Question

我需要绘制一个显示计数的条形图和一个在一个图表中显示速率的折线图，我可以分别做两个，但是当我把它们放在一起时，我是第一层的比例（即{{1 }}）与第二层重叠（即geom_bar）。

我可以将geom_line的轴向右移动吗？

Answer 1

这在ggplot2中是不可能的，因为我相信具有单独y尺度的图（不是彼此变换的y尺度）从根本上是有缺陷的。一些问题：

不可逆：在绘图空间上给定一个点，您无法将其唯一映射回数据空间中的某个点。
与其他选项相比，它们相对难以正确阅读。有关详细信息，请参阅Petra Isenberg，Anastasia Bezerianos，Pierre Dragicevic和Jean-Daniel Fekete的A Study on Dual-Scale Data Charts。
它们很容易操纵误导：没有独特的方法来指定轴的相对比例，使它们处于操作状态。 Junkcharts博客中的两个示例：one，two
它们是任意的：为什么只有2个刻度，而不是3个，4个或10个？

您也可以阅读Stephen Few关于Dual-Scaled Axes in Graphs Are They Ever the Best Solution?主题的冗长讨论。

Answer 2

从ggplot2 2.2.0开始，您可以添加这样的辅助轴（取自ggplot2 2.2.0 announcement）：

ggplot(mpg, aes(displ, hwy)) + 
  geom_point() + 
  scale_y_continuous(
    "mpg (US)", 
    sec.axis = sec_axis(~ . * 1.20, name = "mpg (UK)")
  )

Answer 3

有时客户需要两个y尺度。给他们“有缺陷”的演讲往往毫无意义。但我确实喜欢ggplot2坚持以正确的方式做事。我确信ggplot实际上是在教会普通用户正确的可视化技术。

也许您可以使用分面和缩放比较两个数据系列？ - 例如看这里：https://github.com/hadley/ggplot2/wiki/Align-two-plots-on-a-page

Answer 4

通过以上答案和一些微调（以及任何它的价值），这是通过sec_axis实现两个量表的方法：

假设一个简单的（纯粹是虚构的）数据集dt：五天，它跟踪中断次数VS生产力：

        when numinter prod
1 2018-03-20        1 0.95
2 2018-03-21        5 0.50
3 2018-03-23        4 0.70
4 2018-03-24        3 0.75
5 2018-03-25        4 0.60

（两列的范围相差约5倍）。

以下代码将绘制它们用完整个y轴的两个系列：

ggplot() + 
  geom_bar(mapping = aes(x = dt$when, y = dt$numinter), stat = "identity", fill = "grey") +
  geom_line(mapping = aes(x = dt$when, y = dt$prod*5), size = 2, color = "blue") + 
  scale_x_date(name = "Day", labels = NULL) +
  scale_y_continuous(name = "Interruptions/day", 
    sec.axis = sec_axis(~./5, name = "Productivity % of best", 
      labels = function(b) { paste0(round(b * 100, 0), "%")})) + 
  theme(
      axis.title.y = element_text(color = "grey"),
      axis.title.y.right = element_text(color = "blue"))

这是结果（上面的代码+一些颜色调整）：

要点（除了在指定y_scale时使用sec_axis，要指定系列时，第二个数据系列的每个值为乘以。为了使标签正确在在sec_axis定义中，它需要将除以5（和格式化）。因此上面代码中的一个关键部分是geom_line中的*5和sec_axis中的~./5（公式分割）当前值.乘以5）。

相比之下（我不想在这里判断方法），这就是两个图表彼此重叠的样子：

你可以自己判断哪一个更好地传达信息（“不要扰乱工作中的人！”）。猜猜这是一个公平的决定方式。

这两个图片的完整代码（它不仅仅是上面的内容，只是完整并准备好运行）在这里：https://gist.github.com/sebastianrothbucher/de847063f32fdff02c83b75f59c36a7d这里有更详细的解释：https://sebastianrothbucher.github.io/datascience/r/visualization/ggplot/2018/03/24/two-scales-ggplot-r.html

Answer 5

Kohske 大约3年前[KOHSKE]提供了解决这一挑战的技术支柱。关于其解决方案的主题和技术已经在Stackoverflow上的几个实例中进行了讨论[ID：18989001,29235405,21026598]。因此，我将仅使用上述解决方案提供特定的变体和一些解释性演练。

我们假设我们在 G1 组中有一些数据 y1 ， G2 组中的某些数据 y2 >以某种方式相关，例如范围/比例变换或添加一些噪音。因此，人们希望将数据绘制在一个绘图上，左边是 y1 ，右边是 y2 。

  df <- data.frame(item=LETTERS[1:n],  y1=c(-0.8684, 4.2242, -0.3181, 0.5797, -0.4875), y2=c(-5.719, 205.184, 4.781, 41.952, 9.911 )) # made up!

> df
  item      y1         y2
1    A -0.8684 -19.154567
2    B  4.2242 219.092499
3    C -0.3181  18.849686
4    D  0.5797  46.945161
5    E -0.4875  -4.721973

如果我们现在将数据与

一起绘制

ggplot(data=df, aes(label=item)) +
  theme_bw() + 
  geom_segment(aes(x='G1', xend='G2', y=y1, yend=y2), color='grey')+
  geom_text(aes(x='G1', y=y1), color='blue') +
  geom_text(aes(x='G2', y=y2), color='red') +
  theme(legend.position='none', panel.grid=element_blank())

它没有很好地对齐，因为较小的比例 y1 显然会被更大比例的 y2 折叠。

应对挑战的技巧是在第一个比例 y1 上技术性地绘制两个数据集，但是在第二个轴上报告第二个数据集并显示标签原始比例 y2 。

因此我们构建了第一个辅助函数 CalcFudgeAxis ，它计算并收集要显示的新轴的特征。该函数可以修改为ayones喜欢（这个只是将 y2 映射到 y1 的范围内）。

CalcFudgeAxis = function( y1, y2=y1) {
  Cast2To1 = function(x) ((ylim1[2]-ylim1[1])/(ylim2[2]-ylim2[1])*x) # x gets mapped to range of ylim2
  ylim1 <- c(min(y1),max(y1))
  ylim2 <- c(min(y2),max(y2))    
  yf <- Cast2To1(y2)
  labelsyf <- pretty(y2)  
  return(list(
    yf=yf,
    labels=labelsyf,
    breaks=Cast2To1(labelsyf)
  ))
}

什么产生了一些：

> FudgeAxis <- CalcFudgeAxis( df$y1, df$y2 )

> FudgeAxis
$yf
[1] -0.4094344  4.6831656  0.4029175  1.0034664 -0.1009335

$labels
[1] -50   0  50 100 150 200 250

$breaks
[1] -1.068764  0.000000  1.068764  2.137529  3.206293  4.275058  5.343822


> cbind(df, FudgeAxis$yf)
  item      y1         y2 FudgeAxis$yf
1    A -0.8684 -19.154567   -0.4094344
2    B  4.2242 219.092499    4.6831656
3    C -0.3181  18.849686    0.4029175
4    D  0.5797  46.945161    1.0034664
5    E -0.4875  -4.721973   -0.1009335

现在我在第二个辅助函数 PlotWithFudgeAxis 中包含 Kohske的解决方案（我们将新ggplot对象和新轴的辅助对象抛入其中）：< / p>

library(gtable)
library(grid)

PlotWithFudgeAxis = function( plot1, FudgeAxis) {
  # based on: https://rpubs.com/kohske/dual_axis_in_ggplot2
  plot2 <- plot1 + with(FudgeAxis, scale_y_continuous( breaks=breaks, labels=labels))

  #extract gtable
  g1<-ggplot_gtable(ggplot_build(plot1))
  g2<-ggplot_gtable(ggplot_build(plot2))

  #overlap the panel of the 2nd plot on that of the 1st plot
  pp<-c(subset(g1$layout, name=="panel", se=t:r))
  g<-gtable_add_grob(g1, g2$grobs[[which(g2$layout$name=="panel")]], pp$t, pp$l, pp$b,pp$l)

  ia <- which(g2$layout$name == "axis-l")
  ga <- g2$grobs[[ia]]
  ax <- ga$children[[2]]
  ax$widths <- rev(ax$widths)
  ax$grobs <- rev(ax$grobs)
  ax$grobs[[1]]$x <- ax$grobs[[1]]$x - unit(1, "npc") + unit(0.15, "cm")
  g <- gtable_add_cols(g, g2$widths[g2$layout[ia, ]$l], length(g$widths) - 1)
  g <- gtable_add_grob(g, ax, pp$t, length(g$widths) - 1, pp$b)

  grid.draw(g)
}

现在可以将所有内容组合在一起：下面的代码显示，建议的解决方案如何在日常环境中使用。情节调用现在不再绘制原始数据 y2 ，而是克隆版本 yf （保存在预先计算的辅助对象 FudgeAxis 中），运行 y1 的比例。然后使用 Kohske 辅助函数 PlotWithFudgeAxis 操作原始ggplot对象，以添加第二个轴，保留 y2 的比例。它也描绘了操纵的情节。

FudgeAxis <- CalcFudgeAxis( df$y1, df$y2 )

tmpPlot <- ggplot(data=df, aes(label=item)) +
      theme_bw() + 
      geom_segment(aes(x='G1', xend='G2', y=y1, yend=FudgeAxis$yf), color='grey')+
      geom_text(aes(x='G1', y=y1), color='blue') +
      geom_text(aes(x='G2', y=FudgeAxis$yf), color='red') +
      theme(legend.position='none', panel.grid=element_blank())

PlotWithFudgeAxis(tmpPlot, FudgeAxis)

现在根据需要绘制两个轴，左边是 y1 ，右边是 y2

以上解决方案是，直截了当，一个有限的摇摇欲坠的黑客。当它与ggplot内核一起使用时，它会抛出一些警告，我们交换事后的比例等等。它必须小心处理，并可能在另一个设置中产生一些不希望的行为。同样，可能需要使用辅助函数来调整以获得所需的布局。图例的位置是一个问题（它将放置在面板和新轴之间;这就是我下垂的原因）。 2轴的缩放/对齐也有点挑战性：当两个刻度包含＆＃34; 0＆＃34;时，上面的代码很好地工作，否则一个轴会移位。所以肯定有一些改进的机会...

如果想要保存pic，必须将呼叫包装到设备打开/关闭中：

png(...)
PlotWithFudgeAxis(tmpPlot, FudgeAxis)
dev.off()

Answer 6

有一些常见的用例对决轴，例如climatograph显示每月的温度和降水量。这是一个简单的解决方案，从威震天的解决方案中推广而来，它允许您将变量的下限设置为零以外的其他值：

示例数据：

climate <- tibble(
  Month = 1:12,
  Temp = c(-4,-4,0,5,11,15,16,15,11,6,1,-3),
  Precip = c(49,36,47,41,53,65,81,89,90,84,73,55)
  )

手动设置每个轴的限制：

ylim.prim <- c(0, 180)   # in this example, precipitation
ylim.sec <- c(-4, 18)    # in this example, temperature

以下内容基于这些限制进行必要的计算，并绘制出图本身：

b <- diff(ylim.prim)/diff(ylim.sec)
a <- b*(ylim.prim[1] - ylim.sec[1])

ggplot(climate, aes(Month, Precip)) +
  geom_col() +
  geom_line(aes(y = a + Temp*b), color = "red") +
  scale_y_continuous("Precipitation", sec.axis = sec_axis(~ (. - a)/b, name = "Temperature")) +
  scale_x_continuous("Month", breaks = 1:12) +
  ggtitle("Climatogram for Oslo (1961-1990)")

如果要确保红线与右侧的y轴相对应，可以在代码中添加theme句子：

ggplot(climate, aes(Month, Precip)) +
  geom_col() +
  geom_line(aes(y = a + Temp*b), color = "red") +
  scale_y_continuous("Precipitation", sec.axis = sec_axis(~ (. - a)/b, name = "Temperature")) +
  scale_x_continuous("Month", breaks = 1:12) +
  theme(axis.line.y.right = element_line(color = "red"), 
        axis.ticks.y.right = element_line(color = "red"),
        axis.text.y.right = element_text(color = "red"), 
        axis.title.y.right = element_text(color = "red")
        ) +
  ggtitle("Climatogram for Oslo (1961-1990)")

为右轴着色：

Answer 7

以下文章帮助我将ggplot2生成的两个图组合在一行：

Multiple graphs on one page (ggplot2) by Cookbook for R

以下是代码在这种情况下的样子：

p1 <- 
  ggplot() + aes(mns)+ geom_histogram(aes(y=..density..), binwidth=0.01, colour="black", fill="white") + geom_vline(aes(xintercept=mean(mns, na.rm=T)), color="red", linetype="dashed", size=1) +  geom_density(alpha=.2)

p2 <- 
  ggplot() + aes(mns)+ geom_histogram( binwidth=0.01, colour="black", fill="white") + geom_vline(aes(xintercept=mean(mns, na.rm=T)), color="red", linetype="dashed", size=1)  

multiplot(p1,p2,cols=2)

