为什么我的 lm 模型不显示线性关系，但在 geom_smooth 中显示？答案

【问题标题】：Why does my lm model not show a linear relationship but does in geom_smooth?为什么我的 lm 模型不显示线性关系，但在 geom_smooth 中显示？
【发布时间】：2020-04-30 02:52:41
【问题描述】：

我试图建立一个线性模型来解释粒子浓度和荧光之间的关系。出于某种原因，我无法让模型使用lm 拟合数据，但它确实在 ggplot geom_smooth 函数中工作。

下面是对数荧光和对数粒子浓度的图...

我用下面的代码做了一个模型

Calicurve.M1 <- lm(Fluorescence~Particle.conc,
                na.action = na.exclude,
                data = Calicurve)

但是，当我使用此模型来预测值并添加到我的绘图（在 ggplot2 中）时，它似乎不正确

    ### predict values and put into dataframe
pdat <- expand.grid(Particle.conc = c(5, 50, 500, 5000, 50000, 500000, 5000000,
                                      50000000, 500000000, 5000000000, 50000000000,
                                      500000000000))
pred <- predict(Calicurve.M1, newdata = pdat, na.rm=T,
                type = "response", se.fit = TRUE)
predframe <- data.frame (pdat, preds=pred$fit, se=pred$se.fit)
predframe$upperse <- (predframe$preds+predframe$se)
predframe$lowerse <- (predframe$preds-predframe$se)

### plot calibration curve ###    
plot <- ggplot(data=Calicurve, aes(x=Particle.conc, y=Fluorescence)) +
  geom_point()+
  scale_y_log10(name = "Fluorecence (AFU)",
                     limits = c(1,1200))+
  scale_x_log10(name = "Particle concentration (particles/mL)")+
  theme_bw() +
  theme(panel.grid.major = element_blank(), panel.grid.minor = element_blank(),
        strip.text = element_text(face = "italic"),
        legend.position = c(0.6, 0.75), legend.justification = c(1, 0)) +
  geom_line(data= predframe, aes(x=Particle.conc,y=preds), linetype=1) +
  geom_line(data= predframe, aes(x=Particle.conc,y=upperse), linetype=2) +
  geom_line(data= predframe, aes(x=Particle.conc,y=lowerse), linetype=2)

最后，当我使用 #geom_smooth(method='lm')+ 在 geom_smooth 中使用线性模型时，它会按照我的预期绘制关系...

我想了解第一个模型为什么不起作用以及如何修复它，以便我可以使用该模型来预测荧光值。

根据要求提供样本数据

structure(list(Particle.conc = c(50, 50, 50, 500, 500, 500, 5000, 
5000, 5000, 50000, 50000, 50000, 5e+05, 5e+05, 5e+05, 5e+06, 
5e+06, 5e+06, 5e+07, 5e+07, 5e+07, 5e+08, 5e+08, 5e+08, 5e+09, 
5e+09, 5e+09, 5e+10, 5e+10, 5e+10, 5e+11, 5e+11, 5e+11), Fluorecence = c(2.649, 
2.671, 2.502, 3.926, 3.965, 4, 6.674, 6.337, 6.56, 12.204, 12.168, 
12.209, 24.91, 25.54, 25.384, 38.232, 37.845, 37.979, 80.547, 
80.343, 79.891, 168.693, 168.008, 168.826, 349.318, 351.304, 
355.288, 556.081, 555.348, 554.112, 1105.749, 1103.063, 1097.552
), Average.FL = c(2.607333333, NA, NA, 3.963666667, NA, NA, 6.523666667, 
NA, NA, 12.19366667, NA, NA, 25.278, NA, NA, 38.01866667, NA, 
NA, 80.26033333, NA, NA, 168.509, NA, NA, 351.97, NA, NA, 555.1803333, 
NA, NA, 1102.121333, NA, NA), Fl.Bl = c(1.463, NA, NA, 2.819333333, 
NA, NA, 5.379333333, NA, NA, 11.04933333, NA, NA, 24.13366667, 
NA, NA, 36.87433333, NA, NA, 79.116, NA, NA, 167.3646667, NA, 
NA, 350.8256667, NA, NA, 554.036, NA, NA, 1100.977, NA, NA)), .Names = c("Particle.conc", 
"Fluorecence", "Average.FL", "Fl.Bl"), class = "data.frame", row.names = c(NA, 
-33L))

【问题讨论】：

能否提供一些示例数据？例如将dput(Calicurve) 的输出粘贴到您的问题中
Hi - 已按要求更新了样本数据

标签： r ggplot2 regression glm lm

【解决方案1】：

您必须先转换变量，然后再将它们传递给lm。

df = tibble(x=10^(1:100), y=10^((1:100)*2) + rnorm(100))
lm(log(y, 10)~log(x, 10), df) %>% summary()

# Call:
# lm(formula = log(y, 10) ~ log(x, 10), data = df)
# 
# Residuals:
#        Min         1Q     Median         3Q        Max 
# -6.000e-05 -3.698e-05 -1.480e-05  8.320e-06  1.486e-03 
# 
# Coefficients:
#              Estimate Std. Error   t value Pr(>|t|)    
# (Intercept) 6.291e-05  3.086e-05 2.038e+00   0.0442 *  
# log(x, 10)  2.000e+00  5.305e-07 3.770e+06   <2e-16 ***
# ---
# Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
# 
# Residual standard error: 0.0001531 on 98 degrees of freedom
# Multiple R-squared:      1,   Adjusted R-squared:      1 
# F-statistic: 1.421e+13 on 1 and 98 DF,  p-value: < 2.2e-16

然后要进行预测，您必须像这样反转变换：

df %>% 
    mutate(y_hat = 10^predict(lm(log(y, 10)~log(x,10))))

# # A tibble: 100 x 3
#              x        y   y_hat
#          <dbl>    <dbl>   <dbl>
#  1          10  1.00e 2 1.00e 2
#  2         100  1.00e 4 1.00e 4
#  3        1000  1.00e 6 1.00e 6
#  4       10000 10.00e 7 1.00e 8
#  5      100000  1.00e10 1.00e10
#  6     1000000 10.00e11 1.00e12
#  7    10000000  1.00e14 1.00e14
#  8   100000000  1.00e16 1.00e16
#  9  1000000000  1.00e18 1.00e18
# 10 10000000000  1.00e20 1.00e20
# # ... with 90 more rows

【讨论】：