从 R 中的不同时间步减去值

Question

I am working with some temperature data, and I have to perform a simple subtraction on a data frame where I want to create a new column eg x= df(i)- df(i-1) , so given the following df.

df<- c(5,10,20,30,40)

I should get the following result:

(5,10,10,10,40)

I managed to do it with help of other posts on stack overflow using a for loop as follows:

df<- c(5,10,20,30,40)

x<- array(NA,length(df))
for(i in 2: length(df)){
  x[i]<- Mod(df[i-1]-df[i])
}
print(x)
NA  5 10 10 10 

I have two problems, first and more important, I need to perform this for a each column of a large data frame that has temperatures at different depths this example is just for a depth=0.5m , so with the for loop takes forever... Is there a more efficient way to do it?

Secondly, I would like that the output would be something like this:

(5,10,10,10,40)

Many thanks for your help

Answer 1

Here is base R solution (maybe for your objective)

dfout <- rbind(diff(as.matrix(df)),tail(df,1))

Example

> df
  X1 X2 X3 X4 X5
1  1  2  3  4  5
2  6  7  8  9 10
3 11 12 13 14 15
4 16 17 18 19 20

> dfout
  X1 X2 X3 X4 X5
1  5  5  5  5  5
2  5  5  5  5  5
3  5  5  5  5  5
4 16 17 18 19 20

DATA

df <- structure(list(X1 = c(1L, 6L, 11L, 16L), X2 = c(2L, 7L, 12L, 
17L), X3 = c(3L, 8L, 13L, 18L), X4 = c(4L, 9L, 14L, 19L), X5 = c(5L, 
10L, 15L, 20L)), class = "data.frame", row.names = c(NA, -4L))

Answer 2

foo <- function(x) c(diff(x), x[length(x)])

# Examle
df <- c(5,10,20,30,40)
foo(df)
# [1]  5 10 10 10 40

To apply to each column of data.frame, say X, you can do:

X[] <- lapply(X, foo)
# or to assign to a new object
Y <- data.frame(lapply(df, foo))