有什么方法可以将 gTree 转换回 r 中可用的 ggplot 吗？

Question

这是一些示例数据：

exampledata <- structure(list(x = c(2.93131952459005, 3.21275054434318, 1.36466997175509, 
2.13626543532502, 1.45889556823722, 1.94598707699052, 0.719062322132357, 
2.38139571953234, 2.37813367615963, 3.98126576880209), y = c(7.51581380181603, 
9.77495763943671, 8.9666894018554, 8.62675858853528, 7.89238665417542, 
9.84865061237773, 7.24526820962333, 7.64727218939944, 7.28026738945878, 
8.6913070524479), z = structure(c(1L, 1L, 1L, 2L, 2L, 2L, 2L, 
3L, 3L, 3L), .Label = c("a", "b", "c"), class = "factor"), z2 = structure(c(1L, 
1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L), .Label = c("cat", "dog"), class = "factor")), class = "data.frame", row.names = c(NA, 
-10L))

该数据的图表：

asdf <- ggplot(exampledata, aes(x = x, y = y, color = z, shape = z)) +
  geom_point() + 
  geom_line(aes(color = z, linetype = z2))+
  scale_linetype_manual(values = c(1,2,3)) +
  theme(legend.position = 'top', 
  legend.spacing = unit(2, 'cm'))

下面的代码只是增加了图例 z 分量内点的大小（独立于 z 中的线或图中的实际点）。它的输出是一个 gTree 对象。虽然这部分看起来很武断，但它是最终情节的重要一步

grid::grid.ls(grid::grid.force())

# Set the size of the point in the legend to 2 mm
grid::grid.gedit("key-1-[-0-9]+-1.2-[-0-9]+-2-[-0-9]+", size = unit(4, "mm"))

# save the modified plot to an object
g2 <- grid::grid.grab()
ggsave(g2, filename = 'g2.tiff')

现在想象一下以下数据框的散点图：

datasetb <- structure(list(x = c(2.55279478309192, 0.929375129220925, 1.56509894419863, 
2.48026699500513, 1.18018131012236, 1.79675395182943, 0.817046700547386, 
1.99710482619256, 2.18780091987683, 3.41661353718804), y = c(8.88460717718884, 
9.11053089978428, 7.68492406933585, 8.23110925234432, 7.48154953916593, 
9.0253526297593, 9.41899905471655, 8.54779428609509, 9.17050925351926, 
5.83078691211861)), class = "data.frame", row.names = c(NA, -10L
))

b <- ggplot(data = datasetb, aes(x = x, y = y) +
  geom_point()

我希望能够使用plot_grid将这些地块合二为一。但是我不认为 gTree 可以用作 get_legend 的参数。有什么方法可以将 gTree 转换回 ggplot 对象，或者任何解决方法

以下是我的最终目标。

prow <- plot_grid( asdf + theme(legend.position="none"),
                   b + theme(legend.position="none"),
                   align = 'vh',
                   labels = c("A", "B"),
                   hjust = -1,
                   nrow = 1
)

legend_a <- get_legend(asdf + theme(legend.position="top"))


p <- plot_grid( legend_a, prow, ncol = 1, rel_heights = c(.2, 1))

tiff("BothPlots.tiff", units = 'in', width = 12, height = 7, res = 400)

p
dev.off()

Answer 1

简短的回答是否定的。 ggplot 就像一个食谱。 gTree 就像食谱制作的蛋糕。你不能把蛋糕拆开来取回食谱。

但是，这里的答案是，您可以将图拼接在一起然后修改图例，而不是修改图例然后将其提取并将图拼接在一起。所以如果你按这个顺序做事：

asdf <- ggplot(exampledata, aes(x = x, y = y, color = z, shape = z)) +
  geom_point() +
  geom_line(aes(color = z, linetype = z2)) +
  scale_linetype_manual(values = c(1, 2, 3)) +
  theme(legend.position = 'top', legend.spacing = unit(2, 'cm'))


b <- ggplot(data = datasetb, aes(x = x, y = y)) + geom_point()

prow <- plot_grid(asdf + theme(legend.position="none"),
                  b + theme(legend.position="none"),
                  align = 'vh',
                  labels = c("A", "B"),
                  hjust = -1,
                  nrow = 1)

legend_a <- get_legend(asdf + theme(legend.position = "top"))

plot_grid(legend_a, prow, ncol = 1, rel_heights = c(.2, 1))

grid.ls(grid::grid.force())
grid.gedit("key-1-[-0-9]+-1.2-[-0-9]+-2-[-0-9]+", size = unit(4, "mm"))

# save the modified plot to an object
g2 <- grid::grid.grab()

现在我们可以保存了（我不得不保存为小 png 以允许上传到这里）：

png("BothPlots.png", units = 'in', width = 6, height = 3.5, res = 200)

grid::grid.draw(g2)
dev.off()

你得到：

BothPlots.png

有什么方法可以将 gTree 转换回 r 中可用的 ggplot 吗？

Is there any way to convert a gTree back to a workable ggplot in r?

r

ggplot2

r-grid