时间戳

首页 > 解决方案 > GGPLOT：如何为我的数据点的指定子集绘制黄土曲线？

r - GGPLOT：如何为我的数据点的指定子集绘制黄土曲线？

问题描述

（我在 R 自学并经常使用这个论坛，但这是我的第一篇文章。感谢反馈。）

这应该有一个相对简单的解决方案，但我找不到它，这让我想把我的电脑扔出窗外。说到重点，我有一个简单的数据集：

mydata <- structure(list(Date = c("2020-06-22", "2020-06-22", "2020-06-23", 
"2020-06-23", "2020-06-24", "2020-06-24", "2020-06-25", "2020-06-25", 
"2020-06-26", "2020-06-26", "2020-06-29", "2020-06-29", "2020-06-30", 
"2020-06-30", "2020-07-01", "2020-07-01", "2020-07-02", "2020-07-02", 
"2020-07-06", "2020-07-06", "2020-07-06", "2020-07-06", "2020-07-07", 
"2020-07-07", "2020-07-08", "2020-07-08", "2020-07-08", "2020-07-09", 
"2020-07-09", "2020-07-09"), Location = c("Haskell", "Bustamante", 
"Haskell", "Bustamante", "Haskell", "Bustamante", "Bustamante", 
"Haskell", "Bustamante", "Haskell", "Bustamante", "Haskell", 
"Bustamante", "Haskell", "Bustamante", "Haskell", "Bustamante", 
"Haskell", "Bustamante", "Haskell", "Bustamante", "Haskell", 
"Bustamante", "Haskell", "Bustamante", "Haskell", "Tap Water", 
"Bustamante", "Haskell", "Tap Water"), UVT = c(72.2, 65.6, 70, 
61.8, 71.5, 63.9, 63.9, 71.5, 68.1, 71.5, 68.9, 71.3, 71.3, 72.4, 
68.9, 67.3, 49.4, 49, 39.3, 42.3, 64.2, 70.9, 33.3, 49.3, 46, 
48.8, 88.7, 66, 70.5, 84.7), Source = c("Shawn", "Shawn", "Jesus", 
"Jesus", "Jesus", "Jesus", "Jesus", "Jesus", "Jesus", "Jesus", 
"Jesus", "Jesus", "Jesus", "Jesus", "Jesus", "Jesus", "Jesus", 
"Jesus", "Jesus", "Jesus", "Shawn", "Shawn", "Jesus", "Jesus", 
"Jesus", "Jesus", "Jesus", "Jesus", "Jesus", "Jesus")), row.names = c(NA, 
-30L), class = "data.frame")

首先，我尝试按位置绘制数据分组，但我猜由于“自来水”组只有 2 个数据点，它不符合度数要求：

#Import Packages
library(tidyverse)

#Import Data
mydata <- read.csv("L:\\2019\\19W06195 - EPW HRS and RRB WWTPs Disinfection Study\\Design\\Design Criteria\\R\\UVT Graphs\\UVTdata.csv")

#Plot
p <- ggplot(data=mydata, aes(x=as.Date(mydata[,1], "%Y-%m-%d"), y=mydata[,3], color=mydata[,2])) + geom_point() + geom_smooth(method = "loess", se = FALSE)

p + scale_x_date(date_breaks = "days" , date_labels = "%b-%d")

情节尝试#1

这是我收到的错误：

Warning messages:
1: In simpleLoess(y, x, w, span, degree = degree, parametric = parametric,  ... :
  span too small.   fewer data values than degrees of freedom.
2: In simpleLoess(y, x, w, span, degree = degree, parametric = parametric,  ... :
  at  18451
3: In simpleLoess(y, x, w, span, degree = degree, parametric = parametric,  ... :
  radius  2.5e-005
4: In simpleLoess(y, x, w, span, degree = degree, parametric = parametric,  ... :
  all data on boundary of neighborhood. make span bigger
5: In simpleLoess(y, x, w, span, degree = degree, parametric = parametric,  ... :
  pseudoinverse used at 18451
6: In simpleLoess(y, x, w, span, degree = degree, parametric = parametric,  ... :
  neighborhood radius 0.005
7: In simpleLoess(y, x, w, span, degree = degree, parametric = parametric,  ... :
  reciprocal condition number  1
8: In simpleLoess(y, x, w, span, degree = degree, parametric = parametric,  ... :
  at  18452
9: In simpleLoess(y, x, w, span, degree = degree, parametric = parametric,  ... :
  radius  2.5e-005
10: In simpleLoess(y, x, w, span, degree = degree, parametric = parametric,  ... :
  all data on boundary of neighborhood. make span bigger
11: In simpleLoess(y, x, w, span, degree = degree, parametric = parametric,  ... :
  There are other near singularities as well. 2.5e-005
12: In simpleLoess(y, x, w, span, degree = degree, parametric = parametric,  ... :
  zero-width neighborhood. make span bigger
13: In simpleLoess(y, x, w, span, degree = degree, parametric = parametric,  ... :
  zero-width neighborhood. make span bigger
14: Computation failed in `stat_smooth()`:
NA/NaN/Inf in foreign function call (arg 5)

请注意，运行相同的代码，但指定“method=lm”而不是“method=loess”可以完美运行，但没有显示我想要的趋势。

线性回归模型

为了解决这个问题，我尝试将条件设置为默认为具有太少数据点的数据子集的线性回归：

sProduct <- unique(mydata[,2])
p <- ggplot(mydata, aes(as.Date(mydata[,1], "%Y-%m-%d"), mydata[,3], color = mydata[,2])) + geom_point()

for (i in sProduct){

  sMethod <- ifelse(sum(mydata[,2] == i) <= 5, "lm", "loess")
  p <- p + geom_smooth(data = subset(mydata, mydata[,2] == i), method = sMethod, se = FALSE)
}

p

尽管做出了这些努力，我现在还是遇到了一个审美错误：

Error: Aesthetics must be either length 1 or the same as the data (14): x, y and colour
Run `rlang::last_error()` to see where the error occurred.

我认为这是由于 geom_points 和 geom_smooth 中的数据子集之间的数据点数量不一致，但我不确定。我还尝试设置数据子集以从 geom_smooth 中排除“自来水”，因为无论如何我通常对那里的趋势不感兴趣：

p <- ggplot(data=mydata, aes(x=as.Date(mydata[,1], "%Y-%m-%d"), y=mydata[,3], color=mydata[,2])) + geom_point() + geom_smooth(data=subset(mydata, Location=="Bustamante" | Location=="Haskell"), method = "loess", se = FALSE)
p + scale_x_date(date_breaks = "days" , date_labels = "%b-%d")

这会产生相同的错误。这里的任何帮助将不胜感激！谢谢！

标签： rggplot2regression

解决方案

尝试作为formula='y ~ x'_geom_smooth

geom_smooth(method = "loess", formula='y ~ x', se = FALSE)

然后您将获得以下输出（此处未格式化日期）：

输出

推荐阅读