预测 R 曲线上的值

Question

I am trying to predict the values of a curve in R using this data.

   Distance Waggle_time
1       100        9.45
2       200        7.90
3       300        7.04
4       385        6.49
5       400        6.34
6       500        6.01
7       600        5.59
8       700        5.07
9       800        4.79
10      900        4.73
11     1000        4.62
12     1100        4.34
13     1250        4.33
14     1300        4.30
15     1400        4.10
16     1500        4.06
17     2000        3.31
18     2500        3.13
19     3000        2.77
20     3500        2.65
21     4000        2.52
22     4500        2.30
23     5000        2.22
24     6000        1.93
25     7000        1.71
26     8000        1.62
27     8500        1.46
28     9500        1.36

I have tried to use the predict function by creating a linear regression model and providing x values that should be used to predict the y values, but this gives me data that is completely wrong. I understand that I am probably completely on the wrong path so any help would be very much appreciated.

Answer 1

A few options are shown below. It's not clear to me if you need help with the model fitting (statistical help) or the prediction of new values (technical help). If the former, you should look at Cross Validated .

df <- structure(list(Distance = c(100L, 200L, 300L, 385L, 400L, 500L, 
600L, 700L, 800L, 900L, 1000L, 1100L, 1250L, 1300L, 1400L, 1500L, 
2000L, 2500L, 3000L, 3500L, 4000L, 4500L, 5000L, 6000L, 7000L, 
8000L, 8500L, 9500L), Waggle_time = c(9.45, 7.9, 7.04, 6.49, 
6.34, 6.01, 5.59, 5.07, 4.79, 4.73, 4.62, 4.34, 4.33, 4.3, 4.1, 
4.06, 3.31, 3.13, 2.77, 2.65, 2.52, 2.3, 2.22, 1.93, 1.71, 1.62, 
1.46, 1.36)), class = "data.frame", row.names = c("1", "2", "3", 
"4", "5", "6", "7", "8", "9", "10", "11", "12", "13", "14", "15", 
"16", "17", "18", "19", "20", "21", "22", "23", "24", "25", "26", 
"27", "28"))



splfit <- smooth.spline(x = df$Distance, y = df$Waggle_time, spar = 0.5)
splpred <- data.frame(Distance = seq(min(df$Distance), max(df$Distance), length.out = 100))
splpred$Waggle_time <- predict(splfit, x = newdat$Distance)$y

# lmfit <- glm(Waggle_time ~ Distance, df, family = Gamma(link = "inverse"))
lmfit <- glm(Waggle_time ~ Distance, df, family = inverse.gaussian(link = "1/mu^2"))
glmpred <- splpred
glmpred$Waggle_time <- predict(lmfit, newdata = newdat, type = "response")


plot(Waggle_time ~ Distance, df)
lines(Waggle_time ~ Distance, splpred, col = 2)
lines(Waggle_time ~ Distance, glmpred, col = 3)
legend("topright", legend = c("spline", "inv. Gaussian"), col = 2:3, lty = 1)

Answer 2

Your data are described better by a log than a linear model. Although you can do a formal model comparison to find this out, it's clear just from plotting the data and some candidate best fit lines:

library(tidyverse)

ggplot(df, aes(x = Distance, y = Waggle_time)) + 
  geom_point() + 
  geom_smooth(formula = y ~ log(x), method = "lm", se=F, colour = "blue") +
  geom_smooth(formula = y ~ x, method = "lm", se=F, colour = "red")

As such, you can log transform your data, fit a linear model, use this to predict the your distance values. You can then take the exponent if you'd like the untransformed predictions back.

df <- mutate(df, logdistance = log(Distance))
modelfit <- lm(logdistance ~ Waggle_time, df)
to_predict <- tibble(Waggle_time = c(12, 10, 8))

# Predict values with augment
augment(modelfit, newdata = to_predict) %>%
  mutate(truefitted = exp(.fitted))

Output, with predictions of Distance on your original scale in the column truefitted :

  Waggle_time .fitted .se.fit truefitted
        <dbl>   <dbl>   <dbl>      <dbl>
1          12    2.74  0.116        15.5
2          10    3.90  0.0885       49.6
3           8    5.07  0.0623      159.