问题描述
在此可重现的示例网格图中,3个图具有3种填充色,并且z显示为"col"为蓝色,但是在第四个图中只有1个"col",因此z显示为红色.
In this reproducible example grid plot, 3 plots have 3 fill colours, and z displays with the "col" blue, but in the fourth plot there is only 1 "col", so z displays as red.
我只想显示一个常见的图例(我可以这样做),但是我希望z在所有四个图中都为蓝色..有没有简单的方法可以做到这一点?
I want to show only one common legend (which I can do), but I want z to be blue in all four plots.. Is there a simple way to do that?
#---------------------
# Reproducible example
#---------------------
library(tidyverse)
library(ggplot2)
library(grid)
library(gridExtra)
d0 <- read_csv("x, y, col\na,2,x\nb,2,y\nc,1,z")
d1 <- read_csv("x, y, col\na,2,x\nb,2,y\nc,1,z")
d2 <- read_csv("x, y, col\na,2,x\nb,2,y\nc,1,z")
d3 <- read_csv("x, y, col\na,2,z\nb,2,z\nc,1,z")
p0 <- ggplot(d0) + geom_col(mapping = aes(x, y, fill = col))
p1 <- ggplot(d1) + geom_col(mapping = aes(x, y, fill = col))
p2 <- ggplot(d2) + geom_col(mapping = aes(x, y, fill = col))
p3 <- ggplot(d3) + geom_col(mapping = aes(x, y, fill = col))
grid.arrange(p0, arrangeGrob(p1,p2,p3, ncol=3), ncol=1)
推荐答案
这可以通过使用gtable提取图例并反转col
因素的水平来实现:
This can be achieved using gtable to extract the legend and reversing the levels of col
factor:
library(tidyverse)
library(ggplot2)
library(grid)
library(gridExtra)
library(gtable)
d0 <- read_csv("x, y, col\na,2,x\nb,2,y\nc,1,z")
d1 <- read_csv("x, y, col\na,2,x\nb,2,y\nc,1,z")
d2 <- read_csv("x, y, col\na,2,x\nb,2,y\nc,1,z")
d3 <- read_csv("x, y, col\na,2,z\nb,2,z\nc,1,z")
d0 %>%
mutate(col = factor(col, levels = c("z", "y", "x"))) %>%
ggplot() + geom_col(mapping = aes(x, y, fill = col)) -> p0
d1 %>%
mutate(col = factor(col, levels = c("z", "y", "x"))) %>%
ggplot() + geom_col(mapping = aes(x, y, fill = col))+
theme(legend.position="bottom") -> p1
d2 %>%
mutate(col = factor(col, levels = c("z", "y", "x"))) %>%
ggplot() + geom_col(mapping = aes(x, y, fill = col)) -> p2
d3 %>%
ggplot() + geom_col(mapping = aes(x, y, fill = col)) -> p3
legend = gtable_filter(ggplot_gtable(ggplot_build(p1)), "guide-box")
grid.arrange(p0 + theme(legend.position="none"),
arrangeGrob(p1 + theme(legend.position="none"),
p2 + theme(legend.position="none"),
p3 + theme(legend.position="none"),
nrow = 1),
legend,
heights=c(1.1, 1.1, 0.1),
nrow = 3)
另一种方法是在每个图中使用scale_fill_manual
而不更改因子水平.
Another approach is to use scale_fill_manual
in every plot without changing the factor levels.
示例:
p0 + scale_fill_manual(values = c("x" = "red", "z" = "black", "y" = "green"))
因此,请提取原始数据和图例:
so with your original data and legend extracted:
d0 <- read_csv("x, y, col\na,2,x\nb,2,y\nc,1,z")
d1 <- read_csv("x, y, col\na,2,x\nb,2,y\nc,1,z")
d2 <- read_csv("x, y, col\na,2,x\nb,2,y\nc,1,z")
d3 <- read_csv("x, y, col\na,2,z\nb,2,z\nc,1,z")
p0 <- ggplot(d0) + geom_col(mapping = aes(x, y, fill = col))
p1 <- ggplot(d1) + geom_col(mapping = aes(x, y, fill = col))
p2 <- ggplot(d2) + geom_col(mapping = aes(x, y, fill = col))
p3 <- ggplot(d3) + geom_col(mapping = aes(x, y, fill = col))
legend = gtable_filter(ggplot_gtable(ggplot_build(p1 + theme(legend.position="bottom"))), "guide-box")
grid.arrange(p0 + theme(legend.position="none"),
arrangeGrob(p1 + theme(legend.position="none"),
p2 + theme(legend.position="none"),
p3 + theme(legend.position="none") +
scale_fill_manual(values = c("z" = "#619CFF")),
nrow = 1),
legend,
heights=c(1.1, 1.1, 0.1),
nrow = 3)
这篇关于网格图的常见图例的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!