Answer 8

对我来说，棘手的部分是弄清楚两个轴之间的转换功能。我使用了myCurveFit。

> dput(combined_80_8192 %>% filter (time > 270, time < 280))
structure(list(run = c(268L, 268L, 268L, 268L, 268L, 268L, 268L, 
268L, 268L, 268L, 263L, 263L, 263L, 263L, 263L, 263L, 263L, 263L, 
263L, 263L, 269L, 269L, 269L, 269L, 269L, 269L, 269L, 269L, 269L, 
269L, 261L, 261L, 261L, 261L, 261L, 261L, 261L, 261L, 261L, 261L, 
267L, 267L, 267L, 267L, 267L, 267L, 267L, 267L, 267L, 267L, 265L, 
265L, 265L, 265L, 265L, 265L, 265L, 265L, 265L, 265L, 266L, 266L, 
266L, 266L, 266L, 266L, 266L, 266L, 266L, 266L, 262L, 262L, 262L, 
262L, 262L, 262L, 262L, 262L, 262L, 262L, 264L, 264L, 264L, 264L, 
264L, 264L, 264L, 264L, 264L, 264L, 260L, 260L, 260L, 260L, 260L, 
260L, 260L, 260L, 260L, 260L), repetition = c(8L, 8L, 8L, 8L, 
8L, 8L, 8L, 8L, 8L, 8L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 
9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 5L, 5L, 
5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 
6L, 6L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 4L, 4L, 4L, 4L, 
4L, 4L, 4L, 4L, 4L, 4L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L
), module = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = "scenario.node[0].nicVLCTail.phyVLC", class = "factor"), 
    configname = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    1L, 1L), .Label = "Road-Vlc", class = "factor"), packetByteLength = c(8192L, 
    8192L, 8192L, 8192L, 8192L, 8192L, 8192L, 8192L, 8192L, 8192L, 
    8192L, 8192L, 8192L, 8192L, 8192L, 8192L, 8192L, 8192L, 8192L, 
    8192L, 8192L, 8192L, 8192L, 8192L, 8192L, 8192L, 8192L, 8192L, 
    8192L, 8192L, 8192L, 8192L, 8192L, 8192L, 8192L, 8192L, 8192L, 
    8192L, 8192L, 8192L, 8192L, 8192L, 8192L, 8192L, 8192L, 8192L, 
    8192L, 8192L, 8192L, 8192L, 8192L, 8192L, 8192L, 8192L, 8192L, 
    8192L, 8192L, 8192L, 8192L, 8192L, 8192L, 8192L, 8192L, 8192L, 
    8192L, 8192L, 8192L, 8192L, 8192L, 8192L, 8192L, 8192L, 8192L, 
    8192L, 8192L, 8192L, 8192L, 8192L, 8192L, 8192L, 8192L, 8192L, 
    8192L, 8192L, 8192L, 8192L, 8192L, 8192L, 8192L, 8192L, 8192L, 
    8192L, 8192L, 8192L, 8192L, 8192L, 8192L, 8192L, 8192L, 8192L
    ), numVehicles = c(2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
    2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
    2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
    2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
    2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
    2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
    2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L
    ), dDistance = c(80L, 80L, 80L, 80L, 80L, 80L, 80L, 80L, 
    80L, 80L, 80L, 80L, 80L, 80L, 80L, 80L, 80L, 80L, 80L, 80L, 
    80L, 80L, 80L, 80L, 80L, 80L, 80L, 80L, 80L, 80L, 80L, 80L, 
    80L, 80L, 80L, 80L, 80L, 80L, 80L, 80L, 80L, 80L, 80L, 80L, 
    80L, 80L, 80L, 80L, 80L, 80L, 80L, 80L, 80L, 80L, 80L, 80L, 
    80L, 80L, 80L, 80L, 80L, 80L, 80L, 80L, 80L, 80L, 80L, 80L, 
    80L, 80L, 80L, 80L, 80L, 80L, 80L, 80L, 80L, 80L, 80L, 80L, 
    80L, 80L, 80L, 80L, 80L, 80L, 80L, 80L, 80L, 80L, 80L, 80L, 
    80L, 80L, 80L, 80L, 80L, 80L, 80L, 80L), time = c(270.166006903445, 
    271.173853699836, 272.175873251122, 273.177524313334, 274.182946177105, 
    275.188959464989, 276.189675339937, 277.198250244799, 278.204619457189, 
    279.212562800009, 270.164199199177, 271.168527215152, 272.173072994958, 
    273.179210429715, 274.184351047337, 275.18980754378, 276.194816792995, 
    277.198598277809, 278.202398083519, 279.210634593917, 270.210674322891, 
    271.212395107473, 272.218871923292, 273.219060500457, 274.220486359614, 
    275.22401452372, 276.229646658839, 277.231060448138, 278.240407241942, 
    279.2437126347, 270.283554249858, 271.293168593832, 272.298574288769, 
    273.304413221348, 274.306272082517, 275.309023049011, 276.317805897347, 
    277.324403550028, 278.332855848701, 279.334046374594, 270.118608539613, 
    271.127947700074, 272.133887145863, 273.135726000491, 274.135994529981, 
    275.136563912708, 276.140120735361, 277.144298344151, 278.146885137621, 
    279.147552358659, 270.206015567272, 271.214618077209, 272.216566814903, 
    273.225435592582, 274.234014573683, 275.242949179958, 276.248417809711, 
    277.248800670023, 278.249750333404, 279.252926560188, 270.217182684494, 
    271.218357511397, 272.224698488895, 273.231112784327, 274.238740508457, 
    275.242715184122, 276.249053562718, 277.250325509798, 278.258488063493, 
    279.261141590137, 270.282904173953, 271.284689544638, 272.294220723234, 
    273.299749415592, 274.30628880553, 275.312075103126, 276.31579134717, 
    277.321905523606, 278.326305136748, 279.333056502253, 270.258991527456, 
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    3.593647329558e-20, 1.9750692814982e-12, 0.00019705170257492, 
    1.9748966344895e-06, 1.7515881895994e-12, 2.1868296425817e-06, 
    1.8649940680806e-06, 8.7517439682173e-24, 4.3621551072316e-27, 
    2.553168170837e-28, 3.6469582463164e-20, 1.0032983660212e-15, 
    0.00019385229409318, 1.9830820164805e-06, 1.7760568361323e-12, 
    2.919419915209e-05, 1.8741284335866e-06, 2.8285944348148e-25, 
    4.1960751547207e-27, 7.8468215407139e-29, 8.0407329049747e-16, 
    1.9380328071065e-12, 0.00020004849911333, 1.9393279417733e-06, 
    5.9354475879597e-10, 6.4258355913627e-10, 2.6065221215415e-05
    )), class = "data.frame", row.names = c(NA, -100L), .Names = c("run", 
"repetition", "module", "configname", "packetByteLength", "numVehicles", 
"dDistance", "time", "distanceToTx", "headerNoError", "receivedPower_dbm", 
"snr", "frameId", "packetOkSinr", "snir", "ookSnirBer", "ookSnrBer"
))

查找转换功能

y1 - ＆gt; Y2 该函数用于将次y轴的数据转换为根据第一y轴“标准化”

转换功能：f(y1) = 0.025*x + 2.75

y2 - ＆gt; Y1 此函数用于将第一个y轴的断点转换为第二个y轴的值。请注意，轴现在已交换。

转换功能：f(y1) = 40*x - 110

<强>绘图

注意如何在ggplot调用中使用转换函数来“即时”转换数据

ggplot(data=combined_80_8192 %>% filter (time > 270, time < 280), aes(x=time) ) +
  stat_summary(aes(y=receivedPower_dbm ), fun.y=mean, geom="line", colour="black") +
  stat_summary(aes(y=packetOkSinr*40 - 110 ), fun.y=mean, geom="line", colour="black", position = position_dodge(width=10)) +
  scale_x_continuous() +
  scale_y_continuous(breaks = seq(-0,-110,-10), "y_first", sec.axis=sec_axis(~.*0.025+2.75, name="y_second") )

第一个stat_summary调用是为第一个y轴设置基础的调用。调用第二个stat_summary调用来转换数据。请记住，所有数据都将作为第一个y轴的基础。因此，需要针对第一个y轴对数据进行归一化。为此，我在数据上使用转换函数：y=packetOkSinr*40 - 110

现在转换第二个轴我在scale_y_continuous调用中使用相反的函数：sec.axis=sec_axis(~.*0.025+2.75, name="y_second")。

Answer 9

您可以创建一个缩放系数，该缩放系数将应用于第二个几何图形和右y轴。这源自塞巴斯蒂安的解决方案。

library(ggplot2)

scaleFactor <- max(mtcars$cyl) / max(mtcars$hp)

ggplot(mtcars, aes(x=disp)) +
  geom_smooth(aes(y=cyl), method="loess", col="blue") +
  geom_smooth(aes(y=hp * scaleFactor), method="loess", col="red") +
  scale_y_continuous(name="cyl", sec.axis=sec_axis(~./scaleFactor, name="hp")) +
  theme(
    axis.title.y.left=element_text(color="blue"),
    axis.text.y.left=element_text(color="blue"),
    axis.title.y.right=element_text(color="red"),
    axis.text.y.right=element_text(color="red")
  )

