I have pandas dataframe like this:

d = {'dollar_amount': ['200.25', '350.00', '120.00', '400.50', '1231.25', '700.00', '350.00', '200.25', '2340.00'], 'date': ['22-01-2010','22-01-2010','23-01-2010','15-02-2010','27-02-2010','07-03-2010','14-01-2011','09-10-2011','28-07-2012']}
df = pd.DataFrame(data=d)df['date'] = pd.to_datetime(df['date'], format='%d-%m-%Y')
pd.options.display.float_format = '{:,.4f}'.format
df['dollar_amount'] = df['dollar_amount'].astype(float)
dfdate        dollar_amount
0   22-01-2010  200.25
1   22-01-2010  350.00
2   23-01-2010  120.00
3   15-02-2010  400.50
4   27-02-2010  1231.25
5   07-03-2010  700.00
6   14-01-2011  350.00
7   09-10-2011  200.25
8   11-11-2011  2340.00
9   12-12-2011  144.50
10  12-09-2012  760.00
11  22-10-2012  255.00
12  28-07-2012  650.00

I want to sum amounts for each day in each year. So I am dividing the years like this:

date1 = df[(df['date'] >= '2010-01-01') & (df['date'] < '2011-01-01')]
date2 = df[(df['date'] >= '2011-01-01') & (df['date'] < '2012-01-01')]
date3 = df[(df['date'] >= '2012-01-01') & (df['date'] < '2013-01-01')]

So now I have 3 dataframes with dates from the year 2010 in date1 dataframe, dates from the year 2011 in date2 and dates from 2012 in date3.

Lets look at date1:

print type(date1)
date1<class 'pandas.core.frame.DataFrame'>date        dollar_amount
0   2010-01-22  200.2500
1   2010-01-22  350.0000
2   2010-01-23  120.0000
3   2010-02-15  400.5000
4   2010-02-27  1,231.2500
5   2010-03-07  700.0000

Next I am summing up the amounts date wise, so I am grouping on date using this:

date1 = date1.groupby('date', as_index=False).sum()
date1 = date1[['date','dollar_amount']].sort_values(by=['date'], 
ascending=True)date2 = date2.groupby('date', as_index=False).sum()
date2 = date2[['date','dollar_amount']].sort_values(by=['date'], 
ascending=True)date3 = date3.groupby('date', as_index=False).sum()
date3 = date3[['date','dollar_amount']].sort_values(by=['date'], 
ascending=True)

Let's look at the dateframe date1 now:

date1date        dollar_amount
0   2010-01-22  550.2500
1   2010-01-23  120.0000
2   2010-02-15  400.5000
3   2010-02-27  1,231.2500
4   2010-03-07  700.0000

This is just sorting them in ascending date wise order:

date1 = date1[['date','dollar_amount']].sort_values(by=['date'], 
ascending=True)

Now I have got the date wise sum of dollarAmounts for each year in different dataframes. Then I am plotting traces for each year. Its working fine and fulfilling the task. But this code is very redundant and I am copying the same code and if I have say data from year 2000 to 2017 then I will have to copy and paste the same piece of code 18 times. I think its not very effective way of doing this.

I am sure there must be a better way of doing this but I cant figure out how. Kindly help me. Thanks.

I think you can create MultiIndex by years to output:

df1 = df.groupby('date', as_index=False)['dollar_amount'].sum()
df1 = df1.set_index(df['date'].rename('year').dt.year, append=True).swaplevel(0,1)
print (df1)date  dollar_amount
year                            
2010 0 2010-01-22       550.25001 2010-01-23       120.00002 2010-02-15       400.50003 2010-02-27     1,231.25004 2010-03-07       700.0000
2011 5 2011-01-14       350.00006 2011-10-09       200.2500
2012 7 2012-07-28     2,340.0000print (df1.loc[2010])date  dollar_amount
0 2010-01-22       550.2500
1 2010-01-23       120.0000
2 2010-02-15       400.5000
3 2010-02-27     1,231.2500
4 2010-03-07       700.0000print (df1.loc[2011])date  dollar_amount
5 2011-01-14       350.0000
6 2011-10-09       200.2500print (df1.loc[2012])date  dollar_amount
7 2012-07-28     2,340.0000

If want create dictionary of DataFrames:

d = dict(tuple(df.groupby(df['date'].dt.year)))
print (d)print (d[2010])date  dollar_amount
0 2010-01-22       550.2500
1 2010-01-23       120.0000
2 2010-02-15       400.5000
3 2010-02-27     1,231.2500
4 2010-03-07       700.0000print (d[2011])date  dollar_amount
5 2011-01-14       350.0000
6 2011-10-09       200.2500print (d[2012])date  dollar_amount
7 2012-07-28     2,340.0000