一种总是避开直方图的方法？ [英] A way to always dodge a histogram?

问题描述

使用ggplot2我用水平轴上的一个因子创建一个柱状图，并使用闪避位置创建填充颜色的另一个因子。我的问题是填充因子有时只取一个值作为横向因子的值，而没有任何东西可以闪避，因此它占据了整个宽度。有没有办法让它不闪避，因此所有的条宽都是一样的？或等价地将0绘制出来？

例如

  ggplot（ data = mtcars，aes（x = factor（carb），fill = factor（gear）））+ 
 geom_histogram（position =dodge）
  

这个答案有一些想法。在新版本发布之前也有人问到，所以可能有所改变？使用方面（也显示此处）我不喜欢我的情况，但我想编辑数据并使用 geom_bar 可以工作，但感觉不雅。此外，无论如何，当我尝试使用facetting时，无论如何，b / b

  ggplot（mtcars，aes（x =因子（碳水化合物） ））+ 
 geom_bar（）+ facet_grid（〜factor（carb））
  

错误layout_base（data，cols，drop = drop）中的错误：
至少有一个图层必须包含所有用于构图的变量

我想我可以生成计数的数据框，然后使用 geom_bar ，

  mtcounts <  -  ddply（子集（mtcars，select = c（carb，gear）），
 .fun = count，.variables = c（carb，gear））
  

填写0不存在的关卡。有人知道这是否会奏效，或者有更好的方法吗？

geom_bar needs stat =identity

我不确定如果这对您来说太晚了，但请参阅最近发布的答案这里
也就是说，我会采用Joran的建议来预先计算 ggplot 调用之外的计数，并使用 geom_bar 。与其他文章的答案一样，这些计数分两步获得：首先，使用 dcast 获得计数的交叉表;然后秒，融化交叉制表。

  library（ggplot2）
 library（reshape2）
 
 dat = dcast（mtcars，因子（carb）〜因子（齿轮），fun.aggregate =长度）
 dat.melt = melt（dat，id.vars =factor（carb），measure.vars = c（3， 4），5））
 dat.melt 
 
（p <-ggplot（dat.melt，aes（x =`factor（carb）`，y = value，fill =变量））+ 
 geom_bar（stat =identity，position =dodge））
  

图表：

Using ggplot2 I'm creating a histogram with a factor on the horizontal axis and another factor for the fill color, using a dodged position. My problem is that the fill factor sometimes takes only one value for a value of the horizontal factor, and with nothing to dodge the bar takes up the full width. Is there a way to make it dodge nothing so that all bar widths are the same? Or equivalently to plot the 0's?

For example

ggplot(data = mtcars, aes(x = factor(carb), fill = factor(gear))) +
geom_histogram(position = "dodge")

This answer has a couple ideas. It was also asked before the new version was released, so maybe something changed? Using facets (also shown here) I don't like for my situation, though I suppose editing the data and using geom_bar could work, but it feels inelegant. Moreover, when I tried facetting anyway

ggplot(mtcars, aes(x = factor(carb), fill = factor(gear))) +
    geom_bar() + facet_grid(~factor(carb))

I get the error "Error in layout_base(data, cols, drop = drop): At least one layer must contain all variables used for facetting"

I suppose I could generate a data frame of counts and then use geom_bar,

mtcounts <- ddply(subset(mtcars, select = c("carb", "gear")),
    .fun = count, .variables = c("carb", "gear"))

filling out the levels that aren't present with 0's. Does anyone know if that would work or if there's a better way?

解决方案

Updated geom_bar needs stat = "identity"

I'm not sure if this is too late for you, but see the answer to a recent post here That is, I'd take Joran's advice to pre-calculate the counts outside the ggplot call and to use geom_bar. As with the answer to other post, the counts are obtained in two steps: first, a crosstabulation of counts is obtained using dcast; then second, melt the crosstabulation.

library(ggplot2)
library(reshape2)

dat = dcast(mtcars, factor(carb) ~ factor(gear), fun.aggregate = length)
dat.melt = melt(dat, id.vars = "factor(carb)", measure.vars = c("3", "4", "5"))
dat.melt

(p <- ggplot(dat.melt, aes(x = `factor(carb)`, y = value, fill = variable)) + 
  geom_bar(stat = "identity", position = "dodge"))

The chart:

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