舍入到列表中最接近的任意数字

我基本上是在寻找一种方法来做一些变化这个 Ruby 脚本 https://stackoverflow.com/questions/3160502/ruby-round-number-down-to-nearest-number-based-on-arbitrary-list-of-numbers in R.
我有一个任意数字列表（在这种情况下回归图的主持人的步骤）彼此之间的距离不相等，并且我想将这些数字周围的范围内的值四舍五入到列表中最接近的数字。 范围不重叠。

arbitrary.numbers <- c(4,10,15) / 10
numbers <- c(16:1 / 10, 0.39, 1.45)
range <- 0.1

预期输出：

numbers
## 1.6 1.5 1.4 1.3 1.2 1.1 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.39 1.45
round_to_nearest_neighbour_in_range(numbers,arbitrary.numbers,range)
## 1.5 1.5 1.5 1.3 1.2 1.0 1.0 1.0 0.8 0.7 0.6 0.4 0.4 0.4 0.2 0.1 0.4 1.5

我有一个小辅助函数可以解决我的特定问题，但它不是很灵活并且包含一个循环。我可以将其发布在这里，但我认为真正的解决方案看起来会完全不同。

不同的答案按速度计时（针对一百万个数字）

> numbers = rep(numbers,length.out = 1000000)
> system.time({ mvg.round(numbers,arbitrary.numbers,range) })[3]
elapsed 
  0.067 
> system.time({ rinker.loop.round(numbers,arbitrary.numbers,range) })[3]
elapsed 
  0.289 
> system.time({ rinker.round(numbers,arbitrary.numbers,range) })[3]
elapsed 
  1.403 
> system.time({ nograpes.round(numbers,arbitrary.numbers,range) })[3]
elapsed 
  1.971 
> system.time({ january.round(numbers,arbitrary.numbers,range) })[3]
elapsed 
  16.12 
> system.time({ shariff.round(numbers,arbitrary.numbers,range) })[3]
elapsed 
15.833 
> system.time({ mplourde.round(numbers,arbitrary.numbers,range) })[3]
elapsed 
  9.613 
> system.time({ kohske.round(numbers,arbitrary.numbers,range) })[3]
elapsed 
 26.274 

MvG 的函数是最快的，比 Tyler Rinker 的第二个函数快大约 5 倍。

矢量化解决方案，无需任何apply族函数或循环：

关键是findInterval，它找到“空间”arbitrary.numbers其中每个元素numbers在。。。之间”。所以，findInterval(6,c(2,4,7,8))回报2， 因为6位于第二个和第三个索引之间c(2,4,7,8).

# arbitrary.numbers is assumed to be sorted.
# find the index of the number just below each number, and just above.
# So for 6 in c(2,4,7,8) we would find 2 and 3.
low<-findInterval(numbers,arbitrary.numbers) # find index of number just below
high<-low+1 # find the corresponding index just above.

# Find the actual absolute difference between the arbitrary number above and below.
# So for 6 in c(2,4,7,8) we would find 2 and 1. 
# (The absolute differences to 4 and 7).
low.diff<-numbers-arbitrary.numbers[ifelse(low==0,NA,low)]
high.diff<-arbitrary.numbers[ifelse(high==0,NA,high)]-numbers

# Find the minimum difference. 
# In the example we would find that 6 is closest to 7, 
# because the difference is 1.
mins<-pmin(low.diff,high.diff,na.rm=T) 
# For each number, pick the arbitrary number with the minimum difference.
# So for 6 pick out 7.
pick<-ifelse(!is.na(low.diff) & mins==low.diff,low,high)

# Compare the actual minimum difference to the range. 
ifelse(mins<=range+.Machine$double.eps,arbitrary.numbers[pick],numbers)
# [1] 1.5 1.5 1.5 1.3 1.2 1.0 1.0 1.0 0.8 0.7 0.6 0.4 0.4 0.4 0.2 0.1 0.4 1.5