Как создать недостающие строки в фрейме данных

Идея создана MultiIndex автором DataFrame.set_index и DataFrame.reindex автором MultiIndex автором MultiIndex.from_product:

data = {'id': [100, 100, 100, 100, 100, 200, 200], 
    'year':  [2010, 2013, 2014, 2015, 2016, 2010, 2012],
    'value':  [3000, 1000, 2000, 1200, 1300, 2000, 1500]}

df = pd.DataFrame(data) 

#you can specify minimal and maximal year by scalar
mux = pd.MultiIndex.from_product([df['id'].unique(),
                                  np.arange(df['year'].min(), 2019)],
                                  names=['id','year'])

df1 = df.set_index(['id','year']).reindex(mux, fill_value=0).reset_index()
print (df1)
     id  year  value
0   100  2010   3000
1   100  2011      0
2   100  2012      0
3   100  2013   1000
4   100  2014   2000
5   100  2015   1200
6   100  2016   1300
7   100  2017      0
8   100  2018      0
9   200  2010   2000
10  200  2011      0
11  200  2012   1500
12  200  2013      0
13  200  2014      0
14  200  2015      0
15  200  2016      0
16  200  2017      0
17  200  2018      0

#you can specify minimal and maximal year by min and max functions
mux = pd.MultiIndex.from_product([df['id'].unique(),
                                  np.arange(df['year'].min(), df['year'].max() + 1)], 
                                  names=['id','year'])

df2 = df.set_index(['id','year']).reindex(mux, fill_value=0).reset_index()
print (df2)
     id  year  value
0   100  2010   3000
1   100  2011      0
2   100  2012      0
3   100  2013   1000
4   100  2014   2000
5   100  2015   1200
6   100  2016   1300
7   200  2010   2000
8   200  2011      0
9   200  2012   1500
10  200  2013      0
11  200  2014      0
12  200  2015      0
13  200  2016      0

import datetime
import pandas as pd
year = datetime.datetime.today().year #Today date(year)

data = {'id': [100, 100, 100, 100, 100, 200, 200], 
    'year':  [2010, 2013, 2014, 2015, 2016, 2010, 2012],
    'value':  [3000, 1000, 2000, 1200, 1300, 2000, 1500]}

df = pd.DataFrame(data)
min = df['year'].min() #Find min year in data's

difference = year-min
years = range(year, year -difference , -1)

data['year'] = years

df = pd.DataFrame.from_dict(data, orient='index') #When we add years,it gives an error 'arrays must all be same length' so avoid it with adding 'Nan
df.transpose()
print df

Вот подход, который генерирует строки для «недостающих» лет:

data = {'id': [100, 100, 100, 100, 100, 200, 200], 
    'year':  [2010, 2013, 2014, 2015, 2016, 2010, 2012],
    'value':  [3000, 1000, 2000, 1200, 1300, 2000, 1500]}

df = pd.DataFrame(data) 
#Pick the existing year extremes
min_year = df.year.min()
max_year = df.year.max()

#Look for missing years
missing_years = [x for x in range(min_year, max_year+1) if x not in df.year.values]

#Generate a dataframe
df_missing =  pd.DataFrame([[None, x, None] for x in missing_years], columns = df.columns)

#Append it to original 
df.append(df_missing)

Выход

+----+-------+-------+-------+
|    |  id   | year  | value |
+----+-------+-------+-------+
| 0  | 100   | 2010  | 3000  |
| 1  | 100   | 2013  | 1000  |
| 2  | 100   | 2014  | 2000  |
| 3  | 100   | 2015  | 1200  |
| 4  | 100   | 2016  | 1300  |
| 5  | 200   | 2010  | 2000  |
| 6  | 200   | 2012  | 1500  |
| 0  | None  | 2011  | None  |
+----+-------+-------+-------+

IIUC вы хотите интерполировать данные в максимальный диапазон лет по идентификатору:

# get the min and max year range
# change max_year if needed
min_year, max_year = df.year.min(), df.year.max()

# for merge
new_df = pd.DataFrame({'year':np.arange(min_year,max_year+1)})

# groupby:
new_df = df.groupby('id').apply(lambda x: new_df.merge(x, how='outer')
                                       .drop('id',axis=1)
                               ).reset_index()

new_df.value.fillna(0, inplace=True)

# this will fill other columns with copies of others
new_df.groupby('id').ffill()

Выход:

     id  level_1  year   value
0   100        0  2010  3000.0
1   100        1  2011     0.0
2   100        2  2012     0.0
3   100        3  2013  1000.0
4   100        4  2014  2000.0
5   100        5  2015  1200.0
6   100        6  2016  1300.0
7   200        0  2010  2000.0
8   200        1  2011     0.0
9   200        2  2012  1500.0
10  200        3  2013     0.0
11  200        4  2014     0.0
12  200        5  2015     0.0
13  200        6  2016     0.0

Другой метод с использованием groupby и reindex()

l=[g.set_index('year').reindex(range(2010,2019)) for i,g in df.groupby('id')]
final=pd.concat(l).reset_index()
final=final.assign(**{'id':final.id.ffill(),'value':final.value.fillna(0)})
print(final)

    year     id   value
0   2010  100.0  3000.0
1   2011  100.0     0.0
2   2012  100.0     0.0
3   2013  100.0  1000.0
4   2014  100.0  2000.0
5   2015  100.0  1200.0
6   2016  100.0  1300.0
7   2017  100.0     0.0
8   2018  100.0     0.0
9   2010  200.0  2000.0
10  2011  200.0     0.0
11  2012  200.0  1500.0
12  2013  200.0     0.0
13  2014  200.0     0.0
14  2015  200.0     0.0
15  2016  200.0     0.0
16  2017  200.0     0.0
17  2018  200.0     0.0

Используйте метод pd.merge_ordered:

df=pd.DataFrame({'id':(100,100,100,100,100,200,200),'year':(2010,2013,2014,2015,2016,2010,2012),\
                                                        'value':[3000,1000,2000,1200,1300,2000,1500]})

df_year=pd.DataFrame({'year':(2010,2011,2012,2013,2014,2015,2016,2017,2018,2019,2020)})


df_merged=pd.merge_ordered(df,df_year,left_by='id').fillna(0)