Question

假定

x = c(1, 2, 3.5, 4, 6, 7.5, 8, 9, 10, 11.5, 12) 
y = c(2.5, 6.5) 
I = split(x, findInterval(x, y))
f = function(vec, x) {
        d = pmax(outer(x, vec, "-"), 0)
        colSums(d - d^2/2)
}

I want to calculate the value of f(I[[i]], x) in each values of each interval and then find which I[[i]] actual value has the maximum value of f(I[[i]], x ) in each interval. Is there any other way which is more efficient than this for loop:

for (i in 1:length(I)) {
    max.values[[i]] = I[[i]][which.max(f(I[[i]], x))]
}

这是我想要取得的结果:

 > max.values
 [1]  2  6 10

Answer 1

如果你只是想去掉 for。您可以用一点来取代它。 :

max.values <- unlist( lapply( I, function(v) v[which.max(f(v, x))] ) );

This will only make a difference if length(I) is large. To gain more performance, see if you can simplify f(.) just for the purpose of finding a max. For best optimization, you should consider re-writing the performance critical part in C, C++, or Fortran.

R can be horribly slow when the data vector gets large, when a lengthy loop exists, or when the available data structures are not suited for the task. Just as an anecdote, I wrote a "for-loop"-less R code that was killed after 2 weeks of Wall time (input array: n ~ 1e6). (The R code runs fine on input with n ~ 1e4). A C++ equivalent code took 1 min. A slightly more optimized C++ code took 10 seconds...

Answer 2

你可以做些什么。

mapply( [ , I, lapply(lapply(I, f, x), which.max))
# 0  1  2 
# 2  6 10

下面是中间步骤:

lapply(I, f, x)
# $`0`
# [1] -190.875 -142.375
# 
# $`1`
# [1] -85.75 -70.75 -26.75
# 
# $`2`
# [1] -9.500 -6.125 -1.625  0.375  0.375  0.000

lapply(lapply(I, f, x), which.max)
# $`0`
# [1] 2
# 
# $`1`
# [1] 3
# 
# $`2`
# [1] 4

Answer 3

这更是紧凑的,但我不知道它是否更有效率。

v <- sapply(lapply(I,f,x=x),which.max)
mapply(getElement,I,v)

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