pandas DataFrame: aggregate values within blocks of repeating IDs

Question

Given a DataFrame with an ID column and corresponding values column, how can I aggregate (let's say sum) the values within blocks of repeating IDs?

Example DF:

import numpy as np
import pandas as pd

df = pd.DataFrame(
    {'id': ['a', 'a', 'a', 'a', 'a', 'b', 'b', 'b', 'a', 'a', 'b', 'a', 'b', 'b', 'b'],
     'v': np.ones(15)}
    )

Note that there's only two unique IDs, so a simple groupby('id') won't work. Also, the IDs don't alternate/repeat in a regular manner. What I came up with was to recreate the index, to represent the blocks of changed IDs:

# where id changes:
m = [True] + list(df['id'].values[:-1] != df['id'].values[1:])

# generate a new index from m:
idx, i = [], -1
for b in m:
    if b:
        i += 1
    idx.append(i)

# set as index:
df = df.set_index(np.array(idx))

# now I can use groupby:
df.groupby(df.index)['v'].sum()
# 0    5.0
# 1    3.0
# 2    2.0
# 3    1.0
# 4    1.0
# 5    3.0

This re-creation of the index feels sort-of not how you'd do this in pandas . What did I miss? Is there a better way to do this?

Answer 1

Here is necessary create helper Series with compare shifted values for not equal by ne with cumulative sums and pass to groupby , for id column is possible pass together in list, remove first level of MultiIndex by first reset_index(level=0, drop=True) and then convert index to column id :

print (df['id'].ne(df['id'].shift()).cumsum())
0     1
1     1
2     1
3     1
4     1
5     2
6     2
7     2
8     3
9     3
10    4
11    5
12    6
13    6
14    6
Name: id, dtype: int32

df1 = (df.groupby([df['id'].ne(df['id'].shift()).cumsum(), 'id'])['v'].sum()
          .reset_index(level=0, drop=True)
          .reset_index())
print (df1)
  id    v
0  a  5.0
1  b  3.0
2  a  2.0
3  b  1.0
4  a  1.0
5  b  3.0

Another idea is use GroupBy.agg with dictioanry and aggregate id column by GroupBy.first :

df1 = (df.groupby(df['id'].ne(df['id'].shift()).cumsum(), as_index=False)
         .agg({'id':'first', 'v':'sum'}))