How to find the difference in value in every two consecutive rows in R?
Here's an example of how to use diff() on the built-in mtcars data.frame. You have to select a column to perform the diff over:
mtcars
mpg cyl disp hp drat wt qsec vs am gear carb
Mazda RX4 21.0 6 160.0 110 3.90 2.620 16.46 0 1 4 4
Mazda RX4 Wag 21.0 6 160.0 110 3.90 2.875 17.02 0 1 4 4
Datsun 710 22.8 4 108.0 93 3.85 2.320 18.61 1 1 4 1
Hornet 4 Drive 21.4 6 258.0 110 3.08 3.215 19.44 1 0 3 1
[..snip..]
Ford Pantera L 15.8 8 351.0 264 4.22 3.170 14.50 0 1 5 4
Ferrari Dino 19.7 6 145.0 175 3.62 2.770 15.50 0 1 5 6
Maserati Bora 15.0 8 301.0 335 3.54 3.570 14.60 0 1 5 8
Volvo 142E 21.4 4 121.0 109 4.11 2.780 18.60 1 1 4 2
Calculate the successive differences of e.g. the column "qsec":
diff(mtcars$qsec)
[1] 0.56 1.59 0.83 -2.42 3.20 -4.38 4.16 2.90 -4.60 0.60 -1.50 0.20
[13] 0.40 -0.02 -0.16 -0.40 2.05 -0.95 1.38 0.11 -3.14 0.43 -1.89 1.64
[25] 1.85 -2.20 0.20 -2.40 1.00 -0.90 4.00
You could simply subtract a data.frame consisting of rows 1:(n-1) of the original data.frame from a second one consisting of rows 2:n. (Here n is the number of rows in the original data.frame):
# Example data
df <- data.frame(a=1:4, b=4:1, c=11:14, d=c(2,4,10,0))
# a b c d
# 1 1 4 11 2
# 2 2 3 12 4
# 3 3 2 13 10
# 4 4 1 14 0
# Calculate the differences
diff_df <- df[-1,] - df[-nrow(df),]
diff_df
# a b c d
# 2 1 -1 1 2
# 3 1 -1 1 6
# 4 1 -1 1 -10
You can then rename the rows as you see fit using something like:
row.names(diff_df) <- paste("d", seq_len(nrow(diff_df)), sep="")
diff_df
# a b c d
# d1 1 -1 1 2
# d2 1 -1 1 6
# d3 1 -1 1 -10
This would have be clearer with sample data. Let's assume you mean "numerical difference", and your data can be represented as a matrix, this would do.
set.seed(4871)
m = matrix(sample(1:5,50,TRUE),nrow=10,ncol=5)
m
t(apply(m,1,diff))
We can also use dplyr::lag and data.table::shift
library(dplyr)
df <- mtcars
df %>% mutate(diff_row = mpg - lag(mpg))
# mpg cyl disp hp drat wt qsec vs am gear carb diff_row
#1 21.0 6 160.0 110 3.90 2.620 16.46 0 1 4 4 NA
#2 21.0 6 160.0 110 3.90 2.875 17.02 0 1 4 4 0.0
#3 22.8 4 108.0 93 3.85 2.320 18.61 1 1 4 1 1.8
#4 21.4 6 258.0 110 3.08 3.215 19.44 1 0 3 1 -1.4
#5 18.7 8 360.0 175 3.15 3.440 17.02 0 0 3 2 -2.7
#6 18.1 6 225.0 105 2.76 3.460 20.22 1 0 3 1 -0.6
#7 14.3 8 360.0 245 3.21 3.570 15.84 0 0 3 4 -3.8
#....
As there is no lag for first row, the default value is NA. If we want the first value returned to be 0 instead, we can set the default value to first value of the variable so the subtracting yields to 0.
df %>% mutate(diff_row = mpg - lag(mpg, default = first(mpg)))
# mpg cyl disp hp drat wt qsec vs am gear carb diff_row
#1 21.0 6 160.0 110 3.90 2.620 16.46 0 1 4 4 0.0
#2 21.0 6 160.0 110 3.90 2.875 17.02 0 1 4 4 0.0
#3 22.8 4 108.0 93 3.85 2.320 18.61 1 1 4 1 1.8
#4 21.4 6 258.0 110 3.08 3.215 19.44 1 0 3 1 -1.4
#5 18.7 8 360.0 175 3.15 3.440 17.02 0 0 3 2 -2.7
#6 18.1 6 225.0 105 2.76 3.460 20.22 1 0 3 1 -0.6
#7 14.3 8 360.0 245 3.21 3.570 15.84 0 0 3 4 -3.8
#....
The same process can also be done using shift from data.table whose default type is lag
library(data.table)
setDT(df)[, diff_row := mpg - shift(mpg)]
Or to get 0 value in the first row, here we use fill argument.
setDT(df)[, diff_row := mpg - shift(mpg, fill = first(mpg))]