Subtracting one column by another by the first row in every 5 rows in dataframe

Question

Say I have a dataframe, df, such that:

set.seed(123)
df1 <- 0
df1$visit <- rep(c("scr", "1mo", "3mo", "6mo0", "12mo", "2yr"), 2)
df1 <- as.data.frame(df1)
df1$id <- rep(c("101","102"), each = 6)
df1 <- df1[ ,-c(1)]
df1$x <- sample(0:30, 12, replace = T)

df1$want = c(0, -16, -12, -17, -28, -21, 0, 4, -7, -13, 2, -4)

What I would like to do: Subtract every row (X values) after the screening row (can be negative) to create a change variable from screening visit ONLY. So it's essentially looping through the set to calculate the change from screening visit, then repeats that for each ID/set of visits (this dummy set has essentially two ID's).

I've tried: looking on here for similar answers, and closest I could get to was using mutate() from dplyr. All answers I found either tell me how to subtract lagging or leading rows or mutate when certain conditions match.

I could do this in excel maybe but I will reuse this frequently in future analyses.

edit: added variable that would be exactly the right values.

Answer 1

This will work, we just need to group by id , then take advantage of the first() function to take the difference versus the first value of x for each group.

library(tidyverse)

df1 %>% group_by(id) %>% mutate(new = x - first(x))

# A tibble: 12 x 5
# Groups:   id [2]
   visit id        x  want   new
   <fct> <chr> <int> <dbl> <int>
 1 scr   101      30     0     0
 2 1mo   101      14   -16   -16
 3 3mo   101      18   -12   -12
 4 6mo0  101      13   -17   -17
 5 12mo  101       2   -28   -28
 6 2yr   101       9   -21   -21
 7 scr   102      17     0     0
 8 1mo   102      21     4     4
 9 3mo   102      10    -7    -7
10 6mo0  102       4   -13   -13
11 12mo  102      19     2     2
12 2yr   102      13    -4    -4