Find nearest smaller number

Question

I have a vector of numbers

f <- c(1, 3, 5, 8, 10, 12, 19, 27)

I want to compare the values in the vector to another number, and find the c

Answer 1

In a functional programming style:

f <- c(1, 3, 6, 8, 10, 12, 19, 27)
x <- 18
Position(function(fi) fi <= x, f, right = TRUE)

Answer 2

I think this answer is pretty straightforward:

f <- c(1,3,6,8,10,12,19,27)
x <- 18

# find the value that is closest to x
maxless <- max(f[f <= x])
# find out which value that is
which(f == maxless)

Answer 3

If your vector f is always sorted, then you can do sum(f <= x)

f <- c(1,3,6,8,10,12,19,27)

x <- 18
sum(f <= x)
# [1] 6

x <- 19
sum(f <= x)
# [1] 7

Answer 4

There is findInterval:

findInterval(18:19, f)
#[1] 6 7

And building a more concrete function:

ff = function(x, table)
{
    ot = order(table)
    ans = findInterval(x, table[ot]) 
    ot[ifelse(ans == 0, NA, ans)]
}

set.seed(007); sf = sample(f)
sf
#[1] 27  6  1 12 10 19  8  3
ff(c(0, 1, 18, 19, 28), sf)
#[1] NA  3  4  6  1

Answer 5

You could try:

x <- 18
f <- c(1,3,6,8,10,12,19,27)

ifelse(x %in% f, which(f %in% x), which.min(abs(f - x)) - 1)

That way if x is not in f, it will return the nearest previous index. If x is in f, it will return x index.

Answer 6

Try this (not a perfect solution)

x<-c(1,3,6,8,10,12,19,27)
showIndex<-function(x,input){
 abs.diff<-abs(x-input)
 index.value<-unique(ifelse(abs.diff==0,which.min(abs.diff),which.min(abs.diff)-1))
return(index.value)
 }
 showIndex(x,12)
    [1] 6
showIndex(x,19)
[1] 7