How do I map a vector of values ​​to another vector with my o...

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  • replace values in a vector based on another vector 2 answers

I want a function whose input is a vector of 1s, 2s, and 3s which sends 1 to .2, 2 to .4 and 3 to .5. (The output should be a vector of equal length.) How do I accomplish this?

For example, if

myVector<-c(1,2,3,2,3,3,1)

Then the function

mapVector(myVector)

should return a vector like (.2,.4,.5,.4,.5,.5,.2)

A couple of options, all using:

myVector<-c(1,2,3,2,3,3,1)

Factor

newvals <- c(.2,.4,.5)
newvals[as.factor(myVector)]
#[1] 0.2 0.4 0.5 0.4 0.5 0.5 0.2

Named vector

newvals <- c(`1`=.2,`2`=.4,`3`=.5)
newvals
#  1   2   3
#0.2 0.4 0.5 

newvals[as.character(myVector)]
#  1   2   3   2   3   3   1
#0.2 0.4 0.5 0.4 0.5 0.5 0.2

Lookup table

mapdf <- data.frame(old=c(1,2,3),new=c(.2,.4,.5))
mapdf$new[match(myVector,mapdf$old)]
#[1] 0.2 0.4 0.5 0.4 0.5 0.5 0.2

Benchmarks to quantify @Joe 's comment below and address @Ananda's comment as well.

myVector <- c(1,2,3,2,3,3,1)
# setup for the benchmarking
test <- sample(myVector,1e6,replace=TRUE)
newvals <- c(.2,.4,.5)
newvalsvec <- c(`1`=.2,`2`=.4,`3`=.5)
mapdf <- data.frame(old=c(1,2,3),new=c(.2,.4,.5))

microbenchmark(
  newvals[as.factor(test)],
  newvalsvec[as.character(test)],
  mapdf$new[match(test,mapdf$old)],
  newvals[test],
  times=10L
)

#Unit: milliseconds
#         expr        min         lq     median         uq        max
#factor        1863.40146 1876.04197 1890.99147 1913.13046 2014.23609
#namedvector   1809.26883 1812.76272 1837.18852 1851.42954 1858.44996
#lookup          38.48697   38.83405   39.90146   69.65140   71.75051
#newvals[test]   34.07380   34.55885   50.61287   65.69495   66.08699
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