Vectorized IF statement in R?
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Vectorized IF statement in R?
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?
if
if (condition)
6 Answers
6
x <- seq(0.1,10,0.1)
> 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
This is also very useful for only replacing some values: e.g.
x = ...; x[x < 5] = 1 is ifelse(x < 5, 1, x)– flying sheep
Feb 21 '15 at 17:23
x = ...; x[x < 5] = 1
ifelse(x < 5, 1, x)
What if I want to return the elements of another vector?
– skan
Mar 8 '17 at 13:38
@skan try
y[x < 5]. Assuming lengths of x and y are identical– Roman Luštrik
Mar 8 '17 at 15:12
y[x < 5]
x
y
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 <- if (x < 5) 1 else 2
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
ifelse
if
?"if"
?ifelse
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.
ifelse
> 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
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
– 42-
Oct 31 '10 at 4:31
@Dwin: Very nice solution! thx. But on my machine, it runs only marginally faster (0.47 compared to 0.48)
– Joris Meys
Oct 31 '10 at 19:59
Careful - your two examples are not equivalent if
x contains NA elements (which would remain NA in the first but would be assigned 1 for the second).– jbaums
Jul 13 '14 at 9:31
x
NA
NA
1
@jbaums That's correct. Adding an extra line y[is.na(x)] <- NA still keeps that solution a factor 2 faster than ifelse()
– Joris Meys
Jul 14 '14 at 10:23
You could also just create a logical vector and 1 to it
x <- seq(0.1, 10, 0.1) # Your data set
(x >= 5) + 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 1 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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
# [92] 2 2 2 2 2 2 2 2 2
If would like to compare performance, it would be the fastest solution
set.seed(100)
x <- runif(1e6, 1, 10)
RL <- function(x) y <- ifelse(x < 5,1,2)
JM <- function(x) y <- rep(2, length(x)); y[x < 5] <- 1
DA <- function(x) y <- (x >= 5) + 1
library(microbenchmark)
microbenchmark(RL(x),
JM(x),
DA(x))
# Unit: milliseconds
# expr min lq mean median uq max neval
# RL(x) 331.83448 366.52940 378.89182 374.99741 381.08659 609.21218 100
# JM(x) 38.72894 42.18745 44.36493 43.25086 44.09626 82.76168 100
# DA(x) 10.01644 11.96482 14.21593 13.17825 14.12930 53.76923 100
For a slightly more efficient approach you can modify your code to
function(x) y <- (x >= 5L) + 1L but generally a nice answer and interesting to see how slow, in comparison, ifelse is.– docendo discimus
Dec 30 '14 at 14:21
function(x) y <- (x >= 5L) + 1L
ifelse
Following the above post you can even use and modify the elements of a vector satisfying the criteria. In my opinion if it's not more costly to compute faster one should always do it.
x = seq(0.1,10,0.1)
y <- rep(2,length(x))
y[x<5] <- x[x<5]*2
The code of the previous post is best to answer the question. But if I had to use the code above I would do:
x = seq(0.1,10,0.1)
y <- rep(2,length(x))
y[x<5] <- x[x<5]*0 +1
Answers may move up or down depending on votes so relating to "the above post" might be troublesome.
– Johan Larsson
Aug 31 '16 at 17:40
nzMean <- function(x) mean(x[x!=-1],na.rm=TRUE)
nzMin <- function(x) min(x[x!=-1],na.rm=TRUE)
nzMax <- function(x) max(x[x!=-1],na.rm=TRUE)
nzRange<-function(x) nzMax(x)-nzMin(x)
nzSD <- function(x) SD(x[x!=-1],na.rm=TRUE)
#following function works
nzN1<- function(x) ifelse(x!=-1,(x-nzMin(x))/nzRange(x) ,x)
#following is bad as it returns only 4 not 5 elements of vector
nzN2<- function(x) ifelse(x!=-1,(x[x!=-1]-nzMin(x))/nzRange(x) ,x)
#following is bad as it returns 5 elements of vector but not correct answer
nzN3<- function(x) ifelse(x!=-1,(x[x!=-1]-nzMin(x))/nzRange(x) ,-1)
y<-c(1,-1,-20,2,4)
a<-nzMean(y)
b<-nzMin(y)
c<-nzMax(y)
d<-nzRange(y)
# test the working function
z<-nzN1(y)
print(z)
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Is this also possible with an
if (condition)else (condition) construction? If the yes no arguments get a little trickier, it´s sometimes hard to read. I had the same problem like Christian, used if else just as suggested here which works just fine, but looks ugly. So far I am using expression(yes) which is fine as a work around, but still I wonder if there´s a to do it with if and else.– Matt Bannert
Dec 6 '10 at 9:17