How to get the sum of differences in R dataframe

user_18433340 0
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i am wondering if anyone knows of any functions/tactics for calculating this problem i have:

so if I have the variable x

x <- c(1,2,3,4,5,6,7,8,9) 
dataframe <- data.frame(x)
dataframe$y <- 0

and I want to calculate the sum of the differences between each value and the rest of the values. Then each value in this variable will have a score of the difference. Something along the lines of: y[1]<-sum((x[1]-2) + (x[1]-3) + (x[1]-4)....(x[1]-9))

Something along the lines of this (since in reality I have a big data frame with many conditions i want to run this calculation over)

difference_sum <- 0

for (i in x) {
  value_of_interest <- x[i] 
  difference_sum <- difference_sum  + (x[i] - x[i+1])
  difference_sum <- difference_sum  + (x[i] - x[i+2])
} 
#all the way through the end of the list

thanks!

tidyverse dataframe r
3 Answers

Not sure if these are the values you're looking for, but perhaps

o <- outer(dataframe$x, dataframe$x, `-`)
o[lower.tri(o)] <- NA
dataframe$y <- rowSums(o, na.rm = TRUE)
dataframe
#   x   y
# 1 1 -36
# 2 2 -28
# 3 3 -21
# 4 4 -15
# 5 5 -10
# 6 6  -6
# 7 7  -3
# 8 8  -1
# 9 9   0
user_33582720

You can use imap() from purrr.

library(dplyr)
library(purrr)

df %>%
  mutate(y = imap_dbl(x, ~ sum(.x - x[-(1:.y)])))

#   x   y
# 1 1 -36
# 2 2 -28
# 3 3 -21
# 4 4 -15
# 5 5 -10
# 6 6  -6
# 7 7  -3
# 8 8  -1
# 9 9   0

The method above may break in case x is a named vector or the data is grouped. A safer way is passing the indices 1:n() into map2():

df %>%
  mutate(y = map2_dbl(x, 1:n(), ~ sum(.x - x[-(1:.y)])))
user_100689850

Using data.table:

setDT(dataframe)[, y:=rev(cumsum(rev(x)))]
dataframe[, y:=(.N:1)*x - y]
dataframe
##    x   y
## 1: 1 -36
## 2: 2 -28
## 3: 3 -21
## 4: 4 -15
## 5: 5 -10
## 6: 6  -6
## 7: 7  -3
## 8: 8  -1
## 9: 9   0
user_29850070
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