Rolling average of pandas data frame with multiple id's
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I have a pandas dataframe on which I am calculating the rolling average on over multiple id's.
df:
╔════╦═══════╗
║ id ║ value ║
╠════╬═══════╣
║ 1 ║ 2 ║
║ 1 ║ 5 ║
║ 1 ║ 1 ║
║ 2 ║ 4 ║
║ 2 ║ 1 ║
║ 2 ║ 5 ║
║ 2 ║ 3 ║
║ 3 ║ 6 ║
║ 3 ║ 5 ║
╚════╩═══════╝
Current Resulting df:
╔════╦═══════╦═════════╗
║ id ║ value ║ average ║
╠════╬═══════╬═════════╣
║ 1 ║ 2 ║ ║
║ 1 ║ 5 ║ 3.5 ║
║ 1 ║ 1 ║ 3 ║
║ 2 ║ 4 ║ 2.5 ║
║ 2 ║ 1 ║ 2.5 ║
║ 2 ║ 5 ║ 3 ║
║ 2 ║ 3 ║ 4 ║
║ 3 ║ 6 ║ 4.5 ║
║ 3 ║ 5 ║ 5.5 ║
╚════╩═══════╩═════════╝
Expected Resulting df:
╔════╦═══════╦═════════╗
║ id ║ value ║ average ║
╠════╬═══════╬═════════╣
║ 1 ║ 2 ║ ║
║ 1 ║ 5 ║ 3.5 ║
║ 1 ║ 1 ║ 3 ║
║ 2 ║ 4 ║ ║
║ 2 ║ 1 ║ 2.5 ║
║ 2 ║ 5 ║ 3 ║
║ 2 ║ 3 ║ 4 ║
║ 3 ║ 6 ║ ║
║ 3 ║ 5 ║ 5.5 ║
╚════╩═══════╩═════════╝
Right now my code does not take into account the change in id, so it will still take the average of the last 2 values. Is there anyway to take into account the change in id.
My current code is df['value'] = df['value'].df(window = 2, min_periods = 1).mean()
Any help would be much appreciated
python pandas dataframe
edited Nov 9 at 22:40
Willem Van Onsem
142k16135226
asked Nov 9 at 22:39
Mr.P
154
add a comment |
up vote
0
down vote
favorite
I have a pandas dataframe on which I am calculating the rolling average on over multiple id's.
df:
╔════╦═══════╗
║ id ║ value ║
╠════╬═══════╣
║ 1 ║ 2 ║
║ 1 ║ 5 ║
║ 1 ║ 1 ║
║ 2 ║ 4 ║
║ 2 ║ 1 ║
║ 2 ║ 5 ║
║ 2 ║ 3 ║
║ 3 ║ 6 ║
║ 3 ║ 5 ║
╚════╩═══════╝
Current Resulting df:
╔════╦═══════╦═════════╗
║ id ║ value ║ average ║
╠════╬═══════╬═════════╣
║ 1 ║ 2 ║ ║
║ 1 ║ 5 ║ 3.5 ║
║ 1 ║ 1 ║ 3 ║
║ 2 ║ 4 ║ 2.5 ║
║ 2 ║ 1 ║ 2.5 ║
║ 2 ║ 5 ║ 3 ║
║ 2 ║ 3 ║ 4 ║
║ 3 ║ 6 ║ 4.5 ║
║ 3 ║ 5 ║ 5.5 ║
╚════╩═══════╩═════════╝
Expected Resulting df:
╔════╦═══════╦═════════╗
║ id ║ value ║ average ║
╠════╬═══════╬═════════╣
║ 1 ║ 2 ║ ║
║ 1 ║ 5 ║ 3.5 ║
║ 1 ║ 1 ║ 3 ║
║ 2 ║ 4 ║ ║
║ 2 ║ 1 ║ 2.5 ║
║ 2 ║ 5 ║ 3 ║
║ 2 ║ 3 ║ 4 ║
║ 3 ║ 6 ║ ║
║ 3 ║ 5 ║ 5.5 ║
╚════╩═══════╩═════════╝
Right now my code does not take into account the change in id, so it will still take the average of the last 2 values. Is there anyway to take into account the change in id.
My current code is df['value'] = df['value'].df(window = 2, min_periods = 1).mean()
Any help would be much appreciated
python pandas dataframe
edited Nov 9 at 22:40
Willem Van Onsem
142k16135226
asked Nov 9 at 22:39
Mr.P
154
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have a pandas dataframe on which I am calculating the rolling average on over multiple id's.
df:
╔════╦═══════╗
║ id ║ value ║
╠════╬═══════╣
║ 1 ║ 2 ║
║ 1 ║ 5 ║
║ 1 ║ 1 ║
║ 2 ║ 4 ║
║ 2 ║ 1 ║
║ 2 ║ 5 ║
║ 2 ║ 3 ║
║ 3 ║ 6 ║
║ 3 ║ 5 ║
╚════╩═══════╝
Current Resulting df:
╔════╦═══════╦═════════╗
║ id ║ value ║ average ║
╠════╬═══════╬═════════╣
║ 1 ║ 2 ║ ║
║ 1 ║ 5 ║ 3.5 ║
║ 1 ║ 1 ║ 3 ║
║ 2 ║ 4 ║ 2.5 ║
║ 2 ║ 1 ║ 2.5 ║
║ 2 ║ 5 ║ 3 ║
║ 2 ║ 3 ║ 4 ║
║ 3 ║ 6 ║ 4.5 ║
║ 3 ║ 5 ║ 5.5 ║
╚════╩═══════╩═════════╝
Expected Resulting df:
╔════╦═══════╦═════════╗
║ id ║ value ║ average ║
╠════╬═══════╬═════════╣
║ 1 ║ 2 ║ ║
║ 1 ║ 5 ║ 3.5 ║
║ 1 ║ 1 ║ 3 ║
║ 2 ║ 4 ║ ║
║ 2 ║ 1 ║ 2.5 ║
║ 2 ║ 5 ║ 3 ║
║ 2 ║ 3 ║ 4 ║
║ 3 ║ 6 ║ ║
║ 3 ║ 5 ║ 5.5 ║
╚════╩═══════╩═════════╝
Right now my code does not take into account the change in id, so it will still take the average of the last 2 values. Is there anyway to take into account the change in id.
My current code is df['value'] = df['value'].df(window = 2, min_periods = 1).mean()
Any help would be much appreciated
python pandas dataframe
edited Nov 9 at 22:40
Willem Van Onsem
142k16135226
asked Nov 9 at 22:39
Mr.P
154
I have a pandas dataframe on which I am calculating the rolling average on over multiple id's.
df:
╔════╦═══════╗
║ id ║ value ║
╠════╬═══════╣
║ 1 ║ 2 ║
║ 1 ║ 5 ║
║ 1 ║ 1 ║
║ 2 ║ 4 ║
║ 2 ║ 1 ║
║ 2 ║ 5 ║
║ 2 ║ 3 ║
║ 3 ║ 6 ║
║ 3 ║ 5 ║
╚════╩═══════╝
Current Resulting df:
╔════╦═══════╦═════════╗
║ id ║ value ║ average ║
╠════╬═══════╬═════════╣
║ 1 ║ 2 ║ ║
║ 1 ║ 5 ║ 3.5 ║
║ 1 ║ 1 ║ 3 ║
║ 2 ║ 4 ║ 2.5 ║
║ 2 ║ 1 ║ 2.5 ║
║ 2 ║ 5 ║ 3 ║
║ 2 ║ 3 ║ 4 ║
║ 3 ║ 6 ║ 4.5 ║
║ 3 ║ 5 ║ 5.5 ║
╚════╩═══════╩═════════╝
Expected Resulting df:
╔════╦═══════╦═════════╗
║ id ║ value ║ average ║
╠════╬═══════╬═════════╣
║ 1 ║ 2 ║ ║
║ 1 ║ 5 ║ 3.5 ║
║ 1 ║ 1 ║ 3 ║
║ 2 ║ 4 ║ ║
║ 2 ║ 1 ║ 2.5 ║
║ 2 ║ 5 ║ 3 ║
║ 2 ║ 3 ║ 4 ║
║ 3 ║ 6 ║ ║
║ 3 ║ 5 ║ 5.5 ║
╚════╩═══════╩═════════╝
Right now my code does not take into account the change in id, so it will still take the average of the last 2 values. Is there anyway to take into account the change in id.
My current code is df['value'] = df['value'].df(window = 2, min_periods = 1).mean()
Any help would be much appreciated
python pandas dataframe
python pandas dataframe
edited Nov 9 at 22:40
Willem Van Onsem
142k16135226
asked Nov 9 at 22:39
Mr.P
154
edited Nov 9 at 22:40
Willem Van Onsem
142k16135226
asked Nov 9 at 22:39
Mr.P
154
edited Nov 9 at 22:40
Willem Van Onsem
142k16135226
edited Nov 9 at 22:40
Willem Van Onsem
142k16135226
edited Nov 9 at 22:40
Willem Van Onsem
142k16135226
142k16135226
asked Nov 9 at 22:39
Mr.P
154
asked Nov 9 at 22:39
Mr.P
154
asked Nov 9 at 22:39
Mr.P
154
154
add a comment |
add a comment |
1 Answer
1
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oldest
votes
up vote
0
down vote
concat and groupby
pd.concat([d.rolling(2).mean() for _, d in df.groupby('id')])
id value
0 NaN NaN
1 1.0 3.5
2 1.0 3.0
3 NaN NaN
4 2.0 2.5
5 2.0 3.0
6 2.0 4.0
7 NaN NaN
8 3.0 5.5
answered Nov 9 at 22:43
piRSquared
151k22140282
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
concat and groupby
pd.concat([d.rolling(2).mean() for _, d in df.groupby('id')])
id value
0 NaN NaN
1 1.0 3.5
2 1.0 3.0
3 NaN NaN
4 2.0 2.5
5 2.0 3.0
6 2.0 4.0
7 NaN NaN
8 3.0 5.5
answered Nov 9 at 22:43
piRSquared
151k22140282
add a comment |
up vote
0
down vote
concat and groupby
pd.concat([d.rolling(2).mean() for _, d in df.groupby('id')])
id value
0 NaN NaN
1 1.0 3.5
2 1.0 3.0
3 NaN NaN
4 2.0 2.5
5 2.0 3.0
6 2.0 4.0
7 NaN NaN
8 3.0 5.5
answered Nov 9 at 22:43
piRSquared
151k22140282
add a comment |
up vote
0
down vote
up vote
0
down vote
concat and groupby
pd.concat([d.rolling(2).mean() for _, d in df.groupby('id')])
id value
0 NaN NaN
1 1.0 3.5
2 1.0 3.0
3 NaN NaN
4 2.0 2.5
5 2.0 3.0
6 2.0 4.0
7 NaN NaN
8 3.0 5.5
answered Nov 9 at 22:43
piRSquared
151k22140282
concat and groupby
pd.concat([d.rolling(2).mean() for _, d in df.groupby('id')])
id value
0 NaN NaN
1 1.0 3.5
2 1.0 3.0
3 NaN NaN
4 2.0 2.5
5 2.0 3.0
6 2.0 4.0
7 NaN NaN
8 3.0 5.5
answered Nov 9 at 22:43
piRSquared
151k22140282
answered Nov 9 at 22:43
piRSquared
151k22140282
answered Nov 9 at 22:43
piRSquared
151k22140282
answered Nov 9 at 22:43
piRSquared
151k22140282
151k22140282
add a comment |
add a comment |
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