Reputation: 70003
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?
Upvotes: 28
Views: 36469
Reputation: 581
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
Upvotes: 2
Reputation: 394459
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
Upvotes: 49
Reputation: 1980
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
Upvotes: 2