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I have a time series object grouped of the type <pandas.core.groupby.SeriesGroupBy object at 0x03F1A9F0>. grouped.sum() gives the desired result but I cannot get rolling_sum to work with the groupby object. Is there any way to apply rolling functions to groupby objects? For example:

x = range(0, 6)
id = ['a', 'a', 'a', 'b', 'b', 'b']
df = DataFrame(zip(id, x), columns = ['id', 'x'])
df.groupby('id').sum()
id    x
a    3
b   12

However, I would like to have something like:

  id  x
0  a  0
1  a  1
2  a  3
3  b  3
4  b  7
5  b  12
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4
  • 1
    How exactly do you expect rolling function to work on grouped objects (I mean write out the math you want to do in symbols)? Commented Dec 21, 2012 at 20:06
  • Sorry I should have been more clear. Commented Dec 21, 2012 at 20:28
  • So you want to do a cumsum on each of the groups and then stitch the whole thing back into a single data frame? Commented Dec 21, 2012 at 20:34
  • Yes, ideally cumsum and any rolling function(mean, sum, std). Commented Dec 21, 2012 at 20:43

5 Answers 5

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145

For the Googlers who come upon this old question:

Regarding @kekert's comment on @Garrett's answer to use the new

df.groupby('id')['x'].rolling(2).mean()

rather than the now-deprecated 

df.groupby('id')['x'].apply(pd.rolling_mean, 2, min_periods=1)

curiously, it seems that the new .rolling().mean() approach returns a multi-indexed series, indexed by the group_by column first and then the index. Whereas, the old approach would simply return a series indexed singularly by the original df index, which perhaps makes less sense, but made it very convenient for adding that series as a new column into the original dataframe.

So I think I've figured out a solution that uses the new rolling() method and still works the same:

df.groupby('id')['x'].rolling(2).mean().reset_index(0,drop=True)

which should give you the series

0    0.0
1    0.5
2    1.5
3    3.0
4    3.5
5    4.5

which you can add as a column:

df['x'] = df.groupby('id')['x'].rolling(2).mean().reset_index(0,drop=True)
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9 Comments

I think you can use .transform rather than reset_index?
This actually fails if you're grouping by multiple columns. Dropping the first argument (levels) solves this though as it removes all levels by default. So the line becomes df['x'] = df.groupby('id')['x'].rolling(2).mean().reset_index(drop=True)
As another maddening nuance, use groupby(..., sort=False) if your group variable is not already sorted. I was getting really bizarre results when adding this rolling mean as a new column because the order didn't match the original df.
Very useful information. a) They should add this to their pandas Cookbook b) Can you raise some pandas bugs on the change in functionality? They should consider the consequences better before they deprecate.
Could you elaborate on why we should put .rolling(2), i.e. why window=2 here? Is it because there are 2 groups 'a' and 'b'?
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84

cumulative sum

To answer the question directly, the cumsum method would produced the desired series:

In [17]: df
Out[17]:
  id  x
0  a  0
1  a  1
2  a  2
3  b  3
4  b  4
5  b  5

In [18]: df.groupby('id').x.cumsum()
Out[18]:
0     0
1     1
2     3
3     3
4     7
5    12
Name: x, dtype: int64

pandas rolling functions per group

More generally, any rolling function can be applied to each group as follows (using the new .rolling method as commented by @kekert). Note that the return type is a multi-indexed series, which is different from previous (deprecated) pd.rolling_* methods.

In [10]: df.groupby('id')['x'].rolling(2, min_periods=1).sum()
Out[10]:
id
a   0   0.00
    1   1.00
    2   3.00
b   3   3.00
    4   7.00
    5   9.00
Name: x, dtype: float64

To apply the per-group rolling function and receive result in original dataframe order, transform should be used instead:

In [16]: df.groupby('id')['x'].transform(lambda s: s.rolling(2, min_periods=1).sum())
Out[16]:
0    0
1    1
2    3
3    3
4    7
5    9
Name: x, dtype: int64

deprecated approach

For reference, here's how the now deprecated pandas.rolling_mean behaved:

In [16]: df.groupby('id')['x'].apply(pd.rolling_mean, 2, min_periods=1)
Out[16]: 
0    0.0
1    0.5
2    1.5
3    3.0
4    3.5
5    4.5
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3 Comments

pd.rolling_mean is now deprecated for Series and will be removed, use df.groupby('id')['x'].rolling(2).mean() instead
in case you need it sorted to the original index efficiently: df.reset_index().groupby('id', sort=False)['x'].rolling(2, min_periods=1).mean().sort_index(level=1).reset_index(drop=True)
if original index is already sorted then replace df.reset_index() with df
11

Here is another way that generalizes well and uses pandas' expanding method.

It is very efficient and also works perfectly for rolling window calculations with fixed windows, such as for time series. 

# Import pandas library
import pandas as pd

# Prepare columns
x = range(0, 6)
id = ['a', 'a', 'a', 'b', 'b', 'b']

# Create dataframe from columns above
df = pd.DataFrame({'id':id, 'x':x})

# Calculate rolling sum with infinite window size (i.e. all rows in group) using "expanding"
df['rolling_sum'] = df.groupby('id')['x'].transform(lambda x: x.expanding().sum())

# Output as desired by original poster
print(df)
  id  x  rolling_sum
0  a  0            0
1  a  1            1
2  a  2            3
3  b  3            3
4  b  4            7
5  b  5           12
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5 Comments

Do you have anything to back up that this is "very efficient"? Generally with pandas, doing any sort of iteration (eg. "transform", or "apply") is a major performance hit, compared to doing the same thing with vector operations (which the built-ins of ".sum", ".rolling", etc. will all be). I know Pandas does do some pre-inspection on the iteration loops to see if it can optimize it for you, but in general iteration should be avoided if performance is a concern.
I am sorry I can only give you one upvote, I'm considering creating new accounts to give more credit to this answer. It's the only one that worked for me grouping on multiple columns, thanks!
Cool. This can apply exponential moving average. q['exponential_ave'] = q.groupby('id')['x'].transform(lambda x: x.ewm(com=0.2).mean())
What's the difference between this using expanding vs using rolling?
@liang this article explains it better than I can. In rolling functions the window size remains constant whereas in expanding functions it changes. See this answer as well.
6

If you need to reassign the grouped-rolling-function back to the original Dataframe, while keeping order and groups you can use the transform function.

df.sort_values(by='date', inplace=True)
grpd = df.groupby('group_key')
#using center=false to assign values on window's last row
df['val_rolling_7_mean'] = grpd['val'].transform(lambda x: x.rolling(7, center=False).mean())
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Comments

3

I'm not sure of the mechanics, but this works. Note, the returned value is just an ndarray. I think you could apply any cumulative or "rolling" function in this manner and it should have the same result.

I have tested it with cumprod, cummax and cummin and they all returned an ndarray. I think pandas is smart enough to know that these functions return a series and so the function is applied as a transformation rather than an aggregation.

In [35]: df.groupby('id')['x'].cumsum()
Out[35]:
0     0
1     1
2     3
3     3
4     7
5    12

Edit: I found it curious that this syntax does return a Series:

In [54]: df.groupby('id')['x'].transform('cumsum')
Out[54]:
0     0
1     1
2     3
3     3
4     7
5    12
Name: x
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