Compile data for the same id - r

Question

I want to compile this data frame into a dataframe with only one row for each id, for each column value1, value2, value3, I want to take the value for A- B. I have tried to make a for loop to run over the ids, but I dont know how to continue, any advice? It would be really helpful with a function to run over many many columns, this is just an example of 3 columns. Thanks!
df <- data.frame(
id=rep(1:5,each=2),
trt = rep(c("A","B","A","B", "B","A", "B","A","A","B"),),
value1 = sample(1:100,10),
value2 = sample(1:100,10),
value3 = sample(1:100,10),
stringsAsFactors = F)
> df
id trt value1 value2 value3
1 A 98 64 91
1 B 64 48 16
2 A 26 31 87
2 B 93 62 24
3 B 32 37 71
3 A 83 6 92
4 B 82 68 36
4 A 30 5 66
5 A 87 29 48
5 B 86 47 82
for (i in length(id)) {
d <- df[df$id==i,]
d$value1[d$trt=="A"]- d$value1[d$trt=="B"]
print(i)
}

Answer 1

Here is a base R solution with the split/lapply/combine strategy. See Hadley Wickham's paper.
cols <- grep("value", names(df), value = TRUE)
sp <- split(df[c("trt", cols)], df$id)
res <- lapply(seq_along(sp), function(i){
X <- sp[[i]][order(sp[[i]]$trt), ]
Y <- X[1, cols] - X[2, cols]
cbind(data.frame(id = names(sp)[i]), Y)
})
res <- do.call(rbind, res)
res
# id value1 value2 value3
#1 1 -6 63 28
#3 2 -32 8 -18
#6 3 6 -7 -63
#8 4 -67 -45 2
#9 5 1 20 45

Answer 2

Your Data:
set.seed(42)
df <- data.frame(
id=rep(1:5,each=2),
trt = rep(c("A","B","A","B", "B","A", "B","A","A","B"),),
value1 = sample(1:100,10),
value2 = sample(1:100,10),
value3 = sample(1:100,10),
stringsAsFactors = F)
Solution:
library(tidyr)
library(dplyr)
df %>%
gather(key, value, -trt, -id) %>%
mutate(value = if_else(trt == "B", -value, value)) %>%
group_by(id, key) %>%
summarise(value = sum(value)) %>%
ungroup() %>%
spread(key, value)
The point here is to, at first, transform with gather the dataframe into long form and put all your value in one column.
Seen that values with B must be subtracted, I set values with B as negative.
At that point I'm free to some all values by id and key (which is equal to the label value1, value2, value3).
I can return to the initial configuration of the dataframe with one column per each "value" with spread.
It doesn't matter how many "valueN" columns you have.

Answer 3

You might also want to try the data.tablepackage (using the same reproducible data as #Edo):
library(data.table)
setDT(df, key = c("id","trt"))
df[, lapply(.SD, function(x) diff(x) * -1), by = id, .SDcols = -2]
The result:
id value1 value2 value3
1: 1 -16 17 -58
2: 2 -49 23 -55
3: 3 82 48 -18
4: 4 -23 9 13
5: 5 -18 -79 -7
Alternatively, you could also do:
library(data.table)
setorder(df, id, -trt)
setDT(df)[, lapply(.SD, diff), by = id, .SDcols = -2]