我有一个data.frame
,看起来像:
type id x y label from to polarity group
1 link 3 560 -219 <NA> 2 1 + 1
8 var 2 426 -276 hours worked per week NA NA <NA> 1
7 var 1 610 -226 accomplishments per week NA NA <NA> 1
和函数link_coordinates
,该函数通过查找类型为{{1}的链接元素的坐标来计算类型为from_x/from_y
的所有元素的to_x/to_y
和link
坐标}。链接由var
和from
表示:
to
我可以预先计算这些坐标,然后绘制链接:
link_coordinates <- function(cld) {
cld$from_x <- NA; cld$from_y <- NA; cld$to_x <- NA; cld$to_y <- NA
for(i in 1:nrow(cld)) {
if(cld$type[i] == "link") {
cld$from_x[i] <- cld$x[cld$id == cld$from[i]]
cld$from_y[i] <- cld$y[cld$id == cld$from[i]]
cld$to_x[i] <- cld$x[cld$id == cld$to[i]]
cld$to_y[i] <- cld$y[cld$id == cld$to[i]]
}
}
return(cld)
}
现在,我想在名为library(ggplot2)
cld_linked <- link_coordinates(cld)
ggplot(as.data.frame(cld_linked)) + geom_curve(aes(x = from_x, y = from_y, xend = to_x, yend = to_y))
的定制Geom
中进行该预计算(以及其他工作)。这样做时,我得到一个错误,而不是一个图:
GeomLink
ggplot(as.data.frame(cld)) + geom_link()
为什么,为什么如何必须更改我的Error in FUN(X[[i]], ...) : object 'from_x' not found
和GeomLink
函数,以便它们从上方生成图?
这是我的geom_link
:
data.frame
这是我的cld <- structure(list(type = c("link", "var", "var"), id = c(3, 2, 1
), x = c(560, 426, 610), y = c(-219, -276, -226), label = c(NA,
"hours worked per week", "accomplishments per week"), from = c(2,
NA, NA), to = c(1L, NA, NA), polarity = c("+", NA, NA), group = c(1L,
1L, 1L)), .Names = c("type", "id", "x", "y", "label", "from",
"to", "polarity", "group"), row.names = c(1L, 8L, 7L), class = "data.frame")
和GeomLink
函数:
geom_link