来自`grid`包的`vpList`的参数
来自`grid`包的`vpList`的参数
提问于 2018-12-20 04:44:50
这可能有点傻，但是我如何动态地创建一个vpTree，比如
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？
回答 2
Stack Overflow用户
回答已采纳
发布于 2018-12-20 23:37:21
vpList做的第一件事是将它的参数转换成一个列表。我需要用一个自定义函数覆盖网格中的vpList函数，以绕过列表
> vpList
function (...)
{
vps <- list(...)
vpListFromList(vps)
}
<bytecode: 0x000002e14a815a60>
<environment: namespace:grid>至
vpList <- function (vps)
{
grid:::vpListFromList(vps)
}如下所示：
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))Stack Overflow用户
发布于 2018-12-20 05:05:02
您创建vpList的方式不正确。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)页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/53858896
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