], axis=0, sort=True).
    sort_values(by='time').drop_duplicates().reset_index(drop=True)
acceptable_df 

distance    time

0 0.000 0.000000
1 2.572 1.000000
2 5.000 1.944012
3 5.144 2.000000
4 7.716 3.000000
5 10.000 3.888025

但我没有成功地仅使用 pandas 正确插值：

# Try to interpolate distance (and a bunch of other columns ignored here)
# at every second using pandas time series features
pd_interp_df = df.copy(deep=True) 
pd_interp_df['timedelta'] = pd.to_timedelta(pd_interp_df.time, unit='s')
pd_interp_df.resample(rule='s', on='timedelta').mean().interpolate()

distance    time

时间增量
00:00:00 0.0 0.000000
00:00:01 5.0 1.944012
00:00:02 7.5 2.916019
00:00:03 10.0 3.888025

timedelta 索引和时间列值不一致。 如何仅使用 pandas 构建上述预期或可接受的 dataframe？