How to Divide a Row with Another Row by Pipe Function in R

I am trying to create a new column named 'DB Q' by dividing the value of the two rows and I am not able to do it. I thought I could do it with rowwise() function but it is not working, I am not sure if I am complicating this by keeping the other column as well while performing or the syntax is just simply wrong.

Attached is the code which I wrote, ideally I want to create a lot more rows so passing it through pipe is preferred. Can someone help me out.

library(readxl)
library(dplyr)

Pd_Clean <- Pd_Clean %>% 
  rowwise(SKU) %>% 
  mutate(`DB Q`= `BH Q`/SI)

The data frame I am working with is given below:

structure(list(SKU = c("XYZ_20645", "XYZ_20645"), `Key Figures` = c("BH Q", 
"SI"), `Jan-19` = c(NA, 1), `Feb-19` = c(NA, 1), `Mar-19` = c(NA, 
1), `Apr-19` = c(NA, 1), `May-19` = c(NA, 1), `Jun-19` = c(NA, 
1), `Jul-19` = c(NA, 1), `Aug-19` = c(NA, 1), `Sep-19` = c(NA, 
1), `Oct-19` = c(NA, 1), `Nov-19` = c(NA, 1), `Dec-19` = c(13, 
1), `Jan-20` = c(NA, 1), `Feb-20` = c(14, 1), `Mar-20` = c(NA, 
1), `Apr-20` = c(NA, 1), `May-20` = c(NA, 1), `Jun-20` = c(NA, 
1), `Jul-20` = c(NA, 1), `Aug-20` = c(NA, 1.419196), `Sep-20` = c(NA, 
0.853283), `Oct-20` = c(NA, 0.799518), `Nov-20` = c(24, 0.709845
), `Dec-20` = c(NA, 0.891075), `Jan-21` = c(NA, 1.113528), `Feb-21` = c(NA, 
0.593329), `Mar-21` = c(14, 0.914788), `Apr-21` = c(NA, 1.111516
), `May-21` = c(NA, 1.287797), `Jun-21` = c(110, 1.254029), `Jul-21` = c(NA, 
1.052091), `Aug-21` = c(35, 1.491094), `Sep-21` = c(NA, 0.893486
), `Oct-21` = c(59, 0.831871), `Nov-21` = c(79, 0.738738), `Dec-21` = c(316, 
0.910051)), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA, 
-2L))

The Expected output is given below: The row highlighted in yellow is the new row which needs to be created and the cell highlighted in blue is the mathematical formulae.

One base R approach is to use simple indexing to create a new column then lapply to perform the function across columns:

Pd_Clean[3, -c(1:2)] <- lapply(Pd_Clean[-c(1:2)], 
                               function(x) ifelse(is.na(x[1]), 0, 
                                                  round(x[1] / x[2]))) 
 
Pd_Clean[3, 1:2] <- c(Pd_Clean[1,1], "DB Q")

You could also do the first step without lapply(), but not as clean in my option:

Pd_Clean[3, -c(1:2)] <- ifelse(is.na(Pd_Clean[3, -c(1:2)]), 0, 
                               Pd_Clean[1, -c(1:2)] / Pd_Clean[2, -c(1:2)])

Output:

Pd_Clean[22:38] # columns in image

# `Aug-20` `Sep-20` `Oct-20` `Nov-20` `Dec-20` `Jan-21` `Feb-21` `Mar-21` `Apr-21` `May-21` `Jun-21` `Jul-21` `Aug-21` `Sep-21` `Oct-21` `Nov-21` `Dec-21`
#    <dbl>    <dbl>    <dbl>    <dbl>    <dbl>    <dbl>    <dbl>    <dbl>    <dbl>    <dbl>    <dbl>    <dbl>    <dbl>    <dbl>    <dbl>    <dbl>    <dbl>
#  1    NA      NA       NA       24       NA        NA      NA       14        NA       NA      110       NA       35      NA       59       79      316    
#  2     1.42    0.853    0.800    0.710    0.891     1.11    0.593    0.915     1.11     1.29     1.25     1.05     1.49    0.893    0.832    0.739    0.910
#  3     0       0        0       34        0         0       0       15         0        0       88        0       23       0       71      107      347 

It's a little unclear what you mean by "I want to create a lot more rows" so if you provide more robust example data I am happy to update.

I can't see the image you attached. Is this the output you wanted? I wrote a pretty ugly code and I'm sure there is a better way, maybe with c_across:

Pd_Clean |>
t() |>
data.frame() |>
mutate(across(everything(),as.numeric)) |>
mutate(answer = X1/X2) |>
t() |>
data.frame() |>
suppressWarnings()