CODEWITHSUNDEEP
rweighted-average
Pastor Soto
Pastor Soto

Reputation: 386

Weighted average for different columns in R

I want to calculate the weighted average for different columns in a data frame This is my data:

x.4 <- c(2,3,4,5)
a.4 <- c(2,3,4,5)
x.8 <- c(3,24,2,2)
a.8 <- c(2,3,4,7)
x.12 <- c(3,2,4,5)
a.12 <- c(3,2,4,5)
x.24 <- c(2,4,5,2)
a.24 <- c(2,4,5,2)
x.36 <- c(2,1,3,6)
a.36 <- c(2,4,5,2)
x.50 <- c(2,3,5,2)
a.50 <- c(2,3,5,20)
x.100 <- c(2,3,4,5)
a.100 <- c(2,3,4,5)
x.10000 <- c(2,3,46,2)
a.10000 <- c(2,32,46,2)


name_x <- c("a", "b", "c", "d")
df <- data.frame(name_x, x.4,a.4, x.8, a.8, x.12,a.12,x.24,a.24,x.36,a.36,x.50, a.50,x.100,a.100,x.10000, a.10000)

What I want is to create an "x" and "a" variable containing the weighted average using 8 for those that end on 4, 7 for those that end on 8, 6 for those that end on 12, and so on.

x = (x.4 * 8 + x.8 * 7 + x.12 * 6 + x.24 * 5 .......x.10000 * 1)/36
a = (a.4 * 8 + a.8 * 7 + a.12 * 6 + a.24 * 5 .......a.10000 * 1)/36

I create a variable with each of the weighting values multiply by my columns and then I divided but it takes a lot of time. Is there a more clean approach to calculate the weighted average in this situation?

Upvotes: 0

Views: 167

Answers (2)

r2evans
r2evans

Reputation: 160407

Piece-wise, you can use

Xs <- grep("^x\\.", names(df))
Xs_seq <- rev(seq_along(Xs))
as.matrix(df[Xs]) %*% matrix(Xs_seq, ncol = 1) / sum(Xs_seq)
#          [,1]
# [1,] 2.361111
# [2,] 6.833333
# [3,] 4.888889
# [4,] 3.777778

As <- grep("^a\\.", names(df))
As_seq <- rev(seq_along(As))
as.matrix(df[As]) %*% matrix(As_seq, ncol = 1) / sum(As_seq)
#          [,1]
# [1,] 2.166667
# [2,] 3.888889
# [3,] 5.500000
# [4,] 5.805556

If your pattern of column names is consistent (such as "single letter, a period, then numbers), then

Z <- unique(gsub("\\..*", "", grep("^.\\.[0-9]+$", names(df), value = TRUE)))
Z
# [1] "x" "a"

lapply(setNames(nm = Z), function(z) {
  Zs <- grep(paste0("^", z, "\\."), names(df))
  Zs_seq <- rev(seq_along(Zs))
  as.matrix(df[Zs]) %*% matrix(Zs_seq, ncol = 1) / sum(Zs_seq)
})
# $x
#          [,1]
# [1,] 2.361111
# [2,] 6.833333
# [3,] 4.888889
# [4,] 3.777778
# $a
#          [,1]
# [1,] 2.166667
# [2,] 3.888889
# [3,] 5.500000
# [4,] 5.805556

Upvotes: 1

Leonardo
Leonardo

Reputation: 2485

Try this:

library(tidyverse)

the_weights <- 8:1

df2 <- df %>% 
  gather(var, value, c(x.4, x.8, x.12, x.24, x.36, x.50, x.100, x.10000, 
                       a.4, a.8, a.12, a.24, a.36, a.50, a.100, a.10000)) %>% 
  separate(var, c("var", "number"))

df2 %>% 
  group_by(name_x, var) %>% 
  summarise(sum(the_weights*value)/36)

I preferred to divide the two steps for greater clarity but they can also be joined with a pipe

# A tibble: 8 x 3
# Groups:   name_x [4]
  name_x var   `sum(the_weights * value)/36`
  <chr>  <chr>                         <dbl>
1 a      a                              2.17
2 a      x                              2.36
3 b      a                              3.89
4 b      x                              6.83
5 c      a                              5.5 
6 c      x                              4.89
7 d      a                              5.81
8 d      x                              3.78

Upvotes: 1

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