以下内容结合了Dag Hjermann的基本数据和编程，改进了user4786271创建“转换函数”的策略，以优化组合图和数据轴，并响应了浸信会的提示，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)")

Most noteworthy is that a new "transformation function" works better with just two data points from the data set of each axes—usually the maximum and minimum values of each set. The resulting slopes and intercepts of the two regressions enable ggplot2 to exactly pair the plots of the minimums and maximums of each axis. As user4786271 pointed out, the two regressions transform each data set and plot to the other. One transforms the break points of the first y axis to the values of the second y axis. The second transforms the data of the secondary y axis to be "normalized" according to the first y axis. The following output shows how the axis align the minimums and maximums of each dataset:

使最大值和最小值匹配可能是最合适的;但是，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行轻松地移动与次要轴相关的绘图。下面的输出只是将y1编程行中输入的最小降水量更改为“0”，从而将最小温度水平与“0”降水水平对齐。

从:y1<-(c(max(气候$ precp)， min(气候$ precp)))

到:y1<-(c(max(气候$ precp)， 0))

请注意，生成的新回归和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“基数”对齐。同样，可以很容易地提升Temperature图，使其更加明显。下面的图是通过简单地将上面提到的线更改为:

“日元<——(c(150年,55岁))”

上面的线表示温度曲线的最大值与“150”降水水平相吻合，温度曲线的最小值与“55”降水水平相吻合。再次注意，ggplot2和由此产生的新的回归输出如何使图保持与轴的正确对齐。

以上可能不是理想的输出;然而，这是一个例子，说明了如何容易地操纵图形，并且在图和轴之间仍然有正确的关系。 Dag Hjermann的主题的结合提高了与情节对应的轴的识别。

Hadley的回答参考了Stephen Few的报告《双缩放轴在图中是最好的解决方案吗?》

我不知道OP中的“counts”和“rate”是什么意思，但快速搜索会给我counts和Rates，所以我得到了一些关于北美登山事故的数据:

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Few在上述报告第7页建议的那样绘制图表(并按照OP的要求将计数绘制为柱状图，将率绘制为折线图):

The other less obvious solution, which works only for time series, is to convert all sets of values to a common quantitative scale by displaying percentage differences between each value and a reference (or index) value. For instance, select a particular point in time, such as the first interval that appears in the graph, and express each subsequent value as the percentage difference between it and the initial value. This is done by dividing the value at each point in time by the value for the initial point in time and then multiplying it by 100 to convert the rate to a percentage, as illustrated below.

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:

但我不是很喜欢它，我不能轻易地给它加上一个传奇……

1 威廉森，杰德，等人。2005年北美登山事故。The Mountaineers Books, 2005。

总有办法的。

这里有一个解决方案，允许完全任意轴而不重新缩放。其思想是生成两个除了轴以外完全相同的图，并使用cowplot包中的insert_yaxis_grob和get_y_axis函数将它们组合在一起。

library(ggplot2)
library(cowplot)

## first plot 
p1 <- ggplot(mtcars,aes(disp,hp,color=as.factor(am))) + 
    geom_point() + theme_bw() + theme(legend.position='top', text=element_text(size=16)) +
    ylab("Horse points" )+ xlab("Display size") + scale_color_discrete(name='Transmitter') +
    stat_smooth(se=F)

## same plot with different, arbitrary scale   
p2 <- p1 +
    scale_y_continuous(position='right',breaks=seq(120,173,length.out = 3),
                       labels=c('little','medium little','medium hefty'))

ggdraw(insert_yaxis_grob(p1,get_y_axis(p2,position='right')))

我发现这个答案对我帮助最大，但发现有一些边缘情况，它似乎不能正确处理，特别是消极的情况，以及极限距离为0的情况(如果我们从最大/最小数据中获取极限，就会发生这种情况)。测试似乎表明，这是一致的

我使用以下代码。这里我假设我们有[x1,x2]我们想把它变换成[y1,y2]。我处理这个问题的方法是将[x1,x2]转换为[0,1](一个足够简单的转换)，然后[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])转换次要y轴，然后对实际值y = ylim.prim[1]+(Temp-ylim.sec[1])/b)应用逆。我们还应该确保limits = ylim.prim。

It seemingly appears to be a simple question but it boggles around 2 fundamental questions. A) How to deal with a multi-scalar data while presenting in a comparative chart, and secondly, B) whether this can be done without some thumb rule practices of R programming such as i) melting data, ii) faceting, iii) adding another layer to existing one. The solution given below satisfies both the above conditions as it deals data without having to rescale it and secondly, the techniques mentioned are not used.

这是结果，

如果有兴趣了解更多关于此方法的信息，请点击下面的链接。 如何绘制一个2 y轴图表与条形并排而不重新缩放数据

根据上面的答案和一些微调(无论它有什么价值)，这里有一种通过sec_axis实现两个尺度的方法:

假设有一个简单的(完全虚构的)数据集dt:在五天的时间里，它追踪了被打断的次数VS工作效率:

        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

(两列的范围相差大约5倍)。

下面的代码将画出它们占用整个y轴的两个级数:

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"))

下面是结果(上面的代码+一些颜色调整):

重点(除了在指定y_scale时使用sec_axis之外)是在指定系列时将第二个数据系列的每个值与5相乘。为了在sec_axis定义中获得正确的标签，它需要除以5(并格式化)。因此，上述代码中的关键部分实际上是geom_line和~中的*5。sec_axis中的/5(一个除当前值的公式。5)。

相比之下(我不想在这里判断方法)，这是两个图表叠加在一起的样子:

你可以自己判断哪一个能更好地传递信息(“不要打扰别人工作!”)。我想这是一个公平的决定方式。

这两个图像的完整代码(实际上并没有比上面更多，只是完成并准备运行)在这里:https://gist.github.com/sebastianrothbucher/de847063f32fdff02c83b75f59c36a7d更详细的解释在这里:https://sebastianrothbucher.github.io/datascience/r/visualization/ggplot/2018/03/24/two-scales-ggplot-r.html