Performing lm() and segmented() on multiple columns in R

问题

I am trying to perform lm() and segmented() in R using the same independent variable (x) and multiple dependent response variables (Curve1, Curve2, etc.) one by one. I wish to extract the estimated break point and model coefficients for each response variable. I include an example of my data below.

           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

I am able to do this one curve at a time using the below code. However, my full data set has over 1000 curves, so I would like to be able to repeat this code over every column somehow. I have not been at all successful trying to loop it over every column, so if anyone could show me how to do something like that and create a summary data frame similar to that generated by the below code, but with every column included, I would be extremely grateful. Thanks!

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

回答1:

Here's an approach using tidyverse and broom to return a data frame containing the results for each Curve column:

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

回答2:

You can wrap the whole thing in a function, taking as the arguments the column name and the data, and use lapply on the column names, like this:

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))

Which gives the summary for each fitted curve (not sure which output you actually want).

来源：https://stackoverflow.com/questions/44170937/performing-lm-and-segmented-on-multiple-columns-in-r

标签

regression