Construct new column with first row of a groupby with two columns - Pandas

Question

I am trying to construct a new column that gives a value of 1 if it's the first time that an element of the column "type" has had a specific value of the column "xx", and gives a value of 0 in any other case.

• The original dataframe (df) I am working with is:

  idx = [np.array(['Jan-18', 'Jan-18', 'Feb-18', 'Mar-18', 'Mar-18', 'Mar-18','Apr-18', 'Apr-18', 'May-18', 'Jun-18', 'Jun-18', 'Jun-18','Jul-18', 'Aug-18', 'Aug-18', 'Sep-18', 'Sep-18', 'Oct-18','Oct-18', 'Oct-18', 'Nov-18', 'Dec-18', 'Dec-18',]),np.array(['A', 'B', 'B', 'A', 'B', 'C', 'A', 'B', 'B', 'A', 'B', 'C','A', 'B', 'C', 'A', 'B', 'C', 'A', 'B', 'A', 'B', 'C'])]
  data = [{'xx': 1000}, {'xx': 1000}, {'xx': 1200}, {'xx': 800}, {'xx': 800}, {'xx': 800},{'xx': 1000}, {'xx': 1000}, {'xx': 800}, {'xx': 1200}, {'xx': 1200}, {'xx': 1200},{'xx': 1000}, {'xx': 1000}, {'xx': 1000}, {'xx': 1600}, {'xx': 1600}, {'xx': 1000}, {'xx': 800}, {'xx': 800}, {'xx': 1000}, {'xx': 1600}, {'xx': 1600}]
  df = pd.DataFrame(data, index=idx, columns=['xx'])
  df.index.names=['date','type']
  df=df.reset_index()
  df['date'] = pd.to_datetime(df['date'],format = '%b-%y')
  df=df.set_index(['date','type'])
  df['xx'] = df.xx.astype('float')
  

• The result I am looking for is:

                       xx   yy
  date       type
  2018-01-01 A     1000.0  1.0
             B     1000.0  1.0
  2018-02-01 B     1200.0  1.0
  2018-03-01 A      800.0  1.0
             B      800.0  1.0
             C      800.0  1.0
  2018-04-01 A     1000.0  0.0
             B     1000.0  0.0
  2018-05-01 B      800.0  0.0
  2018-06-01 A     1200.0  1.0
             B     1200.0  0.0
             C     1200.0  1.0
  2018-07-01 A     1000.0  0.0
  2018-08-01 B     1000.0  0.0
             C     1000.0  1.0
  2018-09-01 A     1600.0  1.0
             B     1600.0  1.0
  2018-10-01 C     1000.0  0.0
             A      800.0  0.0
             B      800.0  0.0
  2018-11-01 A     1000.0  0.0
  2018-12-01 B     1600.0  0.0
             C     1600.0  1.0
  

• I tried the following code, but it doesn't work (it gives an errror message):

  df['yy'] = df.assign(zz=(df.groupby(['type','xx']).first())).zz.transform(lambda x: 1)
  

The error message says

ValueError: Wrong number of items passed 0, placement implies 1.

I tried other metodhos, such as nth(0), but it doesn't work either. Any suggestion of how solving this problem is very welcome.

Answer 1

Try:

df['yy'] = (df.groupby(level=1).xx
              .apply(lambda x: (~x.duplicated()).astype(int))
           )

df['yy']

outputs:

date        type
2018-01-01  A       1
            B       1
2018-02-01  B       1
2018-03-01  A       1
            B       1
            C       1
2018-04-01  A       0
            B       0
2018-05-01  B       0
2018-06-01  A       1
            B       0
            C       1
2018-07-01  A       0
2018-08-01  B       0
            C       1
2018-09-01  A       1
            B       1
2018-10-01  C       0
            A       0
            B       0
2018-11-01  A       0
2018-12-01  B       0
            C       1
Name: yy, dtype: int32

Answer 2

Using groupby + cumcount + astype :

df['yy'] = df.reset_index().groupby(['type','xx']).cumcount().eq(0).astype(int).values

Result:

                     xx  yy
date       type            
2018-01-01 A     1000.0   1
           B     1000.0   1
2018-02-01 B     1200.0   1
2018-03-01 A      800.0   1
           B      800.0   1
           C      800.0   1
2018-04-01 A     1000.0   0
           B     1000.0   0
2018-05-01 B      800.0   0
2018-06-01 A     1200.0   1
           B     1200.0   0
           C     1200.0   1
2018-07-01 A     1000.0   0
2018-08-01 B     1000.0   0
           C     1000.0   1
2018-09-01 A     1600.0   1
           B     1600.0   1
2018-10-01 C     1000.0   0
           A      800.0   0
           B      800.0   0
2018-11-01 A     1000.0   0
2018-12-01 B     1600.0   0
           C     1600.0   1

Answer 3

duplicated

• Generate of list of tuples with type and xx
• Wrap it in a pandas.Series because I want to use the pandas.Series.duplicated method
• Use numpy.where to choose between 0 and 1

Note: This does not use groupby and consequently should be more efficient.

---

s = pd.Series([*zip(df.index.get_level_values('type'), df.xx)])
df.assign(id=np.where(s.duplicated(), 0, 1))

                     xx  id
date       type            
2018-01-01 A     1000.0   1
           B     1000.0   1
2018-02-01 B     1200.0   1
2018-03-01 A      800.0   1
           B      800.0   1
           C      800.0   1
2018-04-01 A     1000.0   0
           B     1000.0   0
2018-05-01 B      800.0   0
2018-06-01 A     1200.0   1
           B     1200.0   0
           C     1200.0   1
2018-07-01 A     1000.0   0
2018-08-01 B     1000.0   0
           C     1000.0   1
2018-09-01 A     1600.0   1
           B     1600.0   1
2018-10-01 C     1000.0   0
           A      800.0   0
           B      800.0   0
2018-11-01 A     1000.0   0
2018-12-01 B     1600.0   0
           C     1600.0   1

Answer 4

IIUC

idx=df.groupby([df.index.get_level_values(1),df.xx]).head(1).index
df.loc[:,'new']=0
df.loc[idx,'new']=1
df
Out[869]: 
                     xx  new
date       type             
2018-01-01 A     1000.0    1
           B     1000.0    1
2018-02-01 B     1200.0    1
2018-03-01 A      800.0    1
           B      800.0    1
           C      800.0    1
2018-04-01 A     1000.0    0
           B     1000.0    0
2018-05-01 B      800.0    0
2018-06-01 A     1200.0    1
           B     1200.0    0
           C     1200.0    1
2018-07-01 A     1000.0    0
2018-08-01 B     1000.0    0
           C     1000.0    1
2018-09-01 A     1600.0    1
           B     1600.0    1
2018-10-01 C     1000.0    0
           A      800.0    0
           B      800.0    0
2018-11-01 A     1000.0    0
2018-12-01 B     1600.0    0
           C     1600.0    1