Pandas 分组按月和年

Question

I have the following dataframe:

Date        abc    xyz
01-Jun-13   100    200
03-Jun-13   -20    50
15-Aug-13   40     -5
20-Jan-14   25     15
21-Feb-14   60     80

I need to group the data by year and month. Ie, Group by Jan 2013, Feb 2013, Mar 2013, etc...

I will be using the newly grouped data to create a plot showing abc vs xyz per year/month.

I've tried various combinations of groupby and sum, but I just can't seem to get anything to work. How can I do it?

Answer 1

You can use either resample or Grouper (which resamples under the hood).

First make sure that the datetime column is actually of datetimes (hit it with pd.to_datetime ). It's easier if it's a DatetimeIndex:

In [11]: df1
Out[11]:
            abc  xyz
Date
2013-06-01  100  200
2013-06-03  -20   50
2013-08-15   40   -5
2014-01-20   25   15
2014-02-21   60   80

In [12]: g = df1.groupby(pd.Grouper(freq="M"))  # DataFrameGroupBy (grouped by Month)

In [13]: g.sum()
Out[13]:
            abc  xyz
Date
2013-06-30   80  250
2013-07-31  NaN  NaN
2013-08-31   40   -5
2013-09-30  NaN  NaN
2013-10-31  NaN  NaN
2013-11-30  NaN  NaN
2013-12-31  NaN  NaN
2014-01-31   25   15
2014-02-28   60   80

In [14]: df1.resample("M", how='sum')  # the same
Out[14]:
            abc  xyz
Date
2013-06-30   40  125
2013-07-31  NaN  NaN
2013-08-31   40   -5
2013-09-30  NaN  NaN
2013-10-31  NaN  NaN
2013-11-30  NaN  NaN
2013-12-31  NaN  NaN
2014-01-31   25   15
2014-02-28   60   80

Note: Previously pd.Grouper(freq="M") was written as pd.TimeGrouper("M") . The latter is now deprecated since 0.21.

---

I had thought the following would work, but it doesn't (due to as_index not being respected? I'm not sure.). I'm including this for interest's sake.

If it's a column (it has to be a datetime64 column! as I say, hit it with to_datetime ), you can use the PeriodIndex:

In [21]: df
Out[21]:
        Date  abc  xyz
0 2013-06-01  100  200
1 2013-06-03  -20   50
2 2013-08-15   40   -5
3 2014-01-20   25   15
4 2014-02-21   60   80

In [22]: pd.DatetimeIndex(df.Date).to_period("M")  # old way
Out[22]:
<class 'pandas.tseries.period.PeriodIndex'>
[2013-06, ..., 2014-02]
Length: 5, Freq: M

In [23]: per = df.Date.dt.to_period("M")  # new way to get the same

In [24]: g = df.groupby(per)

In [25]: g.sum()  # dang not quite what we want (doesn't fill in the gaps)
Out[25]:
         abc  xyz
2013-06   80  250
2013-08   40   -5
2014-01   25   15
2014-02   60   80

To get the desired result we have to reindex...

Answer 2

Why not keep it simple?!

GB=DF.groupby([(DF.index.year),(DF.index.month)]).sum()

giving you,

print(GB)
        abc  xyz
2013 6   80  250
     8   40   -5
2014 1   25   15
     2   60   80

and then you can plot like asked using,

GB.plot('abc','xyz',kind='scatter')

Answer 3

There are different ways to do that.

• I created the data frame to showcase the different techniques to filter your data.

 df = pd.DataFrame({'Date':['01-Jun-13','03-Jun-13', '15-Aug-13', '20-Jan-14', '21-Feb-14'],

'abc':[100,-20,40,25,60],'xyz':[200,50,-5,15,80] })

• I separated months/year/day and seperated month-year as you explained.

 def getMonth(s): return s.split("-")[1] def getDay(s): return s.split("-")[0] def getYear(s): return s.split("-")[2] def getYearMonth(s): return s.split("-")[1]+"-"+s.split("-")[2]

• I created new columns: year , month , day and ' yearMonth '. In your case, you need one of both. You can group using two columns 'year','month' or using one column yearMonth

 df['year']= df['Date'].apply(lambda x: getYear(x)) df['month']= df['Date'].apply(lambda x: getMonth(x)) df['day']= df['Date'].apply(lambda x: getDay(x)) df['YearMonth']= df['Date'].apply(lambda x: getYearMonth(x))

Output:

        Date  abc  xyz year month day YearMonth
0  01-Jun-13  100  200   13   Jun  01    Jun-13
1  03-Jun-13  -20   50   13   Jun  03    Jun-13
2  15-Aug-13   40   -5   13   Aug  15    Aug-13
3  20-Jan-14   25   15   14   Jan  20    Jan-14
4  21-Feb-14   60   80   14   Feb  21    Feb-14

• You can go through the different groups in groupby(..) items.

In this case, we are grouping by two columns:

 for key,g in df.groupby(['year','month']): print key,g

Output:

('13', 'Jun')         Date  abc  xyz year month day YearMonth
0  01-Jun-13  100  200   13   Jun  01    Jun-13
1  03-Jun-13  -20   50   13   Jun  03    Jun-13
('13', 'Aug')         Date  abc  xyz year month day YearMonth
2  15-Aug-13   40   -5   13   Aug  15    Aug-13
('14', 'Jan')         Date  abc  xyz year month day YearMonth
3  20-Jan-14   25   15   14   Jan  20    Jan-14
('14', 'Feb')         Date  abc  xyz year month day YearMonth

In this case, we are grouping by one column:

 for key,g in df.groupby(['YearMonth']): print key,g

Output:

Jun-13         Date  abc  xyz year month day YearMonth
0  01-Jun-13  100  200   13   Jun  01    Jun-13
1  03-Jun-13  -20   50   13   Jun  03    Jun-13
Aug-13         Date  abc  xyz year month day YearMonth
2  15-Aug-13   40   -5   13   Aug  15    Aug-13
Jan-14         Date  abc  xyz year month day YearMonth
3  20-Jan-14   25   15   14   Jan  20    Jan-14
Feb-14         Date  abc  xyz year month day YearMonth
4  21-Feb-14   60   80   14   Feb  21    Feb-14

• In case you wanna access to specific item, you can use get_group

print df.groupby(['YearMonth']).get_group('Jun-13')

Output:

        Date  abc  xyz year month day YearMonth
0  01-Jun-13  100  200   13   Jun  01    Jun-13
1  03-Jun-13  -20   50   13   Jun  03    Jun-13

• Similar to get_group . This hack would help to filter values and get the grouped values.

This also would give the same result.

print df[df['YearMonth']=='Jun-13'] 

Output:

        Date  abc  xyz year month day YearMonth
0  01-Jun-13  100  200   13   Jun  01    Jun-13
1  03-Jun-13  -20   50   13   Jun  03    Jun-13

You can select list of abc or xyz values during Jun-13

print df[df['YearMonth']=='Jun-13'].abc.values
print df[df['YearMonth']=='Jun-13'].xyz.values

Output:

[100 -20]  #abc values
[200  50]  #xyz values

You can use this to go through the dates that you have classified as "year-month" and apply cretiria on it to get related data.

for x in set(df.YearMonth): 
    print df[df['YearMonth']==x].abc.values
    print df[df['YearMonth']==x].xyz.values

I recommend also to check this answer as well.

Answer 4

You can also do it by creating a string column with the year and month as follows:

df['date'] = df.index
df['year-month'] = df['date'].apply(lambda x: str(x.year) + ' ' + str(x.month))
grouped = df.groupby('year-month')

However this doesn't preserve the order when you loop over the groups, eg

for name, group in grouped:
    print(name)

Will give:

2007 11
2007 12
2008 1
2008 10
2008 11
2008 12
2008 2
2008 3
2008 4
2008 5
2008 6
2008 7
2008 8
2008 9
2009 1
2009 10

So then, if you want to preserve the order, you must do as suggested by @Q-man above:

grouped = df.groupby([df.index.year, df.index.month])

This will preserve the order in the above loop:

(2007, 11)
(2007, 12)
(2008, 1)
(2008, 2)
(2008, 3)
(2008, 4)
(2008, 5)
(2008, 6)
(2008, 7)
(2008, 8)
(2008, 9)
(2008, 10)

Answer 5

Some of the answers are using Date as an index instead of a column (and there's nothing wrong with doing that).

However, for anyone who has the dates stored as a column (instead of an index), remember to access the column's dt attribute. That is:

# First make sure `Date` is a datetime column
df['Date'] = pd.to_datetime(
  arg=df['Date'],
  format='%d-%b-%y' # Assuming dd-Mon-yy format
)

# Group by year and month
df.groupby(
  [
    df['Date'].dt.year,
    df['Date'].dt.month 
  ]
).sum()