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 closest smaller value.

For example, if the input number is 18, then the closest, smaller value in the vector is 12 (position 6 in the vector). If the input is 19, then the result should be the value 19, i.e. the index 7.

score 15 · Answer 1 · answered May 18 '15 at 23:34

15

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)

answered May 18 '15 at 23:34

John

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the last two lines can be written in one line: which.max(f[f<=x]) – nigelhenry Oct 07 '19 at 17:18
1

@nigelhenry your code will only work when 'f' is sorted and ascending (which it is in the example), since you are giving the index of the element in the truncated vector. – le_andrew Apr 30 '20 at 18:55
@le_andrew - you're right. OK – nigelhenry Aug 25 '22 at 17:32

score 8 · Answer 2 · answered May 18 '15 at 23:16

8

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

answered May 18 '15 at 23:16

Gregor Thomas

136,190
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user227710 · Answer 3 · 2015-05-18T23:32:31.173

4

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

edited May 18 '15 at 23:32

answered May 18 '15 at 22:56

user227710

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the problem is that if the inout is 19...then the output would need to be 7 (the index)... – heinheo May 18 '15 at 22:59
`which.min(abs(x-18))` gives you the index. I am not sure what you are asking. – user227710 May 18 '15 at 23:01
x[which.min(abs(x-19))-1] [1] 18....that is the problem because if the inpout is 19 then it mataches directly the value and there should be no subtraction of 1 – heinheo May 18 '15 at 23:03

Steven Beaupré · Answer 4 · 2015-05-18T23:06:55.620

2

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.

edited May 18 '15 at 23:06

answered May 18 '15 at 23:01

Steven Beaupré

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7
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Beaupre how would I make sure that if I search ftr the nearest value of 19, the thing returns 7 (the index of 19) not 6 (the index of 12)...because of the input is exactly a position in the vector it would need to show that... – heinheo May 18 '15 at 23:05

score 2 · Answer 5 · answered May 18 '15 at 23:19

Another one:

which.min(abs(18 - replace(f, f>18, Inf)))
#[1] 6

f[which.min(abs(18 - replace(f, f>18, Inf)))]
#[1] 12

Or as a function:

minsmaller <- function(x,value) which.min(abs(value - replace(x, x>value, Inf)))
minsmaller(f, 18)
#[1] 6
minsmaller(f, 19)
#[1] 7

Richie Cotton · Answer 6 · 2015-05-19T07:28:09.073

0

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)

edited May 19 '15 at 07:28

answered May 19 '15 at 07:15

Richie Cotton

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score 0 · Answer 7 · answered May 19 '15 at 07:27

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

Find nearest smaller number

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