Question

我试图在R中使用相同的自变量（x）和多个相关响应变量（Curve1，Curve2等）逐个执行lm（）和segmented（）。我希望提取每个响应变量的估计断点和模型系数。我在下面提供了我的数据示例。

           x  Curve1  Curve2   Curve3
1  -0.236422 98.8169 95.6828 101.7910
2  -0.198083 98.3260 95.4185 101.5170
3  -0.121406 97.3442 94.8899 100.9690
4   0.875399 84.5815 88.0176  93.8424
5   0.913738 84.1139 87.7533  93.5683
6   1.795530 73.3582 78.1278  82.9956
7   1.833870 72.8905 77.7093  82.7039
8   1.872200 72.4229 77.3505  82.4123
9   2.907350 59.2070 67.6652  74.5374
10  3.865810 46.4807 58.5158  65.0220
11  3.904150 45.9716 58.1498  64.7121
12  3.942490 45.4626 57.8099  64.4022
13  4.939300 33.3040 48.9742  56.3451
14  4.977640 32.9641 48.6344  56.0352
15  5.936100 24.4682 36.4758  47.0485
16  5.936100 24.4682 36.4758  47.0485
17  6.012780 23.7885 35.9667  46.5002
18  6.971250 20.7387 29.6035  39.6476
19  7.009580 20.6167 29.3490  39.3930
20  8.006390 18.7209 22.7313  32.7753
21  8.121410 18.5022 22.3914  32.1292
22  9.041530 16.4722 19.6728  26.9604
23  9.079870 16.3877 19.5595  26.7450

我可以使用以下代码一次完成一条曲线。但是，我的完整数据集有超过1000条曲线，因此我希望能够以某种方式在每列上重复此代码。我没有成功地试图将它循环遍历每一列，所以如果有人能告诉我如何做这样的事情并创建一个类似于下面代码生成的摘要数据框，但是每个列都包括在内，我会非常感激。谢谢！

model <- lm(Curve1~x, dat) # Linear model
seg_model <- segmented(model, seg.Z = ~x) # Segmented model
breakpoint <- as.matrix(seg_model$psi.history[[5]]) # Extract breakpoint
coefficients <- as.matrix(seg_model$coefficients) # Extract coefficients
summary_curve1 <- as.data.frame(rbind(breakpoint, coefficients)) # combine breakpoint and coefficeints
colnames(summary_curve1) <- "Curve_1" # header name 
summary_curve1 # display summary

Answer 1

以下是使用tidyverse和broom返回包含每个Curve列结果的数据框的方法：

library(broom)
library(tidyverse)

model.results = setNames(names(dat[,-1]), names(dat[,-1])) %>% 
  map(~ lm(paste0(.x, " ~ x"), data=dat) %>% 
        segmented(seg.Z=~x) %>%
        list(model=tidy(.), 
             psi=data.frame(term="breakpoint", estimate=.[["psi.history"]][[5]]))) %>%
  map_df(~.[2:3] %>% bind_rows, .id="Curve")

model.results

    Curve        term   estimate  std.error   statistic      p.value
1  Curve1 (Intercept)  95.866127 0.14972382  640.286416 1.212599e-42
2  Curve1           x -12.691455 0.05220412 -243.112130 1.184191e-34
3  Curve1        U1.x  10.185816 0.11080880   91.922447 1.233602e-26
4  Curve1      psi1.x   0.000000 0.02821843    0.000000 1.000000e+00
5  Curve1  breakpoint   5.595706         NA          NA           NA
6  Curve2 (Intercept)  94.826309 0.45750667  207.267599 2.450058e-33
7  Curve2           x  -9.489342 0.11156425  -85.057193 5.372730e-26
8  Curve2        U1.x   6.532312 1.17332640    5.567344 2.275438e-05
9  Curve2      psi1.x   0.000000 0.23845241    0.000000 1.000000e+00
10 Curve2  breakpoint   7.412087         NA          NA           NA
11 Curve3 (Intercept) 100.027990 0.29453941  339.608175 2.069087e-37
12 Curve3           x  -8.931163 0.08154534 -109.523900 4.447569e-28
13 Curve3        U1.x   2.807215 0.36046013    7.787865 2.492325e-07
14 Curve3      psi1.x   0.000000 0.26319757    0.000000 1.000000e+00
15 Curve3  breakpoint   6.362132         NA          NA           NA

Answer 2

您可以将整个事物包装在一个函数中，将列名称和数据作为参数，并在列名称上使用lapply，如下所示：

library(segmented)
run_mod <- function(varname, data){

  data$Y <- data[,varname]
  model <- lm(Y ~ x, data) # Linear model
  seg_model <- segmented(model, seg.Z = ~x) # Segmented model
  breakpoint <- as.matrix(seg_model$psi.history[[5]]) # Extract breakpoint
  coefficients <- as.matrix(seg_model$coefficients) # Extract coefficients
  summary_curve1 <- as.data.frame(rbind(breakpoint, coefficients)) 
  colnames(summary_curve1) <- varname

return(summary_curve1)
}


lapply(names(dat)[2:ncol(dat)], function(x)run_mod(x, dat))

其中给出了每条拟合曲线的摘要（不确定您实际想要的输出）。

在R中的多个列上执行lm（）和segmented（）

2 个答案: