为什么 do(lm...) 和 geom_smooth(method="lm") 有区别？答案

【问题标题】：Why is there a difference between do(lm...) and geom_smooth(method="lm")?为什么 do(lm...) 和 geom_smooth(method="lm") 有区别？
【发布时间】：2019-08-21 13:32:53
【问题描述】：

我有一条略微进入饱和状态的外部校准曲线。所以我拟合了一个二阶多项式和一个测量样本的数据框，我想知道其中的浓度。

df_calibration=structure(list(dilution = c(0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 
0.8, 0.9, 1, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1), 
    area = c(1000, 2000, 3000, 4000, 5000, 6000, 7000, 7800, 
    8200, 8500, 1200, 2200, 3200, 4200, 5200, 6200, 7200, 8000, 
    8400, 8700), substance = c("A", "A", "A", "A", "A", "A", 
    "A", "A", "A", "A", "b", "b", "b", "b", "b", "b", "b", "b", 
    "b", "b")), row.names = c(NA, 20L), class = "data.frame")

df_samples=structure(list(area = c(1100, 1800, 2500, 3200, 3900, 1300, 2000, 
2700, 3400, 4100), substance = c("A", "A", "A", "A", "A", "b", 
"b", "b", "b", "b")), row.names = c(NA, 10L), class = "data.frame")

现在为了计算测量样品的实际稀释度，我采用了由此拟合生成的参数：

df_fits=df_calibration %>% group_by(substance) %>% 
  do(fit = lm(area ~ poly(dilution,2), data = .))%>%
  tidy(fit) %>% 
  select(substance, term, estimate) %>% 
  spread(term, estimate)

df_fits=df_fits %>% rename(a=`poly(dilution, 2)2`,b=`poly(dilution, 2)1`,c=`(Intercept)`)

#join parameters with sample data
df_samples=left_join(df_samples,df_fits)

还有这个公式

#calculate with general solution for polynomial 2nd order
df_samples$dilution_calc=
  (df_samples$b*(-1)+sqrt(df_samples$b^2-(4*df_samples$a*(df_samples$c-df_samples$area))))/(2*df_samples$a)

但是，当我现在绘制这个时，我注意到一些非常奇怪的东西。计算出的 x 值（稀释）不会出现在 stat_smooth() 的曲线上。附加的虚线与图表中方程的参数（与数据框中的数字匹配）用于物质“A”。所以我的计算应该是正确的（或不正确？）为什么会有差异？我究竟做错了什么？我如何从stat_smooth() 完成的拟合中获取参数？

my.formula=y ~ poly(x,2)
ggplot(df_calibration, aes(x = dilution, y = area)) +
  stat_smooth(method = "lm", se=FALSE, formula = my.formula) +

  stat_function(fun=function(x){5250+(7980*x)+(-905*x^2)},      
              inherit.aes = F,linetype="dotted")+

  stat_poly_eq(formula = my.formula, 
               aes(label = paste(..eq.label.., ..rr.label.., sep = "~~~")), 
               parse = TRUE) +         
  geom_point(shape=17)+
  geom_point(data=df_samples,
           aes(x=dilution_calc,y=area),
           shape=1,color="red")+
  facet_wrap(~substance,scales = "free")

任何建议都将受到高度赞赏:-)

【问题讨论】：

在我的脑海中，我无法弄清楚“二次公式”与二阶多项式的预测有什么关系。因此，您可能应该更详细地描述您正在使用的逻辑。如果您使用来自poly() 的系数，那么您 a) 需要包括截距 b) 需要记住二阶项已与一阶项正交。而且，尽管执行了library(tidyverse)，但我仍然收到“整理错误（., fit）：找不到函数“整理””

标签： r ggplot2 curve-fitting data-fitting model-fitting

【解决方案1】：

默认情况下，poly 计算正交多项式。您可以使用 raw=TRUE 参数关闭正交化。

请注意，该公式出现了两种情况：一次使用原始变量名称拟合回归，然后在stat_smooth 中使用通用变量名称x 和y。但否则它应该是相同的公式，raw=TRUE。

library("tidyverse")

# Define/import your data here....

df_fits <- df_calibration %>%
  group_by(substance) %>%
  do(fit = lm(area ~ poly(dilution, 2, raw = TRUE), data = .)) %>%
  broom::tidy(fit) %>%
  select(substance, term, estimate) %>%
  spread(term, estimate) %>%
  # It is simpler to rename the coefficients here
  setNames(c("substance", "c", "b", "a"))

# join parameters with sample data
df_samples <- left_join(df_samples, df_fits)

# calculate with general solution for polynomial 2nd order
df_samples <- df_samples %>%
  mutate(dilution_calc = (b * (-1) + sqrt(b^2 - (4 * a * (c - area)))) / (2 * a))

my.formula <- y ~ poly(x, 2, raw = TRUE)

df_calibration %>%
  ggplot(aes(x = dilution, y = area)) +
  stat_smooth(method = "lm", se = FALSE, formula = my.formula) +
  geom_point(shape = 17) +
  geom_point(
    data = df_samples,
    aes(x = dilution_calc, y = area),
    shape = 1, color = "red"
  ) +
  facet_wrap(~substance, scales = "free")

^{由reprex package (v0.2.1) 于 2019 年 3 月 31 日创建}

【讨论】：