R data.frame add a column depending on row-values - r

Question

In R, I have a data.frame that looks like this:
X Y
20 7
25 84
15 62
22 12
60 24
40 10
60 60
12 50
11 17
now, i want a new Colum, lets call it "SumX", that adds two following values of X into a new field of that SumX column, and one that does the same to "SumY" column. So the result data.frame would look like this:
X Y SumX SumY
20 7 20 #first row = X 7 #first row = Y
25 84 45 #X0 + X1 91 #Y0 + Y1
15 62 40 #X1 + X2 146 #Y1 + Y2
22 12 37 #X2 + X3 74 #Y2 + Y3
60 24 82 #X3 + X4 36 #Y3 + Y4
40 10 100 #X4 + X5 34 #Y4 + Y5
60 60 100 #and so on 70 #and so on
12 50 72 110
11 17 23 67
I can do simple X + Y into a new column with
myFrame$SumXY <- with(myFrame, X+Y)
but it there a simple way to add two X (n + (n-1)) values into SumX, and two Y (n + (n-1)) into SumY? Even if it is with a while-loop, though i would prefer a simpler way (its a lot of data like this). Any help is much appreciated! (I'm still pretty new to R)

Answer 1

The rollapply function from the zoo package will work here.
The following code block will create the rolling sum of each 2 adjacent values.
require(zoo)
myFrame$SumX <- rollapply(myFrame$X, 2, sum) # this is a rolling sum of every 2 values
You could add by = 2 as an argument to rollapply in order to not have a rolling sum (i.e. it sums values 1+2, then 3+4, then 5+6 etc.).
Look up ?rollapply for more info.

Answer 2

Here's a dplyr approach.
Use mutate() to add a new colum and var + lag(var, default = 0) to compute your variable. Example:
library(dplyr)
d <- data.frame(
x = 1:10,
y = 11:20,
z = 21:30
)
mutate(d, sumx = x + lag(x, default = 0))
#> x y z sumx
#> 1 1 11 21 1
#> 2 2 12 22 3
#> 3 3 13 23 5
#> 4 4 14 24 7
#> 5 5 15 25 9
#> 6 6 16 26 11
#> 7 7 17 27 13
#> 8 8 18 28 15
#> 9 9 19 29 17
#> 10 10 20 30 19
More variables can be handled similarly:
mutate(d, sumx = x + lag(x, default = 0), sumy = y + lag(y, default = 0))
#> x y z sumx sumy
#> 1 1 11 21 1 11
#> 2 2 12 22 3 23
#> 3 3 13 23 5 25
#> 4 4 14 24 7 27
#> 5 5 15 25 9 29
#> 6 6 16 26 11 31
#> 7 7 17 27 13 33
#> 8 8 18 28 15 35
#> 9 9 19 29 17 37
#> 10 10 20 30 19 39
If you know that you want to do this for many, or even EVERY column in your data frame, then here's a standard evaluation approach with mutate_() that uses a custom function I adapted from this blog post (note you need to have the lazyeval package installed). The function gets applied to each column in a for loop (which could probably be optimised).
f <- function(df, col, new_col_name) {
mutate_call <- lazyeval::interp(~ x + lag(x, default = 0), x = as.name(col))
df %>% mutate_(.dots = setNames(list(mutate_call), new_col_name))
}
for (var in names(d)) {
d <- f(d, var, paste0('sum', var))
}
d
#> x y z sumx sumy sumz
#> 1 1 11 21 1 11 21
#> 2 2 12 22 3 23 43
#> 3 3 13 23 5 25 45
#> 4 4 14 24 7 27 47
#> 5 5 15 25 9 29 49
#> 6 6 16 26 11 31 51
#> 7 7 17 27 13 33 53
#> 8 8 18 28 15 35 55
#> 9 9 19 29 17 37 57
#> 10 10 20 30 19 39 59
Just to continue the tidyverse theme, here's a solution using the purrr package (again, works for all columns, but can subset columns if need to):
library(purrr)
# Create new columns in new data frame.
# Subset `d` here if only want select columns
sum_d <- map_df(d, ~ . + lag(., default = 0))
# Set names correctly and
# bind back to original data
names(sum_d) <- paste0("sum", names(sum_d))
d <- cbind(d, sum_d)
d
#> x y z sumx sumy sumz
#> 1 1 11 21 2 22 42
#> 2 2 12 22 4 24 44
#> 3 3 13 23 6 26 46
#> 4 4 14 24 8 28 48
#> 5 5 15 25 10 30 50
#> 6 6 16 26 12 32 52
#> 7 7 17 27 14 34 54
#> 8 8 18 28 16 36 56
#> 9 9 19 29 18 38 58
#> 10 10 20 30 20 40 60

Answer 3

You can use the lag function to achieve something like this:
myFrame$SumX[1] <- X[1]
myFrame$SumX[2:nrow(myFrame)] <- X[2:nrow(myFrame)]+lag(X)[2:nrow(myFrame)]

Answer 4

#SumX
cumsum(df$X) - c(0, 0, cumsum(df$X)[1:(nrow(df)-2)])
#[1] 20 45 40 37 82 100 100 72 23
#SumY
cumsum(df$Y) - c(0, 0, cumsum(df$Y)[1:(nrow(df)-2)])
#[1] 7 91 146 74 36 34 70 110 67