pandas 熊猫的滚动差异
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Rolling difference in Pandas
提问by WBM
Does anyone know an efficient function/method such as pandas.rolling_mean, that would calculate the rolling difference of an array
有谁知道一个有效的函数/方法，例如pandas.rolling_mean，可以计算数组的滚动差异
This is my closest solution:
这是我最接近的解决方案：
roll_diff = pd.Series(values).diff(periods=1)
However, it only calculates single-step rolling difference. Ideally the step size would be editable (i.e. difference between current time step and n last steps).
但是，它只计算单步滚动差异。理想情况下，步长是可编辑的（即当前时间步长和最后 n 步之间的差异）。
I've also written this, but for larger arrays, it is quite slow:
我也写过这个，但是对于较大的数组，它很慢：
def roll_diff(values,step):
diff = []
for i in np.arange(step, len(values)-1):
pers_window = np.arange(i-1,i-step-1,-1)
diff.append(np.abs(values[i] - np.mean(values[pers_window])))
diff = np.pad(diff, (0, step+1), 'constant')
return diff
回答by Pierluigi
What about:
关于什么：
import pandas
x = pandas.DataFrame({
'x_1': [0, 1, 2, 3, 0, 1, 2, 500, ],},
index=[0, 1, 2, 3, 4, 5, 6, 7])
x['x_1'].rolling(window=2).apply(lambda x: x.iloc[1] - x.iloc[0])
in general you can replace the lambdafunction with your own function. Note that in this case the first item will be NaN.
一般来说，您可以lambda用您自己的函数替换该函数。请注意，在这种情况下，第一项将是NaN。
Update
更新
Defining the following:
定义以下内容：
n_steps = 2
def my_fun(x):
return x.iloc[-1] - x.iloc[0]
x['x_1'].rolling(window=n_steps).apply(my_fun)
you can compute the differences between values at n_steps.
您可以计算 处的值之间的差异n_steps。
回答by Dan
You can do the same thing as in https://stackoverflow.com/a/48345749/1011724if you work directly on the underlying numpy array:
如果您直接在底层 numpy 数组上工作，您可以执行与https://stackoverflow.com/a/48345749/1011724相同的操作：
import numpy as np
diff_kernel = np.array([1,-1])
np.convolve(rs,diff_kernel ,'same')
where rsis your pandas series
rs你的Pandas系列在哪里
回答by Manualmsdos
If you got KeyError: 0, try with iloc:
如果有KeyError: 0，请尝试iloc：
import pandas
x = pandas.DataFrame({
'x_1': [0, 1, 2, 3, 0, 1, 2, 500, ],},
index=[0, 1, 2, 3, 4, 5, 6, 7])
x['x_1'].rolling(window=2).apply(lambda x: x.iloc[1] - x.iloc[0])
回答by jpp
This should work:
这应该有效：
import numpy as np
x = np.array([1, 3, 6, 1, -5, 6, 4, 1, 6])
def running_diff(arr, N):
return np.array([arr[i] - arr[i-N] for i in range(N, len(arr))])
running_diff(x, 4) # array([-6, 3, -2, 0, 11])
For a given pd.Series, you will have to define what you want for the first few items. The below example just returns the initial series values.
对于给定的pd.Series，您必须为前几项定义您想要的内容。下面的示例仅返回初始系列值。
s_roll_diff = np.hstack((s.values[:4], running_diff(s.values, 4)))
This works because you can assign a np.arraydirectly to a pd.DataFrame, e.g. for a column s, df.s_roll_diff = np.hstack((df.s.values[:4], running_diff(df.s.values, 4)))
这是有效的，因为您可以将 anp.array直接分配给 a pd.DataFrame，例如对于列s，df.s_roll_diff = np.hstack((df.s.values[:4], running_diff(df.s.values, 4)))
