Question

对于这个问题，我已经能够将数据分成两个直方图，一个是收入高于中位数，另一个是收入低于中位数。到目前为止，以下代码是我所做的：

library(openintro)
data("countyComplete")
attach("countyComplete")

median(median_household_income, na.rm = FALSE)

x<-subset(countyComplete,median_household_income > 42445)
y<-subset(countyComplete,median_household_income < 42445)

par(mfrow=c(1,2))
hist(x$median_household_income, main="Income Above Median" )
hist(y$median_household_income,main = "Income Below Median")

但是，我对如何迫使直方图在y轴上使用相同的限制以及中断感到有些困惑。有人可以指出我正确的方向。我尝试这样做：

par(mfrow=c(1,2))
hist(x$median_household_income,
     breaks=seq(0,100,by=5),
     freq = FALSE,
     ylim=c(0,.15),
     xlim = range(breaks),
     main="Income Above Median")
hist(y$median_household_income, main = "Income Below Median")

但是我的绘图屏幕上只显示一个直方图，控制台说 “ hist.default（x $ median_household_income，breaks = seq（0，100，：一些“ x”不计算在内；也许“中断”不跨越“ x”的范围。”

我该怎么办？

Answer 1

我会忘记breaks的论点。这没有任何意义，您正在绘制中位数以下和上方的值，它们不会相交。

对于直方图，我已经预先计算了中位数和密度的最大值。

library(openintro)
data("countyComplete")

med <- median(countyComplete$median_household_income, na.rm = FALSE)

x <- subset(countyComplete, median_household_income > med)
y <- subset(countyComplete, median_household_income < med)

hx <- hist(x$median_household_income, plot = FALSE)
hy <- hist(y$median_household_income, plot = FALSE)
MaxY <- max(c(hx$density, hy$density))

op <- par(mfrow = c(1, 2))
hist(x$median_household_income, main = "Income Above Median",
     freq = FALSE, ylim = c(0, MaxY))
hist(y$median_household_income, main = "Income Below Median",
     freq = FALSE, ylim = c(0, MaxY))
par(op)

极限与突破

1 个答案: