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Find the nearest smaller quantity

have a vector with the following number

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

this element is closest to 18. So 19 will be the closest element, but the function will need to return 6 (which means the value 12), because the element in the vector must always be less if it is not equal to the input. If the input is 19, then the output should be 7 (index) ...

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7 replies


I think this answer is pretty simple:



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)
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If your vector is f

always sorted you can dosum(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
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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
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You may 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)

So, if x

not in f

, it will return the closest previous index. If x

found f

, it will return the index x

.

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Other:

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
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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)
0


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There is findInterval

:

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

And let's build a more specific 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
0


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