Question

这可能很愚蠢，但是如何动态创建类似vpTree的

library(grid)
grid.newpage()
vpTree( viewport(layout=grid.layout(2,2), name = "body"), vpList(viewport(name= "a"), viewport(name ="b"), viewport(name = "c"), viewport(name = "d")))
#> viewport[body]->(viewport[a], viewport[b], viewport[c], viewport[d])

我无法传递视口列表：

library(grid)
grid.newpage()
n <- c(viewport(name= "a"), viewport(name ="b"), viewport(name = "c"), viewport(name = "d"))
vpTree( viewport(layout=grid.layout(2,2), name = "body"), vpList(n))
#> Error in vpListFromList(vps): only viewports allowed in 'vpList'

有什么想法吗？

Answer 1

创建vpList的方法不正确。试试这个

##Supply same list to variable and then pass it in function

library(grid)
grid.newpage()
n <- vpList(viewport(name= "a"), viewport(name ="b"), viewport(name = "c"), viewport(name = "d"))
vpTree( viewport(layout=grid.layout(2,2), name = "body"), n)

Answer 2

vpList要做的第一件事是将其参数转换为列表。我需要使用自定义函数覆盖网格中的vpList函数以绕过

的列表

> vpList
function (...) 
{
    vps <- list(...)
    vpListFromList(vps)
}
<bytecode: 0x000002e14a815a60>
<environment: namespace:grid>

到

vpList <- function (vps) 
{
  grid:::vpListFromList(vps)
}

赞：

library(grid)
grid.newpage()

vpListX <- function (vps) 
{
  grid:::vpListFromList(vps)
}


n <- list(viewport(name= "a"), viewport(name ="b"), viewport(name = "c"), viewport(name = "d"))
vpTree( viewport(layout=grid.layout(2,2), name = "body"), vpListX(n))

grid包中的vpList的参数

2 个答案: