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add 1 value to new column and perform operation on it with other columns

I have a dataframe,

x y z new_col
Nan NaN Nan 1
1 2 3 4
2 3 4 20
2 1 3 60


So basically formula is,
1 of new_col that we set first 1 * (1 + 3), then 4 * (1 + 4), then 20 * (1+3), so on..
How should I first create a new column (new_col) and then add 1 at first row, then perform calculation.

x

y

z

n

new_col[n] = new_col[n-1] * (1 + z[n])

2 Answers
2

cumprod

df.assign(new_col=df.z.fillna(0).add(1).cumprod())

x y z new_col
0 NaN NaN NaN 1.0
1 1.0 2.0 3.0 4.0
2 2.0 3.0 4.0 20.0
3 2.0 1.0 3.0 80.0


Attempt to preserve dtype

df.assign(new_col=df.z.fillna(0, downcast='infer').add(1).cumprod())

x y z new_col
0 NaN NaN NaN 1
1 1.0 2.0 3.0 4
2 2.0 3.0 4.0 20
3 2.0 1.0 3.0 80


The point of this is to show how to perform a linear path dependent calculation. Numba is very fast and if the calculation has a time complexity of O(N) then you don't have to be afraid of using this loop in Numba.

If you don't have numba installed and don't want to install it, just remove the @njit decorator.

numba

@njit

from numba import njit

@njit
def f(a):
out = np.zeros_like(a)
out[0] = 1
for i, x in enumerate(a[1:], 1):
out[i] = out[i-1] * (1 + x)
return out

df.assign(new_col=f(df.z.values))

x y z new_col
0 NaN NaN NaN 1.0
1 1.0 2.0 3.0 4.0
2 2.0 3.0 4.0 20.0
3 2.0 1.0 3.0 80.0


With int

int

df.assign(new_col=f(df.z.fillna(0).astype(int).values))

x y z new_col
0 NaN NaN NaN 1
1 1.0 2.0 3.0 4
2 2.0 3.0 4.0 20
3 2.0 1.0 3.0 80


If you are looking for a much simpler solution, this will be helpful.

You can just create a new column named "new_column" and initialize all the values to 1 (since the first value should be 1).

df['new_column'] = 1


Then, you can use a for loop to iterate through the rows and update the new column values according to your formula.

for i in range(1, len(df)):
df.loc[i, 'new_column'] = df['new_column'][i-1] * (1 + df['Z'][i])


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