pandas 计算系列的本地时间导数
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Calculate local time derivative of Series
提问by Adam
I have data that I'm importing from an hdf5 file. So, it comes in looking like this:
我有从 hdf5 文件导入的数据。所以，它看起来像这样：
import pandas as pd
tmp=pd.Series([1.,3.,4.,3.,5.],['2016-06-27 23:52:00','2016-06-27 23:53:00','2016-06-27 23:54:00','2016-06-27 23:55:00','2016-06-27 23:59:00'])
tmp.index=pd.to_datetime(tmp.index)
>>>tmp
2016-06-27 23:52:00 1.0
2016-06-27 23:53:00 3.0
2016-06-27 23:54:00 4.0
2016-06-27 23:55:00 3.0
2016-06-27 23:59:00 5.0
dtype: float64
I would like to find the local slope of the data. If I just do tmp.diff() I do get the local change in value. But, I want to get the change in value per second (time derivative) I would like to do something like this, but this is the wrong way to do it and gives an error:
我想找到数据的局部斜率。如果我只是做 tmp.diff() 我确实得到了本地的价值变化。但是，我想获得每秒值的变化（时间导数）我想做这样的事情，但这是错误的方法并给出错误：
tmp.diff()/tmp.index.diff()
I have figured out that I can do it by converting all the data to a DataFrame, but that seems inefficient. Especially, since I'm going to have to work with a large, on disk file in chunks. Is there a better way to do it other than this:
我发现我可以通过将所有数据转换为 DataFrame 来实现，但这似乎效率低下。特别是，因为我将不得不分块处理一个大的磁盘文件。除了这个，还有没有更好的方法来做到这一点：
df=pd.DataFrame(tmp)
df['secvalue']=df.index.astype(np.int64)/1e+9
df['slope']=df['Value'].diff()/df['secvalue'].diff()
回答by piRSquared
Use numpy.gradient
用 numpy.gradient
import numpy as np
import pandas as pd
slope = pd.Series(np.gradient(tmp.values), tmp.index, name='slope')
To address the unequal temporal index, i'd resample over minutes and interpolate. Then my gradients would be over equal intervals.
为了解决不相等的时间索引，我会在几分钟内重新采样并进行插值。然后我的梯度将超过相等的间隔。
tmp_ = tmp.resample('T').interpolate()
slope = pd.Series(np.gradient(tmp_.values), tmp_.index, name='slope')
df = pd.concat([tmp_.rename('values'), slope], axis=1)
df
df.plot()
