How to compute cumulative sum of previous N rows in pandas?

Question

I am working with pandas, but I don't have so much experience. I have the following DataFrame:

          A
0       NaN
1      0.00
2      0.00
3      3.33
4     10.21
5      6.67
6      7.00
7      8.27
8      6.07
9      2.17
10     3.38
11     2.48
12     2.08
13     6.95
14     0.00
15     1.75
16     6.66
17     9.69
18     6.73
19     6.20
20     3.01
21     0.32
22     0.52

and I need to compute the cumulative sum of the previous 11 rows. When there is less than 11 previously, they remaining are assumed to be 0.

        B
0     NaN
1    0.00
2    0.00
3    0.00
4    3.33
5    13.54
6    20.21
7    27.20
8    35.47
9    41.54
10    43.72
11   47.09
12   49.57 
13   51.65
14   58.60
15   58.60
16   57.02
17   53.48
18   56.49
19   56.22
20   54.16
21   51.10
22   49.24

I have tried:

df['B'] = df.A.cumsum().shift(-11).fillna(0)

However, this is not achieving what I want, but this is rotating the result of a cumulative sum. How can I achieve this?

Bushey

Answer 1

Call rolling with min_periods=1 and window=11 and sum:

In [142]:
df['A'].rolling(min_periods=1, window=11).sum()

Out[142]:
0       NaN
1      0.00
2      0.00
3      3.33
4     13.54
5     20.21
6     27.21
7     35.48
8     41.55
9     43.72
10    47.10
11    49.58
12    51.66
13    58.61
14    55.28
15    46.82
16    46.81
17    49.50
18    47.96
19    48.09
20    48.93
21    45.87
22    43.91
Name: A, dtype: float64

Flake

Answer 2

you might have to do it the hard way

B = []
i =0
m_lim = 11
while i<len(A):
    if i<m_lim:
      B.append(sum(A[0:i]))
    if i>=m_lim and i < len(A) -m_lim:
        B.append(sum(A[i-m_lim:i]))
    if i>= len(A) -m_lim:
      B.append(sum(A[i:]))
    i=i+1
df['B'] = B

Layla

Answer 3

Check the pandas.Series.expanding. The series.expanding(min_periods=2).sum()

will do the job for you. And don't forget to set 0-th element, since it is NaN. I mean,

accumulation = series.expanding(min_periods=2).sum()
accumulation[0] = series[0] # or as you like

Jungian