Count occurences for each year in pandas dataframe based on subgroup

Question

Imagine a pandasdataframe that are given by

df = pd.DataFrame({
    \'id\': [1, 1, 1, 2, 2],
    \'location\': [1, 2, 3, 1, 2],
    \'date\': [p         


        

Answer 1

Create helper DataFrame by groupby with size, unstack and year and join to original df:

df1 = df.join(df.groupby(['id', df['date'].dt.year]).size().unstack(fill_value=0), on='id')
print (df1)
    location       date  2015  2016  2017  2018
id                                             
1          1 2015-01-01     2     1     0     0
1          2 2016-01-01     2     1     0     0
1          3 2015-01-01     2     1     0     0
2          1 2017-01-01     0     0     1     1
2          2 2018-01-01     0     0     1     1

Detail:

print (df.groupby(['id', df['date'].dt.year]).size().unstack(fill_value=0))

date  2015  2016  2017  2018
id                          
1        2     1     0     0
2        0     0     1     1

Another solution with crosstab:

df1 = df.join(pd.crosstab(df.index, df['date'].dt.year), on='id')

print (pd.crosstab(df.index, df['date'].dt.year))
date   2015  2016  2017  2018
row_0                        
1         2     1     0     0
2         0     0     1     1

Answer 2

get_dummies

df.join(pd.get_dummies(df.date.dt.year).sum(level=0))

         date  location  2015  2016  2017  2018
id                                             
1  2015-01-01         1     2     1     0     0
1  2016-01-01         2     2     1     0     0
1  2015-01-01         3     2     1     0     0
2  2017-01-01         1     0     0     1     1
2  2018-01-01         2     0     0     1     1

---

factorize

i, r = pd.factorize(df.index)
j, c = pd.factorize(df.date.dt.year)
n, m = shape = len(r), len(c)
b = np.zeros(shape, dtype=np.int64)
np.add.at(b, (i, j), 1)

df.join(pd.DataFrame(b, r, c).rename_axis('id'))

         date  location  2015  2016  2017  2018
id                                             
1  2015-01-01         1     2     1     0     0
1  2016-01-01         2     2     1     0     0
1  2015-01-01         3     2     1     0     0
2  2017-01-01         1     0     0     1     1
2  2018-01-01         2     0     0     1     1