Given a parameter p, be any float or integer.
For example, let p=4
| time | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Numbers | a1 | a1*(0.5)^(1/p)^(2-1) | a1*(0.5)^(1/p)^(2-1) | a1*(0.5)^(1/p)^(3-1) | a1*(0.5)^(1/p)^(4-1) |
| Numbers | nan | a2 | a2*(0.5)^(1/p)^(3-2) | a2*(0.5)^(1/p)^(4-2) | a2*(0.5)^(1/p)^(5-2) |
| Numbers | nan | nan | a3 | a3*(0.5)^(1/p)^(4-3) | a3*(0.5)^(1/p)^(5-3) |
| Numbers | nan | nan | nan | a4 | a4*(0.5)^(1/p)^(5-4) |
| Number | nan | nan | nan | nan | a5 |
| Final Results | a1 | sum of column 2 | sum of column 3 | sum of column 4 | sum of column 5 |
Numbers like a1,a2,a3,a4,a5,…,at is given, our goal is to find the Final Results. Combining the answer provided by mozway, I wrote the following function which works well. It is a matrix way to solve the problem.
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def hl(p,column):2
a = np.arange(len(copy_raw))3
factors = (a[:,None]-a)4
factors = np.where(factors<0, np.nan, factors)5
inter = ((1/2)**(1/p))**factors6
copy_raw[column] = np.nansum(copy_raw[column].to_numpy()*inter, axis=1) 7
However, I don’t think this method will work well if we are dealing with large dataframe. Are there any better way to fix the problem? (In this case, faster = better.)
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Answer
Assuming your number of rows is not too large, you can achieve this with numpy broadcasting:
First create a 2D array of factors:
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a = np.arange(len(df))2
factors = (a[:,None]-a)3
factors = np.where(factors<0, np.nan, factors)4
# array([[ 0., nan, nan, nan, nan],5
# [ 1., 0., nan, nan, nan],6
# [ 2., 1., 0., nan, nan],7
# [ 3., 2., 1., 0., nan],8
# [ 4., 3., 2., 1., 0.]])9
Then map to your data and sum:
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df['number2'] = np.nansum(df['number'].to_numpy()*(1/2)**factors, axis=1) 2
example output:
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Index Time number number22
0 0 1997-WK01 1 1.00003
1 1 1997-WK02 2 2.50004
2 2 1997-WK03 3 4.25005
3 3 1997-WK04 2 4.12506
4 4 1997-WK05 4 6.06257
intermediate:
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df['number'].to_numpy()*(1/2)**factors2
3
# array([[1. , nan, nan, nan, nan],4
# [0.5 , 2. , nan, nan, nan],5
# [0.25 , 1. , 3. , nan, nan],6
# [0.125 , 0.5 , 1.5 , 2. , nan],7
# [0.0625, 0.25 , 0.75 , 1. , 4. ]])8