Question


cell_support_xyz <- function(level, zero)
{
  for(i in 1:level[1]){
    for(j in 1:level[2]){
      for(k in 1:level[3]){
        cat("cell (", i, ", ", j, ", ", k,") --> support set = (", 
            +!(i == zero[1]), ", ", +!(j == zero[2]), ", ", +!(k == zero[3]), ")\n", sep = "")
        
      }      
    }
  }
}


#Example 1
l<-c(2,3,2)
z<-c(1,1,1)
> cell_support_xyz(l,z)

cell (1, 1, 1) --> support set = (0, 0, 0)
cell (1, 1, 2) --> support set = (0, 0, 1)
cell (1, 2, 1) --> support set = (0, 1, 0)
cell (1, 2, 2) --> support set = (0, 1, 1)
cell (1, 3, 1) --> support set = (0, 1, 0)
cell (1, 3, 2) --> support set = (0, 1, 1)
cell (2, 1, 1) --> support set = (1, 0, 0)
cell (2, 1, 2) --> support set = (1, 0, 1)
cell (2, 2, 1) --> support set = (1, 1, 0)
cell (2, 2, 2) --> support set = (1, 1, 1)
cell (2, 3, 1) --> support set = (1, 1, 0)
cell (2, 3, 2) --> support set = (1, 1, 1)

上面的代码可以正常工作。但我想避免 for 循环。在这里，我使用了3个for循环（因为两个参数向量的长度均为3）。如果向量的长度增加或减少，该功能将不起作用（我需要相应地调整循环）；这就是为什么我想用适用于任何长度的高效替代品来代替for-loop的原因。有什么建议吗？

Answer 1

一种删除for循环并使解决方案足够灵活以适应任何长度输入的方法。

我们使用expand.grid创建level的所有可能组合，并使用apply逐行创建要打印的字符串。

cell_support_xyz <- function(level, zero) {
  tmp <- do.call(expand.grid, lapply(level, seq))
  abc <- apply(tmp, 1, function(x) 
               cat(sprintf('cell (%s) --> support set = (%s)\n', 
                   toString(x), toString(+(x != zero)))))
}

l<-c(2,3,2)
z<-c(1,1,1)

cell_support_xyz(l, z)
#cell (1, 1, 1) --> support set = (0, 0, 0)
#cell (2, 1, 1) --> support set = (1, 0, 0)
#cell (1, 2, 1) --> support set = (0, 1, 0)
#cell (2, 2, 1) --> support set = (1, 1, 0)
#cell (1, 3, 1) --> support set = (0, 1, 0)
#cell (2, 3, 1) --> support set = (1, 1, 0)
#cell (1, 1, 2) --> support set = (0, 0, 1)
#cell (2, 1, 2) --> support set = (1, 0, 1)
#cell (1, 2, 2) --> support set = (0, 1, 1)
#cell (2, 2, 2) --> support set = (1, 1, 1)
#cell (1, 3, 2) --> support set = (0, 1, 1)
#cell (2, 3, 2) --> support set = (1, 1, 1)

Answer 2

您可以通过2个步骤进行操作：

l<-c(2,3,2)
z<-c(1,1,1)

cells <- expand.grid(lapply(l, seq))

t(apply(cells, 1, function(x) 1L*!(x == z)))

cells包含所有组合。如果订单很重要，则只需重新订购即可：

cells <- dplyr::arrange(cells, Var1, Var2, Var3)

然后，对于每行（apply(,1,)），您可以使用已经向量化的==，将整行与整个z向量进行比较。

乘以1L使其为整数，与+相同。

R中for循环的替代方法

2 个答案: