图r两个分类变量

我使用下面的命令在R中绘制两个分类变量

性别有2个层次，收入有9个层次

spineplot(main$Gender,main$Income, xlab="Gender", ylab="Income levels: 1 is lowest",xaxlabels=c("Male","Female"))

它生成如下图表

如何用彩色绘制此图表

如何显示每个框中每个收入水平的百分比？例如，女性收入水平1占数据的21%。如何在深色区域内显示

21%

################更新1 添加可复制的示例

fail <- factor(c(2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1,
                 1, 1, 1, 2, 1, 1, 1, 1, 1,2,2,2,2),
               levels = c(1, 2), labels = c("male", "female"))
gender <- factor(rep(c(1:9),3))
spineplot(fail,gender)

---

fail我认为使用
barplot
可能更容易做到这一点，因为
spineplot
不会返回任何有用的信息

默认设置如下，但您可以将条形的宽度调整为其他变量（您可以看到返回的x轴坐标）：


经过最后的润色

tbl <- table(gender, fail)
prp <- prop.table(tbl, 2L)
yat <- prp / 2 + apply(rbind(0, prp[-nrow(prp), ]), 2L, cumsum)

bp <- barplot(prp, width = table(fail), axes = FALSE, col = rainbow(nrow(prp)))

axis(2L, at = yat[, 1L], labels = levels(gender), lwd = 0)
axis(4L)

text(rep(bp, each = nrow(prp)), yat, sprintf('%0.f%%', prp * 100), col = 0)

替代@rawr有趣的解决方案的另一个方法是：

fail <- factor(c(2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1,
                 1, 1, 1, 2, 1, 1, 1, 1, 1,2,2,2,2),
               levels = c(1, 2), labels = c("male", "female"))
gender <- factor(rep(c(1:9),3))

mypalette <- colorRampPalette(c("lightblue","darkblue"))
tbl <- spineplot(fail, gender, xlab="Gender", ylab="Income levels: 1 is lowest",
     xaxlabels=c("Male","Female"), col=mypalette(nlevels(gender)) )
print(tbl)

#        Income levels: 1 is lowest
# Gender   1 2 3 4 5 6 7 8 9
# male   2 1 2 1 3 2 2 2 1
# female 1 2 1 2 0 1 1 1 2

print.perc <- function(k, tbl, ndigits=2, str.pct="%") {
   # These lines of codes are the same used by from spineplot
   # for the calculation of the x-position of the stacked bars
   nx <- nrow(tbl)
   off <- 0.02
   xat <- c(0, cumsum(prop.table(margin.table(tbl, 1)) + off))
   posx <- (xat[1L:nx] + xat[2L:(nx + 1L)] - off)/2
   # Proportions by row (gender)       
   ptbl <- prop.table(tbl,1)
   # Define labels as strings with a given format
   lbl <- paste(format(round(100*ptbl[k,], ndigits), nsmall=ndigits), str.pct, sep="")
   # Print labels
   # cumsum(ptbl[k,])-ptbl[k,]/2 is the vector of y-positions
   # for the centers of each stacked bar
   text(posx[k], cumsum(ptbl[k,])-ptbl[k,]/2, lbl)
}

# Print income levels for males and females
strsPct <- c("%","%")
for (k in 1:nrow(tbl)) print.perc(k, tbl, ndigits=2, str.pct=strsPct[k])

失败而不是彩虹，是否可能得到绿色或蓝色的阴影？@user2543622是的，尽管这取决于你所说的阴影是什么意思。例如，您可以选择一种颜色并应用不同数量的透明度
col=Vectorize（adjustcolor）（'green4'，alpha.f=seq（.3,1，length.out=nrow（prp））
或两种或两种以上完全透明的颜色，它们之间有阴影
col=colorrmppalete（c（'limegreen'，'darkgreen'））（nrow（prp））
是否可以就print.perc功能编写一些说明？这将使我能够将您的解决方案应用到我的数据中。此外，是否可以显示如何以单色阴影绘制？是否可以以粗体显示百分比？此外，请让我知道如何在<代码>文本<代码>中查找颜色关键字，如浅蓝色深蓝色添加
font=2
。对于颜色关键字，在命令提示下键入
colors（）
。
spineplot(fail, gender, col = rainbow(nlevels(gender)))

fail <- factor(c(2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1,
                 1, 1, 1, 2, 1, 1, 1, 1, 1,2,2,2,2),
               levels = c(1, 2), labels = c("male", "female"))
gender <- factor(rep(c(1:9),3))

mypalette <- colorRampPalette(c("lightblue","darkblue"))
tbl <- spineplot(fail, gender, xlab="Gender", ylab="Income levels: 1 is lowest",
     xaxlabels=c("Male","Female"), col=mypalette(nlevels(gender)) )
print(tbl)

#        Income levels: 1 is lowest
# Gender   1 2 3 4 5 6 7 8 9
# male   2 1 2 1 3 2 2 2 1
# female 1 2 1 2 0 1 1 1 2

print.perc <- function(k, tbl, ndigits=2, str.pct="%") {
   # These lines of codes are the same used by from spineplot
   # for the calculation of the x-position of the stacked bars
   nx <- nrow(tbl)
   off <- 0.02
   xat <- c(0, cumsum(prop.table(margin.table(tbl, 1)) + off))
   posx <- (xat[1L:nx] + xat[2L:(nx + 1L)] - off)/2
   # Proportions by row (gender)       
   ptbl <- prop.table(tbl,1)
   # Define labels as strings with a given format
   lbl <- paste(format(round(100*ptbl[k,], ndigits), nsmall=ndigits), str.pct, sep="")
   # Print labels
   # cumsum(ptbl[k,])-ptbl[k,]/2 is the vector of y-positions
   # for the centers of each stacked bar
   text(posx[k], cumsum(ptbl[k,])-ptbl[k,]/2, lbl)
}

# Print income levels for males and females
strsPct <- c("%","%")
for (k in 1:nrow(tbl)) print.perc(k, tbl, ndigits=2, str.pct=strsPct[k])