注意：使用ggplot2 v3.0.0

Answer 10

我们绝对可以使用基本R函数plot构建具有双Y轴的图。

# pseudo dataset
df <- data.frame(x = seq(1, 1000, 1), y1 = sample.int(100, 1000, replace=T), y2 = sample(50, 1000, replace = T))

# plot first plot 
with(df, plot(y1 ~ x, col = "red"))

# set new plot
par(new = T) 

# plot second plot, but without axis
with(df, plot(y2 ~ x, type = "l", xaxt = "n", yaxt = "n", xlab = "", ylab = ""))

# define y-axis and put y-labs
axis(4)
with(df, mtext("y2", side = 4))

Answer 11

这是我关于如何对辅助轴进行转换的两分钱。首先，您希望将主要数据和次要数据的范围结合起来。就用您不想要的变量污染您的全局环境而言，这通常很麻烦。

为了使这更容易，我们将创建一个函数工厂来生成两个函数，其中 scales::rescale() 完成所有繁重的工作。因为这些是闭包，所以它们知道创建它们的环境，因此它们对创建之前生成的 to 和 from 参数有“记忆”。

一个函数进行前向转换：将辅助数据转换为主要数据。
第二个函数执行反向转换：将主要单位的数据转换为次要单位。

library(ggplot2)
library(scales)

# Function factory for secondary axis transforms
train_sec <- function(primary, secondary) {
  from <- range(secondary)
  to   <- range(primary)
  # Forward transform for the data
  forward <- function(x) {
    rescale(x, from = from, to = to)
  }
  # Reverse transform for the secondary axis
  reverse <- function(x) {
    rescale(x, from = to, to = from)
  }
  list(fwd = forward, rev = reverse)
}

这看起来都相当复杂，但是让函数工厂变得更容易。现在，在绘制绘图之前，我们将通过向工厂显示主要和次要数据来生成相关函数。我们将使用经济数据集，其中 unemploy 和 psavert 列的范围非常不同。

sec <- with(economics, train_sec(unemploy, psavert))

然后我们使用 y = sec$fwd(psavert) 将辅助数据重新缩放到主轴，并指定 ~ sec$rev(.) 作为辅助轴的转换参数。这为我们提供了一个图，其中主要和次要范围在图上占据相同的空间。

ggplot(economics, aes(date)) +
  geom_line(aes(y = unemploy), colour = "blue") +
  geom_line(aes(y = sec$fwd(psavert)), colour = "red") +
  scale_y_continuous(sec.axis = sec_axis(~sec$rev(.), name = "psavert"))

工厂比这稍微灵活一些，因为如果你只是想重新调整最大值，你可以传入下限为 0 的数据。

# Rescaling the maximum
sec <- with(economics, train_sec(c(0, max(unemploy)),
                                 c(0, max(psavert))))

ggplot(economics, aes(date)) +
  geom_line(aes(y = unemploy), colour = "blue") +
  geom_line(aes(y = sec$fwd(psavert)), colour = "red") +
  scale_y_continuous(sec.axis = sec_axis(~sec$rev(.), name = "psavert"))

^{由 reprex package (v0.3.0) 于 2021 年 2 月 5 日创建}

我承认这个例子中的区别不是很明显，但是如果你仔细观察，你会发现最大值是一样的，而且红线比蓝线低。

Answer 12

我承认并同意hadley（以及其他人），单独的y尺度“基本上是有缺陷的”。话虽如此 - 我经常希望ggplot2具有此功能 - 特别是当数据位于wide-format时，我很快想要查看或检查数据（即仅供个人使用）。

虽然tidyverse库可以很容易地将数据转换为长格式（这样facet_grid()会起作用），但这个过程仍然不是很简单，如下所示：

library(tidyverse)
df.wide %>%
    # Select only the columns you need for the plot.
    select(date, column1, column2, column3) %>%
    # Create an id column – needed in the `gather()` function.
    mutate(id = n()) %>%
    # The `gather()` function converts to long-format. 
    # In which the `type` column will contain three factors (column1, column2, column3),
    # and the `value` column will contain the respective values.
    # All the while we retain the `id` and `date` columns.
    gather(type, value, -id, -date) %>%
    # Create the plot according to your specifications
    ggplot(aes(x = date, y = value)) +
        geom_line() +
        # Create a panel for each `type` (ie. column1, column2, column3).
        # If the types have different scales, you can use the `scales="free"` option.
        facet_grid(type~., scales = "free")

Answer 13

您可以对变量使用facet_wrap(~ variable, ncol= )来创建新的比较。它不在同一轴上，但它是相似的。

Answer 14

以下结合了Dag Hjermann的基础数据和编程，改进了user4786271的策略，创建了一个“转换函数”，以优化组合绘图和数据轴，并响应{{3 }} 注意，这样的函数可以在 R 中创建。

#Climatogram for Oslo (1961-1990)
climate <- tibble(
  Month = 1:12,
  Temp = c(-4,-4,0,5,11,15,16,15,11,6,1,-3),
  Precip = c(49,36,47,41,53,65,81,89,90,84,73,55))

#y1 identifies the position, relative to the y1 axis, 
#the locations of the minimum and maximum of the y2 graph.
#Usually this will be the min and max of y1.
#y1<-(c(max(climate$Precip), 0))
#y1<-(c(150, 55))
y1<-(c(max(climate$Precip), min(climate$Precip)))

#y2 is the Minimum and maximum of the secondary axis data.
y2<-(c(max(climate$Temp), min(climate$Temp)))

#axis combines y1 and y2 into a dataframe used for regressions.
axis<-cbind(y1,y2)
axis<-data.frame(axis)

#Regression of Temperature to Precipitation:
T2P<-lm(formula = y1 ~ y2, data = axis)
T2P_summary <- summary(lm(formula = y1 ~ y2, data = axis))
T2P_summary   

#Identifies the intercept and slope of regressing Temperature to Precipitation:
T2PInt<-T2P_summary$coefficients[1, 1] 
T2PSlope<-T2P_summary$coefficients[2, 1] 


#Regression of Precipitation to Temperature:
P2T<-lm(formula = y2 ~ y1, data = axis)
P2T_summary <- summary(lm(formula = y2 ~ y1, data = axis))
P2T_summary   

#Identifies the intercept and slope of regressing Precipitation to Temperature:
P2TInt<-P2T_summary$coefficients[1, 1] 
P2TSlope<-P2T_summary$coefficients[2, 1] 


#Create Plot:
ggplot(climate, aes(Month, Precip)) +
  geom_col() +
  geom_line(aes(y = T2PSlope*Temp + T2PInt), color = "red") +
  scale_y_continuous("Precipitation", sec.axis = sec_axis(~.*P2TSlope + P2TInt, name = "Temperature")) +
  scale_x_continuous("Month", breaks = 1:12) +
  theme(axis.line.y.right = element_line(color = "red"), 
        axis.ticks.y.right = element_line(color = "red"),
        axis.text.y.right = element_text(color = "red"), 
        axis.title.y.right = element_text(color = "red")) +
  ggtitle("Climatogram for Oslo (1961-1990)")

最值得注意的是，一个新的“转换函数”在每个轴的数据集中只有两个数据点时效果更好——通常是每个集的最大值和最小值。两个回归的结果斜率和截距使 ggplot2 能够精确配对每个轴的最小值和最大值的图。正如 baptist 指出的那样，这两个回归将每个数据集和绘图转换为另一个。一个将第一个 y 轴的断点转换为第二个 y 轴的值。第二个将第二个 y 轴的数据转换为根据第一个 y 轴进行“归一化”。以下输出显示了轴如何对齐每个数据集的最小值和最大值：

user4786271

让最大值和最小值匹配可能是最合适的；然而，这种方法的另一个好处是，如果需要，可以通过更改与主轴数据相关的编程线轻松移动与辅助轴相关的绘图。下面的输出只是将 y1 编程行中的最小降水输入更改为“0”，从而将最小温度水平与“0”降水水平对齐。

来自：y1<-(c(max(climate$Precip), min(climate$Precip)))

至：y1<-(c(max(climate$Precip), 0))

请注意由此产生的新回归和 ggplot2 如何自动调整绘图和轴，以正确地将最低温度与“0”降水水平的新“基础”对齐。同样，可以很容易地提升温度图，使其更加明显。下图是通过简单地将上述行更改为：

"y1<-(c(150, 55))"

上面的线告诉温度图的最大值与“150”降水水平重合，温度线的最小值与“55”降水水平重合。再次注意 ggplot2 和由此产生的新回归输出如何使图形与轴保持正确对齐。

以上可能不是理想的输出；然而，它是一个示例，说明如何可以轻松地操作图形并在图和轴之间仍然具有正确的关系。主题的加入提高了对与情节对应的轴的识别。

Dag Hjermann

Answer 15

The answer by Hadley对Stephen Few的报告Dual-Scaled Axes in Graphs Are They Ever the Best Solution?提供了一个有趣的参考。

我不知道OP的含义是“计数”和“费率”，但快速搜索给了我Counts and Rates，所以我得到一些关于北美登山事故¹的数据：

Years<-c("1998","1999","2000","2001","2002","2003","2004")
Persons.Involved<-c(281,248,301,276,295,231,311)
Fatalities<-c(20,17,24,16,34,18,35)
rate=100*Fatalities/Persons.Involved
df<-data.frame(Years=Years,Persons.Involved=Persons.Involved,Fatalities=Fatalities,rate=rate)
print(df,row.names = FALSE)

 Years Persons.Involved Fatalities      rate
  1998              281         20  7.117438
  1999              248         17  6.854839
  2000              301         24  7.973422
  2001              276         16  5.797101
  2002              295         34 11.525424
  2003              231         18  7.792208
  2004              311         35 11.254019

然后我尝试按照前面提到的报告第7页提出的图表（并按照OP的要求将计数图表绘制成条形图并将费率作为折线图）：

另一个不太明显的解决方案，仅适用于时间序列，是将所有值集转换为通用的量化标准显示每个值与参考之间的百分比差异（或索引）值。例如，选择一个特定的时间点，例如图中出现的第一个区间，并表示每个后续值作为它与之间的百分比差异初始值。这是通过除以每个点的值来完成的时间乘以初始时间点的值，然后乘以它将100转换为百分比，如下图所示。

df2<-df
df2$Persons.Involved <- 100*df$Persons.Involved/df$Persons.Involved[1]
df2$rate <- 100*df$rate/df$rate[1]
plot(ggplot(df2)+
  geom_bar(aes(x=Years,weight=Persons.Involved))+
  geom_line(aes(x=Years,y=rate,group=1))+
  theme(text = element_text(size=30))
  )

这就是结果：

但是我不喜欢它并且我不能轻易地把它放在它上面......

1 <子> WILLIAMSON，Jed，et al。 2005年北美登山事故。 The Mountaineers Books，2005。

Answer 16

这似乎是一个简单的问题，但它围绕着两个基本问题感到困惑。 A）如何在比较图中显示时如何处理多标量数据；其次，B）是否可以在不使用R编程的一些经验法则的情况下完成此操作，例如i）合并数据，ii）分面，iii）添加现有的另一层。下面给出的解决方案满足了以上条件，因为它无需重新缩放即可处理数据，其次，不使用所提到的技术。

这是结果，

对于有兴趣更多地了解这种方法的人，请点击下面的链接。 How to plot a 2- y axis chart with bars side by side without re-scaling the data

Answer 17

我发现这个answer对我的帮助最大，但是发现有些边缘情况似乎无法正确处理，特别是负面情况，还有我的极限距离为0的情况（如果我们要从最大/最小数据中获取限制，可能会发生这种情况。测试似乎表明此方法始终有效

我使用以下代码。在这里，我假设我们要转换为[y1，y2]的[x1，x2]。我处理此问题的方法是将[x1，x2]转换为[0,1]（足够简单的转换），然后将[0,1]转换为[y1，y2]。

climate <- tibble(
  Month = 1:12,
  Temp = c(-4,-4,0,5,11,15,16,15,11,6,1,-3),
  Precip = c(49,36,47,41,53,65,81,89,90,84,73,55)
)
#Set the limits of each axis manually:

  ylim.prim <- c(0, 180)   # in this example, precipitation
ylim.sec <- c(-4, 18)    # in this example, temperature



  b <- diff(ylim.sec)/diff(ylim.prim)

#If all values are the same this messes up the transformation, so we need to modify it here
if(b==0){
  ylim.sec <- c(ylim.sec[1]-1, ylim.sec[2]+1)
  b <- diff(ylim.sec)/diff(ylim.prim)
}
if (is.na(b)){
  ylim.prim <- c(ylim.prim[1]-1, ylim.prim[2]+1)
  b <- diff(ylim.sec)/diff(ylim.prim)
}


ggplot(climate, aes(Month, Precip)) +
  geom_col() +
  geom_line(aes(y = ylim.prim[1]+(Temp-ylim.sec[1])/b), color = "red") +
  scale_y_continuous("Precipitation", sec.axis = sec_axis(~((.-ylim.prim[1]) *b  + ylim.sec[1]), name = "Temperature"), limits = ylim.prim) +
  scale_x_continuous("Month", breaks = 1:12) +
  ggtitle("Climatogram for Oslo (1961-1990)")

这里的关键部分是我们用~((.-ylim.prim[1]) *b + ylim.sec[1])变换辅助y轴，然后将反值应用于实际值y = ylim.prim[1]+(Temp-ylim.sec[1])/b)。我们还应确保limits = ylim.prim。

ggplot，每侧有2个y轴，不同的刻度

17 个答案: