我有以下数据框:
# Dummy data frame
df <- expand.grid(x = 1:3, y = 1:3)
我想像这样使用geom_tile
将其绘制为ggplot2
:
# Tile plot
ggplot(df) +
geom_tile(aes(x = x, y = y),
fill = NA, colour = "red", size = 3, width = 0.7, height = 0.7)
给出
但是,请注意,在每个图块的左上角都缺少一个缺口,该缺口的边界与边界不完全吻合。如果使用geom_rect
,则会得到相同的结果。有一种解决方法可以避免这种情况?
R version 3.5.1 (2018-07-02)
Platform: x86_64-w64-mingw32/x64 (64-bit)
Running under: Windows 7 x64 (build 7601) Service Pack 1
Matrix products: default
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] ggplot2_3.1.0 kitchendraw_0.1.0
loaded via a namespace (and not attached):
[1] Rcpp_1.0.0 rstudioapi_0.8 bindr_0.1.1 magrittr_1.5 tidyselect_0.2.5 munsell_0.5.0 colorspace_1.3-2
[8] R6_2.3.0 rlang_0.3.0.1 plyr_1.8.4 dplyr_0.7.7 tools_3.5.1 grid_3.5.1 gtable_0.2.0
[15] withr_2.1.2 yaml_2.2.0 lazyeval_0.2.1 assertthat_0.2.0 tibble_1.4.2 crayon_1.3.4 bindrcpp_0.2.2
[22] purrr_0.2.5 glue_1.3.0 labeling_0.3 compiler_3.5.1 pillar_1.3.0 scales_1.0.0 pkgconfig_2.0.2
答案 0 :(得分:4)
我认为这是因为每一行的起点就是这样:一个点。并且由于行的大小使行更粗,因此此起点使该空格成为空白。此图使用四个geom_segment
来构成一个正方形,结果表明您遇到了相同的问题:
ggplot(df) +
geom_segment(x = 1, y = 1, xend = 2, yend = 1, size = 3) +
geom_segment(x = 1, y = 1, xend = 1, yend = 2, size = 3) +
geom_segment(x = 1, y = 2, xend = 2, yend = 2, size = 3) +
geom_segment(x = 2, y = 2, xend = 2, yend = 1, size = 3) +
scale_x_continuous(limits = c(0, 3)) +
scale_y_continuous(limits = c(0, 3))
我能想到的唯一解决方案是使x或y轴之一的起点和终点稍稍落后(起点)和前方(终点)。这种解决方案远非理想,但我只能想到。对于size = 3
行,我发现减去和向起点和终点添加0.01可以填充空白:
ggplot(df) +
geom_segment(x = 1-0.01, y = 1, xend = 2+0.01, yend = 1, size = 3) +
geom_segment(x = 1, y = 1, xend = 1, yend = 2, size = 3) +
geom_segment(x = 1-0.01, y = 2, xend = 2+0.01, yend = 2, size = 3) +
geom_segment(x = 2, y = 2, xend = 2, yend = 1, size = 3) +
scale_x_continuous(limits = c(0, 3)) +
scale_y_continuous(limits = c(0, 3))
但是,该解决方案也不理想,因为该值应根据行的大小和所显示图形的比例而改变。
编辑:geom_path()
连接正方形的角而没有留出空格,但是问题仍然存在于直线与原点相交的地方:
df <- data.frame(
x = c(1, 1, 2, 2, 1),
y = c(1, 2, 2, 1, 1)
)
ggplot(df, aes(x, y)) +
geom_path(size = 3, linejoin = "mitre") +
scale_x_continuous(limits = c(0, 3)) +
scale_y_continuous(limits = c(0, 3))
答案 1 :(得分:4)
正如其他人所指出的,这归因于行尾规范,可以在environment(GeomTile$draw_panel)$f
中找到:
function (self, data, panel_params, coord)
{
if (!coord$is_linear()) {
... #omitted for space
}
else {
coords <- coord$transform(data, panel_params)
ggname("geom_rect",
rectGrob(coords$xmin, coords$ymax,
width = coords$xmax - coords$xmin, height = coords$ymax -
coords$ymin, default.units = "native", just = c("left", "top"),
gp = gpar(col = coords$colour,
fill = alpha(coords$fill, coords$alpha),
lwd = coords$size * .pt,
lty = coords$linetype,
lineend = "butt"))) # look here
}
}
geom_tile
层的创建由rectGrob
支持,其硬编码的lineend
参数值为“ butt”。下图(找到https://docs.scipy.org/doc/scipy/reference/generated/scipy.signal.peak_widths.html)很好地说明了三个lineend
值之间的差异:
如果您想深入研究底层GeomTile
的功能并更改代码中所有geom_tile
层的图形参数,则可以这样做。 (我最近通过该解决方案回答了类似的。)但是,对于单个绘图,我只是将ggplot转换为grob对象,并使用那里的gp
参数搞乱了:>
library(grid)
gp <- ggplotGrob(p)
grid.draw(gp)
# this "sharpens" the top left corner
gp$grobs[[which(grepl("panel", gp$layout$name))]]$children[[3]]$gp$lineend <- "square"
grid.draw(gp)
# this further "sharpens" the other three corners
gp$grobs[[which(grepl("panel", gp$layout$name))]]$children[[3]]$gp$linejoin <- "mitre"
grid.draw(gp)
注意:与geom_tile
相对应的正确grob的实际位置不一定是gp$grobs[[which(grepl("panel", gp$layout$name))]]$children[[3]]$gp$linejoin
。这里是children[[3]]
,但是ggplot对象中的其他geom层位于geom_tile
层之下或之上,可以移动其相对位置。在这种情况下,您可能需要检查控制台中gp$grobs[[which(grepl("panel", gp$layout$name))]]$children
的输出,以识别正确的位置编号。
答案 2 :(得分:1)
此问题已由this update修复为ggplot2
。