Subtracting similar column names R - r

Question

I have a dataframe with columns that have 'x1' and 'x1_fit' with the numbers going up to 5 in some cases.
date <- seq(as.Date('2019-11-04'), by = "days", length.out = 7)
x1 <- c(100,120,111,152,110,112,111)
x1_fit <- c(150,142,146,148,123,120,145)
x2 <- c(110,130,151,152,150,142,161)
x2_fit <- c(170,172,176,178,173,170,175)
df <- data.frame(date,x1,x1_fit,x2,x2_fit)
How can I do x1_fit - x1 and so on. The number of x's will change every time.

Answer 1

You can select those columns with regular expressions (surppose the columns are in appropriate order):
> df[, grep('^x\\d+_fit$', colnames(df))] - df[, grep('^x\\d+$', colnames(df))]
x1_fit x2_fit
1 50 60
2 22 42
3 35 25
4 -4 26
5 13 23
6 8 28
7 34 14
If you want to assign the differences to the original df:
df[, paste0(grep('^x\\d+$', colnames(df), value = TRUE), '_diff')] <-
df[, grep('^x\\d+_fit$', colnames(df))] - df[, grep('^x\\d+$', colnames(df))]
# > df
# date x1 x1_fit x2 x2_fit x1_diff x2_diff
# 1 2019-11-04 100 150 110 170 50 60
# 2 2019-11-05 120 142 130 172 22 42
# 3 2019-11-06 111 146 151 176 35 25
# 4 2019-11-07 152 148 152 178 -4 26
# 5 2019-11-08 110 123 150 173 13 23
# 6 2019-11-09 112 120 142 170 8 28
# 7 2019-11-10 111 145 161 175 34 14

Answer 2

Solution from #mt1022 is straightforward, however since you have tagged this as dplyr, here is one approach following it where we convert the data to long format, subtract the corresponding values and get the data in wide format again.
library(dplyr)
library(tidyr)
df %>%
pivot_longer(cols = -date) %>%
mutate(name = sub('_.*', '', name)) %>%
group_by(date, name) %>%
summarise(diff = diff(value)) %>%
pivot_wider(names_from = name, values_from = diff) %>%
rename_at(-1, ~paste0(., "_diff")) %>%
left_join(df, by = "date")
# date x1_diff x2_diff x1 x1_fit x2 x2_fit
# <date> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
#1 2019-11-04 50 60 100 150 110 170
#2 2019-11-05 22 42 120 142 130 172
#3 2019-11-06 35 25 111 146 151 176
#4 2019-11-07 -4 26 152 148 152 178
#5 2019-11-08 13 23 110 123 150 173
#6 2019-11-09 8 28 112 120 142 170
#7 2019-11-10 34 14 111 145 161 175

Answer 3

In base R, you could loop over the unique column names and diff on the the fitted column using
> lapply(setNames(nm = unique(gsub("_.*", "", names(df)))), function(nm) {
fit <- paste0(nm, "_fit")
diff <- df[, nm] - df[, fit]
})
# $x1
# [1] -50 -22 -35 4 -13 -8 -34
#
# $x2
# [1] -60 -42 -25 -26 -23 -28 -14
Here, I set the Date column as the row names and removed the column using
df <- data.frame(date,x1,x1_fit,x2,x2_fit)
row.names(df) <- df$date
df$date <- NULL
but you could just loop over the the column names without the Date column.

Answer 4

We can also do with a split in base R
out <- sapply(split.default(df[-1], sub("_.*", "", names(df)[-1])),
function(x) x[,2] - x[1])
df[sub("\\..*", "_diff", names(lst1))] <- out
df
# date x1 x1_fit x2 x2_fit x1_diff x2_diff
#1 2019-11-04 100 150 110 170 50 60
#2 2019-11-05 120 142 130 172 22 42
#3 2019-11-06 111 146 151 176 35 25
#4 2019-11-07 152 148 152 178 -4 26
#5 2019-11-08 110 123 150 173 13 23
#6 2019-11-09 112 120 142 170 8 28
#7 2019-11-10 111 145 161 175 34 14