generate column values with multiple conditions in R

Question

I have a dataframe z and I want to create the new column based on the values of two old columns of z. Following is the process:

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Answer 1

Based on the suggestion of Señor :

> z$q <- ifelse(z$x == 2, z$t * 2,
         ifelse(z$x == 4, z$t * 4,
         ifelse(z$x == 7, z$t * 3,
                          z$t * 1)))
> z
    x  y  t  q
1   1 11 21 21
2   2 12 22 44
3   3 13 23 23
4   4 14 24 96
5   5 15 25 25
6   6 16 26 26
7   7 17 27 81
8   8 18 28 28
9   9 19 29 29
10 10 20 30 30

Answer 2

I really liked the answer "dinre" posted to flodel's blog:

for (i in 1:length(data_Array)){
data_Array[i] <- switch(data_Array[i], banana="apple", orange="pineapple", "fig")
}

With warnings about reading the help page for switch carefully for integer arguments.

Answer 3

By building a nested ifelse functional by recursion, you can get the benefits of both solutions offered so far: ifelse is fast and can work with any type of data, while @Matthew's solution is more functional yet limited to integers and potentially slow.

decode <- function(x, search, replace, default = NULL) {

   # build a nested ifelse function by recursion
   decode.fun <- function(search, replace, default = NULL)
      if (length(search) == 0) {
         function(x) if (is.null(default)) x else rep(default, length(x))
      } else {
         function(x) ifelse(x == search[1], replace[1],
                                            decode.fun(tail(search, -1),
                                                       tail(replace, -1),
                                                       default)(x))
      }

   return(decode.fun(search, replace, default)(x))
}

Note how the decode function is named after the SQL function. I wish a function like this made it to the base R package... Here are a couple examples illustrating its usage:

decode(x = 1:5, search = 3, replace = -1)
# [1]  1  2 -1  4  5
decode(x = 1:5, search = c(2, 4), replace = c(20, 40), default = 3)
# [1] 3 20  3  40  3

For your particular problem:

transform(z, q = decode(x, search = c(2,4,7), replace = c(2,4,3), default = 1) * t)

#    x  y  t  q
# 1   1 11 21 21
# 2   2 12 22 44
# 3   3 13 23 23
# 4   4 14 24 96
# 5   5 15 25 25
# 6   6 16 26 26
# 7   7 17 27 81
# 8   8 18 28 28
# 9   9 19 29 29
# 10 10 20 30 30

Answer 4

Here is an easy solution with just one ifelse command:

Calculate the multiplier of t:

ifelse(z$x == 7, 3, z$x ^ (z$x %in% c(2, 4)))

The complete command:

transform(z, q = t * ifelse(x == 7, 3, x ^ (x %in% c(2, 4))))

    x  y  t  q
1   1 11 21 21
2   2 12 22 44
3   3 13 23 23
4   4 14 24 96
5   5 15 25 25
6   6 16 26 26
7   7 17 27 81
8   8 18 28 28
9   9 19 29 29
10 10 20 30 30

Answer 5

You can also use match to do this. I tend to use this a lot while assigning parameters like col, pch and cex to points in scatterplots

searchfor<-c(2,4,7)
replacewith<-c(2,4,3)

# generate multiplier column
# q could also be an existing vector where you want to replace certain entries
q<-rep(1,nrow(z))
#
id<-match(z$x,searchfor)
id<-replacewith[id]
# Apply the matches to q
q[!is.na(id)]<-id[!is.na(id)]
# apply to t
z$q<-q*z$t

Answer 6

You can do it in

• base R
• with one line
• in which the mapping is pretty clear to read in the code
• no helper functions (ok, an anonymous function)
• approach works with negatives
• approach works with any atomic vector (reals, characters)

like this:

> transform(z,q=t*sapply(as.character(x),function(x) switch(x,"2"=2,"4"=4,"7"=3,1)))
    x  y  t  q
1   1 11 21 21
2   2 12 22 44
3   3 13 23 23
4   4 14 24 96
5   5 15 25 25
6   6 16 26 26
7   7 17 27 81
8   8 18 28 28
9   9 19 29 29
10 10 20 30 30