Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
581 views
in Technique[技术] by (71.8m points)

python - Pandas equivalent of Oracle Lead/Lag function

First I'm new to pandas, but I'm already falling in love with it. I'm trying to implement the equivalent of the Lag function from Oracle.

Let's suppose you have this DataFrame:

Date                   Group      Data
2014-05-14 09:10:00        A         1
2014-05-14 09:20:00        A         2
2014-05-14 09:30:00        A         3
2014-05-14 09:40:00        A         4
2014-05-14 09:50:00        A         5
2014-05-14 10:00:00        B         1
2014-05-14 10:10:00        B         2
2014-05-14 10:20:00        B         3
2014-05-14 10:30:00        B         4

If this was an oracle database and I wanted to create a lag function grouped by the "Group" column and ordered by the Date I could easily use this function:

 LAG(Data,1,NULL) OVER (PARTITION BY Group ORDER BY Date ASC) AS Data_lagged

This would result in the following Table:

Date                   Group     Data    Data lagged
2014-05-14 09:10:00        A        1           Null
2014-05-14 09:20:00        A        2            1
2014-05-14 09:30:00        A        3            2
2014-05-14 09:40:00        A        4            3
2014-05-14 09:50:00        A        5            4
2014-05-14 10:00:00        B        1           Null
2014-05-14 10:10:00        B        2            1
2014-05-14 10:20:00        B        3            2
2014-05-14 10:30:00        B        4            3

In pandas I can set the date to be an index and use the shift method:

db["Data_lagged"] = db.Data.shift(1)

The only issue is that this doesn't group by a column. Even if I set the two columns Date and Group as indexes, I would still get the "5" in the lagged column.

Is there a way to implement the equivalent of the Lead and lag functions in Pandas?

Question&Answers:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

You could perform a groupby/apply (shift) operation:

In [15]: df['Data_lagged'] = df.groupby(['Group'])['Data'].shift(1)

In [16]: df
Out[16]: 
                Date Group  Data  Data_lagged
2014-05-14  09:10:00     A     1          NaN
2014-05-14  09:20:00     A     2            1
2014-05-14  09:30:00     A     3            2
2014-05-14  09:40:00     A     4            3
2014-05-14  09:50:00     A     5            4
2014-05-14  10:00:00     B     1          NaN
2014-05-14  10:10:00     B     2            1
2014-05-14  10:20:00     B     3            2
2014-05-14  10:30:00     B     4            3

[9 rows x 4 columns]

To obtain the ORDER BY Date ASC effect, you must sort the DataFrame first:

df['Data_lagged'] = (df.sort_values(by=['Date'], ascending=True)
                       .groupby(['Group'])['Data'].shift(1))

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...