how to apply lme function to each column of dataframe?

Question

I'm using the lme function from the nlme package and having a hard time trying to apply it to each column of my tibble. I can successfully run it 'manually' with a single column, but fail when trying to use purrr::map. I feel like something is right under my nose, but I'm not seeing it. Any help would be appreciated.

library(tidyverse)
library(nlme)

# Create Tibble
df <- tibble(
   id = rep(1:5, length = 30), 
   ko = as_factor(rep(c('ctrl', 'ko1', 'ko2', 'ko3', 'ko4', 'ko5'), each = 5)), 
   marker1 = rnorm(30), 
   marker2 = rnorm(30)
)

# 'Manual' calculation
df %>% lme(marker1 ~ ko, random = ~1|id, data = .)

# Attempt at calculating lme for each column
df %>% map_if(is_double, ~ lme(.x ~ ko, random = ~ 1|id, data = .))

# Ideal would be to get the tibble of each like:
df %>% summarize(broom.mixed::tidy(lme(marker1 ~ ko, random = ~1|id, data = .)))

jdobres · Accepted Answer · 2022-03-26T02:27:38.260

Because of how lme() interprets its formula argument, I don't think this can be done across columns. However, we could pivot the data into a longer format so that the dependent variable is in one column, with another column indicating that these should be tested separately:

library(nlme)
library(tidyverse)

tests <- df %>% 
  pivot_longer(marker1:marker2, names_to = 'marker') %>% 
  nest(data = -marker) %>% 
  rowwise() %>% 
  mutate(
    lme = list(lme(value ~ ko, random = ~ 1|id, data = data)),
    result = list(broom.mixed::tidy(lme))
  ) %>% 
  unnest(result)

   marker  data     lme    effect   group    term   estimate std.error    df statistic p.value
   <chr>   <list>   <list> <chr>    <chr>    <chr>     <dbl>     <dbl> <dbl>     <dbl>   <dbl>
 1 marker1 <tibble> <lme>  fixed    fixed    (Inte…  2.00e-1     0.394    20    0.506    0.618
 2 marker1 <tibble> <lme>  fixed    fixed    koko1   3.46e-1     0.558    20    0.621    0.542
 3 marker1 <tibble> <lme>  fixed    fixed    koko2  -7.70e-1     0.558    20   -1.38     0.183
 4 marker1 <tibble> <lme>  fixed    fixed    koko3   1.56e-2     0.558    20    0.0281   0.978
 5 marker1 <tibble> <lme>  fixed    fixed    koko4   1.76e-1     0.558    20    0.316    0.755
 6 marker1 <tibble> <lme>  fixed    fixed    koko5  -2.27e-1     0.558    20   -0.408    0.688
 7 marker1 <tibble> <lme>  ran_pars id       sd_(I…  2.03e-5    NA        NA   NA       NA    
 8 marker1 <tibble> <lme>  ran_pars Residual sd_Ob…  8.82e-1    NA        NA   NA       NA    
 9 marker2 <tibble> <lme>  fixed    fixed    (Inte…  1.75e-1     0.414    20    0.423    0.677
10 marker2 <tibble> <lme>  fixed    fixed    koko1  -1.46e-1     0.542    20   -0.269    0.791
11 marker2 <tibble> <lme>  fixed    fixed    koko2  -4.89e-1     0.542    20   -0.902    0.378
12 marker2 <tibble> <lme>  fixed    fixed    koko3  -6.79e-1     0.542    20   -1.25     0.225
13 marker2 <tibble> <lme>  fixed    fixed    koko4  -5.70e-1     0.542    20   -1.05     0.306
14 marker2 <tibble> <lme>  fixed    fixed    koko5  -5.57e-1     0.542    20   -1.03     0.317
15 marker2 <tibble> <lme>  ran_pars id       sd_(I…  3.48e-1    NA        NA   NA       NA    
16 marker2 <tibble> <lme>  ran_pars Residual sd_Ob…  8.58e-1    NA        NA   NA       NA

score 0 · Answer 2 · answered Mar 26 '22 at 02:15

This is a little bit tricky because the formulas inside lme use their own version of non-standard evaluation. I would suggest

vars <- names(df)[map_lgl(df, is_double)]
(vars
  %>% setNames(vars)  ## trick for naming results 
  %>% map(~reformulate("ko", response = .))
  %>% map(lme, random = ~1|id, data = df)
  %>% map_dfr(broom.mixed::tidy, effects = "fixed", .id = "var")
)

(I'm not sure what summarize() is doing there?) If you were using the lme4 package you could use the refit() function to do this a bit more efficiently ...

Thank you Ben for your answer and context. Will need to check out that `lme4` package. — phdj, Mar 26 '22 at 16:24

how to apply lme function to each column of dataframe?

2 Answers2