从所有后续行中减去第一行

Question

From the following data set I want to subtract the values in the row labeled "BLK" from all subsequent rows (regardles of their lables).

 label   Int_A Int_B Int_C
 BLK     10    20    30    
 SMP     12.5  22.5  35    
 STD     15    25    40    

Desired output:

 label   Int_A Int_B Int_C 
 BLK     10    20    30    
 SMP     2.5   2.5   5    
 STD     5     5     10   

It does not matter if the BLK-row remains unchanged or will be set to zero.

Unfortunately, all answers I found consider only one variable, not all. I tried using the dplyr package, especially rowwise() and transmute() (there is no need to keep the old values) but failed also in calling an entire row and not only a certain variable. In basic RI also tried (and failed) but working with dplyr would be preferable, as the entire data set could exist of more than one of these sections and subsetting is easy with group_by() using a separate column.

I would be very glad if you could give me some advise or the required code.

Answer 1

If we need to subtract the first row from others, use mutate_at to specify the index of numeric columns or mutate_if and then subtract the first element from the all the elements of the column

library(dplyr)
df1 %>% 
    mutate_at(2:4, funs(c(first(.), (. - first(.))[-1])) )

---

Or with mutate_if

df1 %>%
    mutate_if(is.numeric, funs(c(first(.), (. - first(.))[-1])) )

Answer 2

df[-1, -1] <- df[-1 , -1] - df[rep(1, nrow(df) - 1), -1]
df
#   label Int_A Int_B Int_C
# 1   BLK  10.0  20.0    30
# 2   SMP   2.5   2.5     5
# 3   STD   5.0   5.0    10

From relevant rows and columns df[-1 , -1] , subtract the first row rep eated to the necessary number of rows ( nrow(df) - 1 ).

Answer 3

As an alternative using no packages:

diff is a matrix of the same size like df, with first row 0 and the rest equal to the 1st row in df. Then simply subtract.

df <- mtcars

diff <- matrix(c(rep(0,ncol(df)),rep(as.numeric(df[1,]),nrow(df)-1)),nrow=nrow(df),ncol=ncol(df),byrow=T)

df - diff