R 根据同一列中的前一个单元格值生成一个单元格值

Question

I'm currently in the process of running simulations based on certain data. The endgame is to generate a column where the first value is based on one formula, and then the second, third and fourth values are based on the previous value. (eg entry n°2 is dependent on n°1, n°3 on n°2) I've solved this by running the mutate function 3 times over. However, with tidiness in mind, I would like to either have a short loop or use one of the apply functions to execute all 3 repeats at once. Any suggestions?

Here's an example:

sampleframe <- data.frame("value1" = c(15,18,22,19),
                          "value2" = c(12,14,13,12),
                          "parameter" = c(0.8,NA,NA,NA))

sampleframe <- sampleframe %>%
  mutate("value3" = value2 * parameter)

This generatese the dataframe with the first row of the "value3" column, based on one formula. Then I would like to generate the last 3 rows. I run this line:

sampleframe <- sampleframe %>%
  mutate(`value3`= ifelse(is.na(value3) == FALSE,  value3,lag(value3) * value2))

which generates the second row value whilst retaining the first row value. I then have to run the same command an extra two times to get the last 2 rows to fill. It works in the sense that it preserves previous values while always generating the next one, but it seems remarkably inefficient. Back to my question, is there a better way to do this? (I assume there is)

Edit: Given the purrr solution, I ran into the following problem when expanding my above example. If I want to add a constant in the expression, the solution doesn't work anymore:

sampleframe <- sampleframe %>%
  mutate(`value3`= ifelse(is.na(value3) == FALSE,  value3,lag(value3) * value2 + value 1))

In the purr solution:

sampleframe %>% 
  mutate(
    value3 = if_else(row_number() == 1, value2*parameter, value2),
    value3 = accumulate(value3, prod)
  )

Each term in value3 will multiply value 2. The problem is that adding the constant after value 2:

sampleframe %>% 
  mutate(
    value3 = if_else(row_number() == 1, value2*parameter, value2 + value1),
    value3 = accumulate(value3, prod)
  )

Doesn't yield the desired result, since I don't want value1 to be multiplied by value2. Adding it in the second term:

sampleframe %>% 
  mutate(
    value3 = if_else(row_number() == 1, value2*parameter, value2),
    value3 = accumulate(value3, prod) + value1
  )

also doesn't work, because it adds value1 as a block at the very end, meaning that line 1 and 2 are computed correctly, but 3 and 4 are not. I tried any way I could think of to make this command work, but I'm not familiar enough with the purrr package to find a fix. Any ideas?

Answer 1

Limiting my answer to your current approach, you can make things more efficient by using a for loop:

number_iterations = 3

# setup
sampleframe <- data.frame("value1" = c(15,18,22,19),
                          "value2" = c(12,14,13,12),
                          "parameter" = c(0.8,NA,NA,NA))

sampleframe <- sampleframe %>%
  mutate("value3" = value2 * parameter)

# run
for(ii = 1:number_iterations){
  sampleframe <- sampleframe %>%
    mutate(`value3`= ifelse(is.na(value3) == FALSE,  value3,lag(value3) * value2))
}

The four loop will handle the running of your code as many times at you spcify in number_iterations .

However, I would usually recommend operations like mutate for working on an entire column at once, rather than updating one value at a time. So you will likely get further improvements in efficiency from investigating different data structures and solution approaches.

Answer 2

You can use accumulate() from {purrr} and multiply the numbers sequentially.

sampleframe %>% 
  mutate(
    value3 = if_else(row_number() == 1, value2*parameter, value2),
    value3 = accumulate(value3, prod)
  )


#   value1 value2 parameter  value3
# 1     15     12       0.8     9.6
# 2     18     14        NA   134.4
# 3     22     13        NA  1747.2
# 4     19     12        NA 20966.4