CODEWITHSUNDEEP
r
srb633
srb633

Reputation: 813

Have x round down to the nearest specified numeric value from series?

Wondering how x can round down to the nearest below value based on a separate numeric series (y). This is what I had in mind.

Let's say x is in 5s

> y
[1]  5 10 15 20 25 30 35 40 45

And y from a different object is equal to

> x
[1] 39

I'd like to see it round down to the nearest lower specified value from y

> x
[1] 35

Upvotes: 1

Views: 38

Answers (2)

rg255
rg255

Reputation: 4169

You can use %/% and multiply it by y (if you set y to be your multiple, e.g. if 5's):

y <- 5
x <- 39
(x %/% y) * y

If you must use a vector of length > 1 for y then you could use a few different functions like max() or tail(..., 1) to get the largest or last value in y:

y <- seq(0, 50, 5)
tail(y[x > y], 1)
max(y[x > y])

Note that x > y is creating a logical vector which is being used for a linear search, returning a vector of the values of y less than x and passing that to the function.

Upvotes: 4

Ronak Shah
Ronak Shah

Reputation: 388972

You could use findInterval/cut

y <- seq(5, 45, 5)
x <- 39
y[findInterval(x, y)]
#[1] 35

With cut : 

y[cut(x, y, labels = FALSE)]

Upvotes: 2

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