R: Create multiple new columns based upon other columns

Question

Let's say I have a data-frame that looks like this

dd <- read.table(header = TRUE, text = "ID week1_t week1_a  week2_t week2_a
  1      12      22       17       4   
  1      15      32       18       5   
  1      24      12       29       6   
  2      45      11       19       8   
  2      23      33       20      10")

Is there a straightforward way to create a week1_d column, a week2_d column, and so on for every week, that is based on the difference between week1_t and week1_a? Or do I have to manually construct the "difference" columns?

Expected output looks like this:

dd <- read.table(header = TRUE, text = "ID week1_t week1_a  week2_t week2_a week1_d week2_d 
  1      12      22       17       4       10       -13                 
  1      15      32       18       5       17       -13   
  1      24      12       29       6       -12      -23 
  2      45      11       19       8       -34      -11
  2      23      33       20      10       10       -10      ")

In actuality, there are around 30 weeks, so I am trying to avoid manually doing this. I was thinking a for loop the runs through each week, and grepping columns that match week+(index of loop). Is there a better way of doing this?

Answer 1

From a "tidy data" perspective, your problem is that you're encoding (multiple!) pieces of data in your column names: the week number and whatever the letter stands for. I would convert to a long format where week is a column, define d = a - t , and (if necessary) convert back to wide format. But probably I'd keep it in the long format because if there are any other operations you want to do they'll probably be easier to implement on the long data (more manipulation, modeling, plotting...).

library(tidyr)
library(dplyr)

long = dd %>% 
    mutate(real_id = 1:n()) %>%
    gather(key = key, value = value, starts_with("week")) %>%
    separate(key, into = c("week", "letter")) %>% 
    spread(key = letter, value = value) %>%
    mutate(d = a - t)

head(long)
#   ID real_id  week  a  t   d
# 1  1       1 week1 22 12  10
# 2  1       1 week2  4 17 -13
# 3  1       2 week1 32 15  17
# 4  1       2 week2  5 18 -13
# 5  1       3 week1 12 24 -12
# 6  1       3 week2  6 29 -23

wide = gather(long, key = letter, value = value, a, t, d) %>%
    mutate(key = paste(week, letter, sep = "_")) %>%
    select(-week, -letter) %>%
    spread(key = key, value = value)

wide
#   ID real_id week1_a week1_d week1_t week2_a week2_d week2_t
# 1  1       1      22      10      12       4     -13      17
# 2  1       2      32      17      15       5     -13      18
# 3  1       3      12     -12      24       6     -23      29
# 4  2       4      11     -34      45       8     -11      19
# 5  2       5      33      10      23      10     -10      20

Answer 2

We split the 'week' columns ( dd[-1] ) by the names of the dataset after removing the suffix with sub into a list , get the difference between the two columns and assign the list elements to create new columns in 'dd'.

lst <-  lapply(split.default(dd[-1], 
           sub("_.*", "", names(dd)[-1])), function(x) x[2]-x[1])
dd[paste0("week_", seq_along(lst), "d")] <- lapply(lst, unlist, use.names=FALSE)
dd
#    ID week1_t week1_a week2_t week2_a week1_d week2_d
#1  1      12      22      17       4      10     -13
#2  1      15      32      18       5      17     -13
#3  1      24      12      29       6     -12     -23
#4  2      45      11      19       8     -34     -11
#5  2      23      33      20      10      10     -10

---

If the columns are alternating ie 'week1_t' followed by 'week1_a', then 'week2_t', followed by 'week2_a', etc.

Un1 <- unique(sub("_.*", "", names(dd)[-1]))
i1 <-  c(TRUE, FALSE)
dd[paste0(Un1, "_d")] <-  dd[-1][!i1]- dd[-1][i1]
dd
#  ID week1_t week1_a week2_t week2_a week1_d week2_d
#1  1      12      22      17       4      10     -13
#2  1      15      32      18       5      17     -13
#3  1      24      12      29       6     -12     -23
#4  2      45      11      19       8     -34     -11
#5  2      23      33      20      10      10     -10