in code R:
for (i in 1:10) {
if (!i %% 2){
next
}
print(i)
}
I have as output
[1] 1
[1] 3
[1] 5
[1] 7
[1] 9
But I need the vector of i´s that satisfies that condition, ie (1,3,5,7,9). How can I get the vector (without knowing in advance the vector dimension "length")?
The above is an example of my problem:
x=c(0.1,0.4,0.5)
y=c(0.2,0.3,0.6,0.7)
cont1=0
for(i in 1:(length(x)+1)){
for(j in 1:(length(y)+1)){
if(round((abs((j-1)-(i-1)*(length(y)/length(x))) ),3) < round( max.v.d,3) ) {
cont1=cont1+1
print(paste(i-1,j-1))
}
}
}
output
[1] "0 0"
[1] "0 1"
[1] "1 1"
[1] "1 2"
[1] "2 2"
[1] "2 3"
[1] "3 3"
[1] "3 4"
but, I need a matrix with those elements.
in many ways. Here's one:
i <- 1:10
i[as.logical(i%%2)]
an alternative:
i[(i%%2)==1]
a version of that first one for people who hate to type:
i=1:10
i[!!(i%%2)]
If you'll need to do this sort of thing often you should probably write a function, something like this:
odd <- function(x) x%%2 != 0
or
even <- function(x) x%%2 == 0
... or both. Then you can do stuff like
i[odd(i)]
[1] 1 3 5 7 9
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