r - 如何在点阵布局中制作热图风格的二元直方图？

Question

以以下示例数据为例：

x <- rnorm(10000)
y <- rnorm(10000) * x
z <- rnorm(10000) * y
df <- data.frame(x,y,z)

我们可以生成一个散点图矩阵，如下所示：

splom(df)

但由于大量重叠点，很难衡量密度。

有没有一种简单的方法可以用二元直方图热图替换每个图，就像squash产生的那样？

library(squash)
hist2(df$x, df$y)

Answer 1

这panel.hexbinplot对于大型数据集很方便。

library(hexbin)
splom(df, panel=panel.hexbinplot)

您可以像这样自定义面板功能：

library(hexbin)
splom(df,
      panel = function(x, y, ...){
        panel.hexbinplot(x, y, style = "nested.lattice", 
                      type = c("g", "smooth"),col='blue', ...)
      },
      pscale=0, varname.cex=0.7)

您可以使用 tehstyle参数。

Answer 2

这是另一个更符合您的原始要求的选项

# run the code you've provided
library(lattice)
x <- rnorm(10000)
y <- rnorm(10000) * x
z <- rnorm(10000) * y
df <- data.frame(x,y,z)

# look at each of these options one-by-one..  go slowly!

# here's your original
splom(df)


# here each point has been set to very transparent
splom(df , col="#00000005" )

# here each point has been set to moderately transparent
splom(df , col="#00000025" )

# here each point has been set to less transparent
splom(df , col="#00000050" )

Answer 3

这不是您要求的方法，但确实可以帮助您解决您描述的基本问题:)

# run the code you've provided
library(lattice)
x <- rnorm(10000)
y <- rnorm(10000) * x
z <- rnorm(10000) * y
df <- data.frame(x,y,z)

# figure out what ten percent of the total records are
ten.percent <- nrow( df ) / 10

# create a new data frame `df2` containing
# a randomly-sampled ten percent of the original data frame
df2 <- df[ sample( nrow( df ) , ten.percent  ) , ]

# now `splom` that.. and notice it's easier to see densities
splom(df2)