Question

尽管这个问题已经“解决”了很多次，但事实证明总是存在另一个问题。

没有打印功能，它可以正常运行，但是得到以下信息：

Error in .subset2(x, i) : recursive indexing failed at level 2

我要说的是它不喜欢在两层迭代中创建图形吗？将方法更改为“ qplot（whatever：whatever）”存在完全相同的问题。

它旨在为我正在查看的每对变量打印一个图形。他们太多了，无法容纳单张图片，例如pairs函数，我需要能够在轴上看到实际的变量名称。

load("Transport_Survey.RData")

variables <- select(Transport, "InfOfReceievingWeather", "InfOfReceievingTraffic", "InfOfSeeingTraffic", "InfWeather.Ice", "InfWeather.Rain", "InfWeather.Wind", "InfWeather.Storm", "InfWeather.Snow", "InfWeather.Cold", "InfWeather.Warm", "InfWeather.DarkMorn", "InfWeather.DarkEve", "HomeParking", "WorkParking", "Disability", "Age", "CommuteFlexibility", "Gender", "PassionReduceCongest")
varnames <- list("InfOfReceivingWeather", "InfOfReceivingTraffic", "InfOfSeeingTraffic", "InfWeather.Ice", "InfWeather.Rain", "InfWeather.Wind", "InfWeather.Storm", "InfWeather.Snow", "InfWeather.Cold", "InfWeather.Warm", "InfWeather.DarkMorn", "InfWeather.DarkEve", "HomeParking", "WorkParking", "Disability", "Age", "CommuteFlexibility", "Gender", "PassionReduceCongest")

counterx = 1
countery = 1

for (a in variables) {
  for (b in variables) {
    print(ggplot(variables, mapping=aes(x=variables[[a]], y=variables[[b]],
    xlab=varnames[counterx], ylab=varnames[countery]))+
      geom_point())
    countery = countery+1
  counterx = counterx+1
  }
}

#variables2 <- select(Transport, one_of(InfOfReceivingWeather, InfOfReceivingTraffic, InfOfSeeingTraffic, InfWeather.Ice, InfWeather.Rain, InfWeather.Wind, InfWeather.Storm, InfWeather.Snow, InfWeather.Cold, InfWeather.Warm, InfWeather.DarkMorn, InfWeather.DarkEve, HomeParking, WorkParking, Disability, Age, CommuteFlexibility, Gender, PassionReduceCongest))

这是一个供参考的小型数据框，是从我使用的列中采样的：

structure(list(InfOfReceievingWeather = c(1, 1, 1, 1, 4), InfOfReceievingTraffic = c(1, 
1, 1, 1, 4), InfOfSeeingTraffic = c(1, 1, 1, 1, 4), InfWeather.Ice = c(3, 
1, 3, 5, 5), InfWeather.Rain = c(1, 1, 2, 2, 4), InfWeather.Wind = c(1, 
1, 2, 2, 4), InfWeather.Storm = c(1, 1, 1, 2, 5), InfWeather.Snow = c(1, 
1, 2, 5, 5), InfWeather.Cold = c(1, 1, 1, 2, 5), InfWeather.Warm = c(1, 
1, 1, 1, 3), InfWeather.DarkMorn = c(1, 1, 1, 1, 1), InfWeather.DarkEve = c(1, 
1, 1, 1, 1), HomeParking = c(1, 1, 3, 1, 1), WorkParking = c(1, 
4, 4, 5, 4), Disability = c(1, 1, 1, 1, 1), Age = c(19, 45, 35, 
40, 58), CommuteFlexibility = c(2, 1, 5, 1, 2), Gender = c(2, 
2, 2, 2, 1), PassionReduceCongest = c(0, 0, 2, 0, 2)), row.names = c(NA, 
-5L), class = c("tbl_df", "tbl", "data.frame"))

Answer 1

您在分配b和a时出错。基本上，当在b中定义variables和variables时，它们成为aes列中包含的值的向量。因此，在您的variables[[a]]映射中，当您调用a时，基本上就是在写（对于variables中variables[[c(1, 1, 1, 1, 4)]]的第一次迭代）： variables[["InfOfReceievingWeather"]]，而不是for (a in variables) { for (b in variables) { print(ggplot(variables, mapping=aes(x=a, y=b)) ...。因此，它不起作用。

要解决此问题，您必须选择以下一项：

for (a in 1:ncol(variables)) {
  for (b in 1:ncol(variables)) {
    print(ggplot(variables, mapping=aes(x=variables[[a]], y=variables[[b]])) ...

或

尽管第一个选项似乎更简单，但我宁愿选择第二个选项，因为它可以循环使用b和variables作为列指示符来提取xlab的同名ylab和for (a in 1:ncol(variables)) { for (b in 1:ncol(variables)) { print(ggplot(variables, mapping=aes(x=variables[[a]], y=variables[[b]])) + xlab(colnames(variables)[a])+ ylab(colnames(variables)[b])+ geom_point()) } }。最后，编写这样的内容应该可以：

ff = function(x, margin=c("all", "line", "column", "cell", 0, 1, 2)){
  if(!is.numeric(margin)){
    if(is.null(margin)) {
      margin=0:2 #defaulting 
    } else {
      marginopts = list(all = 0:2,
                        line = 1,
                        column = 2,
                        cell = 0)
      margin <- unname(unlist(marginopts[margin]))
    }
  }
  margin
}
ff(NULL, margin=1:2) # [1] 1 2
ff(NULL, margin="all") # [1] 0 1 2
ff(NULL, margin="cell") # [1] 0
ff(NULL, margin=c("column", "cell")) # [1] 2 0
ff(NULL, margin=NULL) # [1] 0 1 2

它回答了您的问题吗？

如何通过两个循环绘制图形

1 个答案: