对于示例数据框:
df <- structure(list(year = c(1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3,
3, 3, 4, 4, 4, 4, 4), imd.quintile = c(1, 2, 3, 4, 5, 1, 2, 3,
4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5), average_antibiotic = c(1.17153515458827,
1.11592565388857, 1.09288449967773, 1.07442652168281, 1.06102887394413,
1.0560582933182, 1.00678980505929, 0.992997489072538, 0.978343676071694,
0.967900478870214, 1.02854157116164, 0.98339099101476, 0.981198852494798,
0.971392872980818, 0.962289579742817, 1.00601488964457, 0.951187417739673,
0.950706064156994, 0.939174499710836, 0.934948233015044)), .Names = c("year",
"imd.quintile", "average_antibiotic"), row.names = c(NA, -20L
), vars = "year", drop = TRUE, class = c("grouped_df", "tbl_df",
"tbl", "data.frame"))
我正在制作一张图表,详细说明按年份和按位数计算的抗生素处方差异:
ggplot(plot_data.quintiles) +
geom_col(aes(x = year, y = average_antibiotic, group=imd.quintile, fill=imd.quintile), position = "dodge") +
ylab("Antibiotic STAR-PU") +
xlab("Year") +
theme_bw() +
ylim(0, 1.5)+
scale_colour_brewer("clarity")
蓝色的选择不符合我的口味,因为imd.quintiles之间的差异不是很明显。我已经阅读了here,here和here的各种帖子,但似乎都没有一个回答我的问题。
我尝试使用“清晰度”颜色来获得更广泛的颜色选择。如何正确更改ggplot2图形中的填充颜色?我有什么选择?
答案 0 :(得分:3)
这是您想要的吗?使用factor(imd.quintile)
创建离散(分类)数据,否则ggplot
将数字/整数imd.quintile
视为连续数据。
df <- data.frame(
year = c(1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4,
4, 4),
imd.quintile = c(1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3,
4, 5),
average_antibiotic = c(1.17153515458827, 1.11592565388857, 1.09288449967773,
1.07442652168281, 1.06102887394413, 1.0560582933182,
1.00678980505929, 0.992997489072538, 0.978343676071694,
0.967900478870214, 1.02854157116164, 0.98339099101476,
0.981198852494798, 0.971392872980818,
0.962289579742817, 1.00601488964457, 0.951187417739673,
0.950706064156994, 0.939174499710836, 0.934948233015044)
)
library(ggplot2)
p1 <- ggplot(df) +
geom_col(aes(
x = year, y = average_antibiotic,
group = imd.quintile, fill = factor(imd.quintile)), position = "dodge") +
ylab("Antibiotic STAR-PU") +
xlab("Year") +
theme_bw() +
ylim(0, 1.5)
p1 +
scale_fill_brewer(palette = "Set2") # use scale_fill_xxx to chose the desired color palette
如果您喜欢连续的(顺序的)颜色图,则viridis或scico是不错的选择:
p1 +
scale_fill_viridis_c(option = 'E', direction = -1)
# install.packages('scico')
library(scico)
p1 +
scale_fill_scico()
由reprex package(v0.2.1.9000)于2018-11-29创建
答案 1 :(得分:2)
scale_####_brewer
使用RColorBrewer
中的调色板,没有称为“清晰度”的调色板。
使用RColorBrewer::display.brewer.all()
查看可用的调色板,然后使用palette
arg通过名称进行调用。另外,您需要将imd.quintile
变量更改为字符或因子。您还通过填充而非颜色来映射美学,因此您需要使用scale_fill_brewer
。
ggplot(df) +
geom_col(aes(x = year, y = average_antibiotic, group=imd.quintile, fill=imd.quintile), position = "dodge") +
ylab("Antibiotic STAR-PU") +
xlab("Year") +
theme_bw() +
ylim(0, 1.5) +
scale_fill_brewer(palette = "Spectral")