tags:

views:

110

answers:

3
x = seq(0.1,10,0.1)
y <- if (x < 5) 1 else 2

I would want the if to operate on every single case instead of operating on the whole vector. What do I have to change?

+5  A: 
ifelse(x < 5, 1, 2)

> x
  [1]  0.1  0.2  0.3  0.4  0.5  0.6  0.7  0.8  0.9  1.0  1.1  1.2  1.3  1.4  1.5
 [16]  1.6  1.7  1.8  1.9  2.0  2.1  2.2  2.3  2.4  2.5  2.6  2.7  2.8  2.9  3.0
 [31]  3.1  3.2  3.3  3.4  3.5  3.6  3.7  3.8  3.9  4.0  4.1  4.2  4.3  4.4  4.5
 [46]  4.6  4.7  4.8  4.9  5.0  5.1  5.2  5.3  5.4  5.5  5.6  5.7  5.8  5.9  6.0
 [61]  6.1  6.2  6.3  6.4  6.5  6.6  6.7  6.8  6.9  7.0  7.1  7.2  7.3  7.4  7.5
 [76]  7.6  7.7  7.8  7.9  8.0  8.1  8.2  8.3  8.4  8.5  8.6  8.7  8.8  8.9  9.0
 [91]  9.1  9.2  9.3  9.4  9.5  9.6  9.7  9.8  9.9 10.0

> ifelse(x < 5, 1, 2)
  [1] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
 [38] 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
 [75] 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
Roman Luštrik
+4  A: 

y <- if (x < 5) 1 else 2 does not operate on the whole vector (the warning you receive tells you only the first element of the condition will be used). You want ifelse:

y <- ifelse(x < 5, 1, 2)

ifelse operates on the whole logical vector, element-by-element. if only accepts one logical value. See ?"if" and ?ifelse

Joshua Ulrich
+6  A: 

For completeness: In big vectors, you can use the indices to speed things up (we do that often in simulations, where functions typically run 1000 to 10000 times). But as long as it isn't necessary, just use ifelse. This reads a lot easier.

> set.seed(100)
> x <- runif(1000,1,10)

> system.time(replicate(10000,{
+     y <- ifelse(x < 5,1,2)
+ }))
   user  system elapsed 
   2.56    0.08    2.64 

> system.time(replicate(10000,{
+   y <- rep(2,length(x))
+   y[x < 5]<- 1
+ }))
   user  system elapsed 
   0.48    0.00    0.48 
Joris Meys
You can cut that time even further. My machine did the second method in 0.436 (although it was slower on the first method), but this improved it by another 200%: system.time(replicate(10000,{ y <- (y < 5) + 2*!(y<5) })) user system elapsed 0.101 0.021 0.128
DWin
@Dwin: Very nice solution! thx. But on my machine, it runs only marginally faster (0.47 compared to 0.48)
Joris Meys