我经常不得不制作带有标签的堆叠条形图。我编码标签的方式非常耗时,我想知道是否有一种方法可以更有效地编码内容。我希望标签位于条形图的每个部分的中心。我更喜欢基础R解决方案。
stemdata <- structure(list( #had to round some nums below for 100% bar
A = c(7, 17, 76),
B = c(14, 10, 76),
C = c( 14, 17, 69),
D = c( 4, 10, 86),
E = c( 7, 17, 76),
F = c(4, 10, 86)),
.Names = c("Food, travel, accommodations, and procedures",
"Travel itinerary and dates",
"Location of the STEM Tour stops",
"Interactions with presenters/guides",
"Duration of each STEM Tour stop",
"Overall quality of the STEM Tour"
),
class = "data.frame",
row.names = c(NA, -3L)) #4L=number of numbers in each letter vector#
# attach(stemdata)
print(stemdata)
par(mar=c(0, 19, 1, 2.1)) # this sets margins to allow long labels
barplot(as.matrix(stemdata),
beside = F, ylim = range(0, 10), xlim = range(0, 100),
horiz = T, col=colors, main="N=29",
border=F, las=1, xaxt='n', width = 1.03)
text(7, 2, "14%")
text(19, 2, "10%")
text(62, 2, "76%")
text(7, 3.2, "14%")
text(22.5, 3.2, "17%")
text(65.5, 3.2, "69%")
text(8, 4.4, "10%")
text(55, 4.4, "86%")
text(3.5, 5.6, "7%")
text(15, 5.6, "17%")
text(62, 5.6, "76%")
text(9, 6.9, "10%")
text(55, 6.9, "86%")
答案 0 :(得分:2)
根据OP要求保留基数R,我们可以轻松地在一个小函数内自动完成内部标签的定位(即x
坐标)。
xFun <- function(x) x/2 + c(0, cumsum(x)[-length(x)])
现在,很高兴知道barplot
会隐式拖曳y
坐标,我们可以通过赋值(此处为byc <- barplot(.)
)来捕获它们。
最终,只需在数据帧labs
中组合坐标和标签,然后通过text
中的sapply
调用进行“循环”。 (按照other question的要求,将col="white"
或col=0
用于白色标签。)
# barplot
colors <- c("gold", "orange", "red")
par(mar=c(2, 19, 4, 2) + 0.1) # expand margins
byc <- barplot(as.matrix(stemdata), horiz=TRUE, col=colors, main="N=29", # assign `byc`
border=FALSE, las=1, xaxt='n')
# labels
labs <- data.frame(x=as.vector(sapply(stemdata, xFun)), # apply `xFun` here
y=rep(byc, each=nrow(stemdata)), # use `byc` here
labels=as.vector(apply(stemdata, 1:2, paste0, "%")),
stringsAsFactors=FALSE)
invisible(sapply(seq(nrow(labs)), function(x) # `invisible` prevents unneeded console output
text(x=labs[x, 1:2], labels=labs[x, 3], cex=.9, font=2, col=0)))
# legend (set `xpd=TRUE` to plot beyond margins!)
legend(-55, 8.5, legend=c("Medium","High", "Very High"), col=colors, pch=15, xpd=TRUE)
par(mar=c(5, 4, 4, 2) + 0.1) # finally better reset par to default
结果
数据
stemdata <- structure(list(`Food, travel, accommodations, and procedures` = c(7,
17, 76), `Travel itinerary and dates` = c(14, 10, 76), `Location of the STEM Tour stops` = c(14,
17, 69), `Interactions with presenters/guides` = c(4, 10, 86),
`Duration of each STEM Tour stop` = c(7, 17, 76), `Overall quality of the STEM Tour` = c(4,
10, 86)), class = "data.frame", row.names = c(NA, -3L))
答案 1 :(得分:1)
您会考虑使用tidyverse
解决方案吗?
library(tidyverse) # for dplyr, tidyr, tibble & ggplot2
stemdata %>%
rownames_to_column(var = "id") %>%
gather(Var, Val, -id) %>%
group_by(Var) %>%
mutate(id = factor(id, levels = 3:1)) %>%
ggplot(aes(Var, Val)) +
geom_col(aes(fill = id)) +
coord_flip() +
geom_text(aes(label = paste0(Val, "%")),
position = position_stack(0.5))
结果: