Pandas percentage change using group by

user_10260243 0
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Suppose I have the following DataFrame:

df = pd.DataFrame({'city': ['a', 'a', 'a', 'b', 'b', 'c', 'd', 'd', 'd'], 
                   'year': [2013, 2014, 2016, 2015, 2016, 2013, 2016, 2017, 2018],
                  'value': [10, 12, 16, 20, 21, 11, 15, 13, 16]})

And I want to find, for each city and year, what was the percentage change of value compared to the year before. My final dataframe would be:

city  year  value
   a  2013    NaN
   a  2014   0.20
   a  2016    NaN
   b  2015    NaN
   b  2016   0.05
   c  2013    NaN
   d  2016    NaN
   d  2017  -0.14
   d  2018   0.23

I tried to use a group in city and then use apply but it didn't work:

df.groupby('city').apply(lambda x: x.sort_values('year')['value'].pct_change()).reset_index()

It didn't work because I couldn't get the year and also because this way I was considereing that I had all years for all cities, but that is not true.

EDIT: I'm not very concerned with efficiency, so any solution that solves the problem is valid for me.

pandas dataframe python
2 Answers

Let's try lazy groupby(), use pct_change for the changes and diff to detect year jump:

groups = df.sort_values('year').groupby(['city'])

df['pct_chg'] = (groups['value'].pct_change()
                    .where(groups['year'].diff()==1)
                )

Output:

  city  year  value   pct_chg
0    a  2013     10       NaN
1    a  2014     12  0.200000
2    a  2016     16       NaN
3    b  2015     20       NaN
4    b  2016     21  0.050000
5    c  2013     11       NaN
6    d  2016     15       NaN
7    d  2017     13 -0.133333
8    d  2018     16  0.230769
user_42384080

Although @Quang's answer is much more elegantly written and concise, I just add another approach using indexing.

sorted_df = df.sort_values(by=['city', 'year'])
sorted_df.loc[((sorted_df.year.diff() == 1) & 
              (sorted_df.city == sorted_df.city.shift(1))), 'pct_chg'] = sorted_df.value.pct_change()

my approach is faster as you can see below run on your df, but the syntax is not as pretty.

%timeit #mine
1.44 ms ± 2.91 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit #@Quang's
2.23 ms ± 40.1 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
user_126639010
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