此代码生成下面的第一个图:
water.height <- seq(0, 5, 1)
y <- seq(0, 1500, length.out = 6)
df <- data.frame(water.height, y)
library(ggplot2)
ggplot(df, aes(water.height, y)) + geom_blank()+ theme_bw()
我在这个蓝色背景下拍照:
我可以使用R代码生成相同的蓝色背景吗?
答案 0 :(得分:13)
评论中提供了the ggplot2 approach的相关链接。从那里复制:
library(grid)
g <- rasterGrob(blues9, width=unit(1,"npc"), height = unit(1,"npc"),
interpolate = TRUE)
# grid.draw(g)
library(ggplot2)
ggplot(mtcars, aes(factor(cyl))) + # add gradient background
annotation_custom(g, xmin=-Inf, xmax=Inf, ymin=-Inf, ymax=Inf) +
geom_bar() # add data layer
我自己的方法:
像往常一样,我无法与baptiste的网格图形问题解决方案的简单优雅竞争,但这是我的方法,因为我去了所有的工作:
gg.background.fill <- function(gg.plot, cols = "white", which = "x") {
#does not work with facets
stopifnot(which %in% c("x", "y"))
which1 <- if (which == "x") "width" else "height"
require(gridExtra)
g <- ggplotGrob(gg.plot)
#g <- ggplotGrob(p)
gg <- g$grobs
findIt <- vapply(gg, function(x) grepl("GRID.gTree", x$name, fixed = TRUE), TRUE)
n1 <- getGrob(gg[findIt][[1]], "grill.gTree", grep=TRUE)$name
n2 <- getGrob(gg[findIt][[1]], "panel.background.rect", grep=TRUE)$name
gg[findIt][[1]]$children[[n1]]$children[[n2]]$gp$fill <- cols
x <- gg[findIt][[1]]$children[[n1]]$children[[n2]][[which]]
w <- gg[findIt][[1]]$children[[n1]]$children[[n2]][[which1]]
attr <- attributes(x)
x <- seq(0 + c(w)/length(cols)/2, 1 - c(w)/length(cols)/2, length.out = length(cols))
attributes(x) <- attr
gg[findIt][[1]]$children[[n1]]$children[[n2]][[which]] <- x
w <- c(w)/length(cols)
attributes(w) <- attr
gg[findIt][[1]]$children[[n1]]$children[[n2]][[which1]] <- w
g$grobs <- gg
class(g) = c("arrange", "ggplot", class(g))
g
}
p1 <- gg.background.fill(p, colorRampPalette(c("red", "blue"))(100))
print(p1)
p2 <- gg.background.fill(p, colorRampPalette(c("red", "blue"))(100), "y")
print(p2)
这会修改可能被视为优势的现有背景,但与annotation_custom
方法相比,它并不适用于分面。为此需要做更多的工作。
答案 1 :(得分:1)
我们要使用线性渐变作为图的背景。
让我们开始制作一个介于0和1之间的数字的矩阵。
# The angle of our linear gradient
deg <- 45
rad <- deg / (180 / pi)
# A 5x5 matrix
n <- 5
mat <- matrix(data = 0, ncol = n, nrow = n)
# Let's fill in the matrix.
for (i in 1:n) {
for (j in 1:n) {
mat[i, j] <- (i / n) * cos(rad) + (j / n) * sin(rad)
}
}
我们得到了什么?
mat
#> [,1] [,2] [,3] [,4] [,5]
#> [1,] 0.2828427 0.4242641 0.5656854 0.7071068 0.8485281
#> [2,] 0.4242641 0.5656854 0.7071068 0.8485281 0.9899495
#> [3,] 0.5656854 0.7071068 0.8485281 0.9899495 1.1313708
#> [4,] 0.7071068 0.8485281 0.9899495 1.1313708 1.2727922
#> [5,] 0.8485281 0.9899495 1.1313708 1.2727922 1.4142136
看起来很接近我们想要的东西。
现在,让我们将值限制在0到1之间。
mat <- mat - min(mat)
mat <- mat / max(mat)
mat
#> [,1] [,2] [,3] [,4] [,5]
#> [1,] 0.000 0.125 0.250 0.375 0.500
#> [2,] 0.125 0.250 0.375 0.500 0.625
#> [3,] 0.250 0.375 0.500 0.625 0.750
#> [4,] 0.375 0.500 0.625 0.750 0.875
#> [5,] 0.500 0.625 0.750 0.875 1.000
好多了!
