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rdataframeapplysapply

how can I apply a formula for each row

I have data like this 

df<-structure(list(data = structure(c(8L, 2L, 3L, 2L, 2L, 2L, 2L, 
1L, 7L, 5L, 6L, 5L, 4L), .Label = c("1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0", 
"2, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0", 
"2, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0", 
"2, 2, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0", 
"2, 2, 2, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0", 
"3, 2, 2, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0", 
"M1yrtr", "Mitered"), class = "factor")), row.names = c(NA, -13L), class = "data.frame")

I am trying to calculate the following for each row

for example for the second row which is 

2, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

I want to calculate this 

n =5
(-(2/n)*log2(2/n)) + (-(1/n)*log2(1/n)) +(-(1/n)*log2(1/n))+ (-(1/n)*log2(1/n))

for the third one which is 

2, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

I will calculate this 

(-(2/n)*log2(2/n)) + (-(2/n)*log2(2/n)) + (-(1/n)*log2(1/n))

so the output looks like this 

dfout<- structure(list(data = structure(c(8L, 2L, 3L, 2L, 2L, 2L, 2L, 
1L, 7L, 5L, 6L, 5L, 4L), .Label = c("1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0", 
"2, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0", 
"2, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0", 
"2, 2, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0", 
"2, 2, 2, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0", 
"3, 2, 2, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0", 
"M1yrtr", "Mitered"), class = "factor"), X = structure(c(8L, 
3L, 2L, 3L, 3L, 3L, 3L, 1L, 7L, 6L, 4L, 6L, 5L), .Label = c("0.2604594", 
"1.03563", "1.168964", "2.020935", "2.077468", "2.204594", "M1yrtr", 
"Mitered"), class = "factor")), class = "data.frame", row.names = c(NA, 
-13L))
Solution

  • In R all basic operations (addition subtraction, multiplication, logarithms,...) are vectorized. This means that for example if x is a vector then log(x) is just the componentwise log function, ore 1 / x is just component wise division.

    Therefore, you can do the following:

    x <- as.numeric(str_split(df[2, ], ", ", simplify = T))
n <- 5
sum((-(x[x > 0]/n)*log2(x[x > 0]/n)))
[1] 1.921928

    If you want to apply this for all rows you can use the sapply function like this:

    myfun <- function(x){
 if (! grepl(",", x)) return(as.character(x))
  n <- 5
  y <- as.numeric(str_split(x, ", ", simplify = T))
  as.character(sum((-(y[y > 0]/n)*log2(y[y > 0]/n))))
}

df$newcol <- sapply(df[,1], myfun)