这是我的功能:
g <- function(x,y){
x <- (x-y):x
y <- 1:30 # ------> (y is always fixed 1:30)
z<- outer(x,y,fv) # ---->(fv is a previous function)
s <- colSums(z)
which(s==max(s),arr.ind=T)
}
它告诉我s中最大值的位置。我在选择y时基本上有一个问题,因为给定一个小y,max(s)在s中出现不止一次。例如:
#given x=53
> g(53,1)
[1] 13 16 20 22 25 26 27
> g(53,2)
[1] 20 25 26
> g(53,3)
[1] 20 25 26
> g(53,4)
[1] 20 25 26
> g(53,5)
[1] 20 25
> g(53,6)
[1] 25 -----> This is the only result i would like from my function (right y=6)
另一个例子:
# given x=71
> g(71,1)
[1] 7 9 14
> g(71,2)
[1] 7 14
> g(71,3)
[1] 14 -----> my desired result (right y=3)
因此,我希望一个函数能够得到尽可能小的第一个唯一解(ex:g(53)=25 , g(71)=14, ...
)。有帮助吗?感谢
这是一个简化的例子。我希望在提问中更清楚:
#The idea is the same:
n <- 1:9
e <- rep(nn,500)
p<- sample(e) # --->(Need to sample in order to have more max later (mixed matrix)
mat <- matrix(p,90)
g <- function(x,y){
x <- (x-y):x
k <- rowSums(mat[,x])
which(k==max(k), arr.ind=T)
}
#In my sample matrix :
k <- rowSums(mat[,44:45])
which(k==max(k), arr.ind=T)
[1] 44 71 90
#In fact
g(45,1)
[1] 44 71 90 # ---> more than one solution
g(45,2)
[1] 90 # ----> I would like to pick up this value wich is the first unique solution given x=45
因此,我想要一个函数,在给定x的情况下,y的第一个唯一解决方案尽可能小(在这个新的例外:g(45)=90
...)。
答案 0 :(得分:1)
我明白了。这有点长,但我认为是对的。
考虑第二个简化示例:
g <- function(x,y){
x <- (x-y):x
k <- rowSums(mat[,x])
q <- which(k==max(k), arr.ind=T)
length(q)
}
gv <- Vectorize(g)
l <- function(x){
y<- 1:30 # <- (until 30 to be sure)
z<- outer(x,y,gv)
y <- which.min(z) # <- (min is surely length=1 and which.min takes the first)
x <- (x-y):x
k <- rowSums(mat[,x])
q <- which(k==max(k), arr.ind=T)
q
}
l(45)
[1] 90
答案 1 :(得分:0)
好像你可以用recursive function来做这件事。请考虑以下事项:
set.seed(42)
n = 1:9
e = rep(n, 500)
p = sample(e)
mat = matrix(p, 90)
g <- function(x, y=1) {
xv <- (x-y):x
k <- rowSums(mat[, xv])
i <- which(k == max(k), arr.ind=T)
n <- length(i)
if (n == 1) {
return(y) # want to know the min y that solves the problem, right?
} else {
y <- y + 1 # increase y by 1
g(x,y) # run our function again with a new value of y
}
}
您现在应该能够运行g(45)
并获得1作为结果,因为这是解决问题的y
的值,而g(33)
得到2。