如何在前瞻性的基础上使用 Pandas 滚动_* 函数
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How to use Pandas rolling_* functions on a forward-looking basis
提问by user2543645
Suppose I have a time series:
假设我有一个时间序列：
In[138] rng = pd.date_range('1/10/2011', periods=10, freq='D')
In[139] ts = pd.Series(randn(len(rng)), index=rng)
In[140]
Out[140]:
2011-01-10 0
2011-01-11 1
2011-01-12 2
2011-01-13 3
2011-01-14 4
2011-01-15 5
2011-01-16 6
2011-01-17 7
2011-01-18 8
2011-01-19 9
Freq: D, dtype: int64
If I use one of the rolling_* functions, for instance rolling_sum, I can get the behavior I want for backward looking rolling calculations:
如果我使用rolling_*函数之一，例如rolling_sum，我可以获得我想要的向后滚动计算的行为：
In [157]: pd.rolling_sum(ts, window=3, min_periods=0)
Out[157]:
2011-01-10 0
2011-01-11 1
2011-01-12 3
2011-01-13 6
2011-01-14 9
2011-01-15 12
2011-01-16 15
2011-01-17 18
2011-01-18 21
2011-01-19 24
Freq: D, dtype: float64
But what if I want to do a forward-looking sum? I've tried something like this:
但是如果我想做一个前瞻性的总结呢？我试过这样的事情：
In [161]: pd.rolling_sum(ts.shift(-2, freq='D'), window=3, min_periods=0)
Out[161]:
2011-01-08 0
2011-01-09 1
2011-01-10 3
2011-01-11 6
2011-01-12 9
2011-01-13 12
2011-01-14 15
2011-01-15 18
2011-01-16 21
2011-01-17 24
Freq: D, dtype: float64
But that's not exactly the behavior I want. What I am looking for as an output is:
但这不完全是我想要的行为。我正在寻找的输出是：
2011-01-10 3
2011-01-11 6
2011-01-12 9
2011-01-13 12
2011-01-14 15
2011-01-15 18
2011-01-16 21
2011-01-17 24
2011-01-18 17
2011-01-19 9
ie - I want the sum of the "current" day plus the next two days. My current solution is not sufficient because I care about what happens at the edges. I know I could solve this manually by setting up two additional columns that are shifted by 1 and 2 days respectively and then summing the three columns, but there's got to be a more elegant solution.
即 - 我想要“当前”天加上接下来两天的总和。我目前的解决方案还不够，因为我关心边缘会发生什么。我知道我可以通过设置两个分别移动 1 天和 2 天的额外列然后对三列求和来手动解决这个问题，但是必须有一个更优雅的解决方案。
回答by Andy Hayden
Why not just do it on the reversed Series (and reverse the answer):
为什么不直接在反向系列上做（并反转答案）：
In [11]: pd.rolling_sum(ts[::-1], window=3, min_periods=0)[::-1]
Out[11]:
2011-01-10 3
2011-01-11 6
2011-01-12 9
2011-01-13 12
2011-01-14 15
2011-01-15 18
2011-01-16 21
2011-01-17 24
2011-01-18 17
2011-01-19 9
Freq: D, dtype: float64
回答by Tom
Maybe you can try bottleneckmodule. When tsis large, bottleneckis much faster than pandas
也许你可以试试bottleneck模块。当ts很大时，bottleneck比pandas
import bottleneck as bn
result = bn.move_sum(ts[::-1], window=3, min_count=1)[::-1]
And bottleneckhas other rolling functions, such as move_max, move_argmin, move_rank.
并bottleneck具有其他滚动功能，如 move_max、move_argmin、move_rank。
回答by MitchellRosenthal256
I struggled with this then found an easy way using shift.
我为此苦苦挣扎，然后找到了一种使用 shift 的简单方法。
If you want a rolling sum for the next 10 periods, try:
如果您想要接下来 10 个期间的滚动总和，请尝试：
df['NewCol'] = df['OtherCol'].shift(-10).rolling(10, min_periods = 0).sum()
We use shift so that "OtherCol" shows up 10 rows ahead of where it normally would be, then we do a rolling sum over the previous 10 rows. Because we shifted, the previous 10 rows are actually the future 10 rows of the unshifted column. :)
我们使用 shift 以便“OtherCol”在正常位置之前显示 10 行，然后我们对前 10 行进行滚动求和。因为我们移位了，前 10 行实际上是未移位列的未来 10 行。:)
