Multiply constant by previous row and accumulate result in R

user_3880480 0
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I'm trying to compute a new column in a dataframe that multiply a constant by the value of the previous row (in the new created column B) and accumulate the result.

To clarify, here is an example of my original data:

A  
---
100
200
300
400
500

An here is what I'm trying to accomplish:

A    B
-----------
100  100.00
200  302.00
300  608.04
400  1020.20
500  1540.60

(The constant being in this case: 0.02)

The (approximate) formula that I am using in excel is something similar to this:

=(B1*(1+constant)+A2) [In Cell B2]

I've tried using lag, mutate and cumsum from dplyr to achieve this, but so far I've had no luck. Here a reproducible example of what I've managed to achieve thus far:

rate = .02
A <- c(100,200,300,400,500)

df <- data.frame(A)

df = df %>%
  mutate(B = lag(A,default=0)*(1+rate)) %>% 
  mutate(B = cumsum(B)+df$A)

This produce the following output (different than what I'm trying to get):

A    B
100  100.00
200  302.00
300  606.00
400  1012.00
500  1520.00

Can anyone help me with some code to compute this? Thanks for the help!

dplyr r
2 Answers

We can use accumulate

library(dplyr)
library(purrr)
df %>%
    mutate(B = accumulate(A, ~ .x*(1 + rate) + .y))

-output

  A        B
1 100  100.000
2 200  302.000
3 300  608.040
4 400 1020.201
5 500 1540.605

Or the same option in base R using Reduce

 Reduce(function(.x, .y) .x * (1 + rate)  + .y, df$A, accumulate = TRUE)
[1]  100.000  302.000  608.040 1020.201 1540.605
user_37322710

Here's a way to do it using a recursive function:

f = function(x) if (x == 1) 100 else 100 * x + (1 + rate) * f(x - 1)
res <- c()
for(i in 1:5) res <- c(res, f(i))

df %>% 
  mutate(B = res)
    A        B
1 100  100.000
2 200  302.000
3 300  608.040
4 400 1020.201
5 500 1540.605
user_134606020
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