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:

>z<-cbind(x=1:10,y=11:20,t=21:30)
> z<-as.data.frame(z)
>z
    x  y  t
1   1 11 21
2   2 12 22
3   3 13 23
4   4 14 24
5   5 15 25
6   6 16 26
7   7 17 27
8   8 18 28
9   9 19 29
10 10 20 30

# generate the column q which is equal to the values of column t times 4 if x=3 and for other values of x, it is equal to the values of column t.

for (i in 1:nrow(z)){
  z$q[i]=if (z$x[i]==4) 4*z$t[i] else z$t[i]}

But, my problem is that I want to apply multiple conditions:

For example, I want to get something like this:

(If x=2, q=t*2; x=4, q=t*4; x=7, q=t*3; for other it is equal to t) 

> 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

How do I get the second output using the loops or any other method?

Answer 1

Here's a version of an SQL decode in R for character vectors (untested with factors) that operates just like the SQL version. i.e. it takes an arbitrary number of target/replacement pairs, and optional last argument that acts as a default value (note that the default won't overwrite NAs).

I can see it being pretty useful in conjunction with dplyr's mutate operation.

> x <- c("apple","apple","orange","pear","pear",NA)

> decode(x, apple, banana)
[1] "banana" "banana" "orange" "pear"   "pear"   NA      

> decode(x, apple, banana, fruit)
[1] "banana" "banana" "fruit"  "fruit"  "fruit"  NA      

> decode(x, apple, banana, pear, passionfruit)
[1] "banana"       "banana"       "orange"       "passionfruit" "passionfruit" NA            

> decode(x, apple, banana, pear, passionfruit, fruit)
[1] "banana"       "banana"       "fruit"        "passionfruit" "passionfruit" NA  

Here's the code I'm using, with a gist I'll keep up to date here (link).

decode <- function(x, ...) {

  args <- as.character((eval(substitute(alist(...))))

  replacements <- args[1:length(args) %% 2 == 0]
  targets      <- args[1:length(args) %% 2 == 1][1:length(replacements)]

  if(length(args) %% 2 == 1)
    x[! x %in% targets & ! is.na(x)] <- tail(args,1)

  for(i in 1:length(targets))
    x <- ifelse(x == targets[i], replacements[i], x)

  return(x)

}

Answer 2

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

Answer 3

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 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

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 7

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 8

Generate a multipler vector:

tt <- rep(1, max(z$x))
tt[2] <- 2
tt[4] <- 4
tt[7] <- 3

And here is your new column:

> z$t * tt[z$x]
 [1] 21 44 23 96 25 26 81 28 29 30

> z$q <- z$t * tt[z$x]
> 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

This will not work if there are negative values in z$x.

Edited

Here is a generalization of the above, where a function is used to generate the multiplier vector. In fact, we create a function based on parameters.

We want to transform the following values:

2 -> 2
4 -> 4
7 -> 3

Otherwise a default of 1 is taken.

Here is a function which generates the desired function:

f <- function(default, x, y) {
  x.min <- min(x)
  x.max <- max(x)
  y.vals <- rep(default, x.max-x.min+1)
  y.vals[x-x.min+1] <- y

  function(z) {
    result <- rep(default, length(z))
    tmp <- z>=x.min & z<=x.max
    result[tmp] <- y.vals[z[tmp]-x.min+1]
    result
  }
}

Here is how we use it:

x <- c(2,4,7)
y <- c(2,4,3)

g <- f(1, x, y)

g is the function that we want. It should be clear that any mapping can be supplied via the x and y parameters to f.

g(z$x)
## [1] 1 2 1 4 1 1 3 1 1 1

g(z$x)*z$t
## [1] 21 44 23 96 25 26 81 28 29 30

It should be clear this only works for integer values.