Divide data frame by vector, by rows and not by columns [duplicate]

Answer 1

I like David's solution but I would modify it to convert it to a data frame since that keeps the factor instead of getting a matrix.

data.frame(type=df[, 1], mapply("/", df[, -1], div))

#  type  V1   V2    V3
#1    A 0.1 0.01 0.001
#2    B 0.1 0.01 0.001
#3    C 0.1 0.01 0.001

which is what I would like to get.

Answer 2

Another method:

for(r in 1:nrow(df))    
    df[r,2:4]= df[r,2:4]/div
df
  type  V1   V2    V3
1    A 0.1 0.01 0.001
2    B 0.1 0.01 0.001
3    C 0.1 0.01 0.001

Answer 3

You can just use the transpose function:

> df[,2:4] <- t(t(df[,2:4]) / div)
> df
  type  V1   V2    V3
1    A 0.1 0.01 0.001
2    B 0.1 0.01 0.001
3    C 0.1 0.01 0.001

Answer 4

The function you are looking for is sweep():

sweep(df[, -1], MARGIN = 2, div, FUN = "/")

> sweep(df[, -1], MARGIN = 2, div, FUN = "/")
   V1   V2    V3
1 0.1 0.01 0.001
2 0.1 0.01 0.001
3 0.1 0.01 0.001

Hence

cbind(type = df[,1], sweep(df[, -1], 2, div, FUN = "/"))

> cbind(type = df[,1], sweep(df[, -1], 2, div, FUN = "/"))
  type  V1   V2    V3
1    A 0.1 0.01 0.001
2    B 0.1 0.01 0.001
3    C 0.1 0.01 0.001

gets you the desired output.

Note that here, the argument MARGIN doesn't refer to the rows (1) or columns (2) like it does in apply(). In sweep() it refers to the margin(s) of the array that correspond to STATS, the vector you wish to sweep out (of divide by in this case). In other words, the first element of STATS (div in your case) is the value to sweep out of column 1, the second element of STATS is the value to sweep out of column 2, and so on.

Answer 5

df[,-1] <- df[,-1]/div[col(df)]
 df
#  type  V1   V2    V3
#1    A 0.1 0.01 0.001
#2    B 0.1 0.01 0.001
#3    C 0.1 0.01 0.001

str(df)
#'data.frame':  3 obs. of  4 variables:
# $ type: Factor w/ 3 levels "A","B","C": 1 2 3
# $ V1  : num  0.1 0.1 0.1
# $ V2  : num  0.01 0.01 0.01
# $ V3  : num  0.001 0.001 0.001

Benchmarks

 set.seed(454)
 dat <- as.data.frame(matrix(sample(200, 1e3*1e2, replace=TRUE), ncol=1e2))
 set.seed(29)
 div <- sample(40, 1e2, replace=TRUE)


 f1 <- function() {sweep(dat, MARGIN = 2, div, FUN = "/")}
 f2 <- function() {t(t(dat) / div)}
 f3 <- function() { mapply("/", dat, div)}
 f4 <- function() {dat/div[col(dat)]}
 f5 <- function() {for(r in 1:nrow(dat)){    
             dat[r,]/div}}

 library(microbenchmark)
 microbenchmark(f1(), f2(), f3(), f4(), f5(), unit="relative")
 #Unit: relative
 # expr         min          lq      median          uq        max neval
 # f1()    6.765024    6.724991    6.434463    5.124457   10.91735   100
 # f2()    1.000000    1.000000    1.000000    1.000000    1.00000   100
 # f3()   18.028441   18.551529   16.742279   14.239107   13.72168   100
 # f4()    6.315330    6.577099    6.333656    5.052068   10.13038   100
 # f5() 4211.839669 3908.555985 3945.130154 2962.534518 1655.12268   100

Answer 6

Or try mapply

cbind(df[, 1], mapply("/", df[, -1], div))

##        V1   V2    V3
## [1,] 1 0.1 0.01 0.001
## [2,] 2 0.1 0.01 0.001
## [3,] 3 0.1 0.01 0.001