让我们使用grid::rasterGrob()
来制作图形对象并
画出来。
library(grid)
g <- rasterGrob(
image = mat,
width = unit(1, "npc"),
height = unit(1, "npc"),
interpolate = TRUE
)
grid.newpage()
grid.draw(g)
由于我们有一个grob,我们可以将其添加到ggplot2图中,
ggplot2::annotation_custom()
。
library(ggplot2)
ggplot(mtcars, aes(factor(cyl))) +
annotation_custom(
grob = g, xmin = -Inf, xmax = Inf, ymin = -Inf, ymax = Inf
) +
geom_bar()
万岁!我们做到了。但是我们还没有完成。
一些注意事项:
请随时复制下面的make_gradient()
函数,然后
改善它。
library(ggplot2)
library(grid)
library(RColorBrewer)
make_gradient <- function(deg = 45, n = 100, cols = blues9) {
cols <- colorRampPalette(cols)(n + 1)
rad <- deg / (180 / pi)
mat <- matrix(
data = rep(seq(0, 1, length.out = n) * cos(rad), n),
byrow = TRUE,
ncol = n
) +
matrix(
data = rep(seq(0, 1, length.out = n) * sin(rad), n),
byrow = FALSE,
ncol = n
)
mat <- mat - min(mat)
mat <- mat / max(mat)
mat <- 1 + mat * n
mat <- matrix(data = cols[round(mat)], ncol = n)
grid::rasterGrob(
image = mat,
width = unit(1, "npc"),
height = unit(1, "npc"),
interpolate = TRUE
)
}
示例1
g <- make_gradient(
deg = 45, n = 500, cols = brewer.pal(9, "Spectral")
)
ggplot(mtcars, aes(factor(cyl))) +
annotation_custom(
grob = g, xmin = -Inf, xmax = Inf, ymin = -Inf, ymax = Inf
) +
geom_bar()
示例2
g <- make_gradient(
deg = 180, n = 500, cols = brewer.pal(9, "RdBu")
)
ggplot(mtcars, aes(factor(cyl))) +
annotation_custom(
grob = g, xmin = -Inf, xmax = Inf, ymin = -Inf, ymax = Inf
) +
geom_bar()
由reprex package(v0.2.1)于2019-02-06创建
答案 2 :(得分:0)
我使用Kamil Slowikowski的示例构建了一个更简单的函数,该函数根据一系列值生成线性梯度。如果您在三个变量之间有某种关系(例如y~x*z
,其中z
在x
上也有所不同),这可能会有用。然后,您只需绘制y~x
并在背景中使用z~x
作为颜色渐变。
water.height <- seq(0, 5, 1)
y <- seq(0, 1500, length.out = 6)
z <- rnorm(6, 10, 1)
df <- data.frame(water.height, y, z)
grad_by_val <- function(x, y, cols = blues9) {
require(grid)
y <- y[order(x)]
ys <- (y - min(y)) / diff(range(y))
cols <- colorRamp(cols)(ys) / 256
colnames(cols) <- c("red", "green", "blue")
cols <- apply(cols, 1, function(z) do.call(rgb, as.list(z)))
mat <- matrix(cols, ncol = length(x))
rasterGrob(
image = mat,
width = unit(1, "npc"),
height = unit(1, "npc"),
interpolate = TRUE
)
}
library(ggplot2)
ggplot(df, aes(water.height, y)) + geom_blank() + theme_bw() +
annotation_custom(
grob = grad_by_val(df$water.height, df$z),
xmin = -Inf,
xmax = Inf,
ymin = -Inf,
ymax = Inf
) +
geom_point(
size = 5,
color = "#FFFFFF",
fill = "#000000",
shape = 21
)
要添加图例,请参见here。