Error using 'segmented' with lm extracted from output of tidyverse 'map' in R

问题

I am using the 'segmented' package to find break points in linear regressions in R

library(tidyverse)
library(segmented)

df <- data.frame(x = c(1:10), y = c(1,1,1,1,1,6:10))
lm_model <- lm(y ~ x, data = df)
seg_model <- segmented(obj = lm_model, seg.Z = ~ x)

But if I run the same model within a purrr:map, segmented fails.

map_test <- df %>% 
  nest() %>%
  mutate(map_lm = map(data, ~lm(y ~ x, data = .)),
         param_map_lm = map(map_lm, tidy))

map_lm_model <- map_test[[2]][[1]]

map_seg_model <- segmented(obj = map_lm_model, seg.Z = ~ x)

"Error in is.data.frame(data) : object '.' not found"

When taking the lm obj from the lm extracted from the map output, segmented fails to find the underlying data.

The two linear model objects, appear, however, identical.

What I actually need to do is a more useful map to run lm over multiple sub-sets of a dataframe, then run 'segmented' on the resulting lm.

回答1:

This is basically the same issue as the interaction between glm() and purrr::map().

lm() captures the expression supplied to it, which works well as a stand-alone case. However, when called by map(), the supplied expression is ., which has no meaning outside the immediate context of that map() call and results in the error you are observing.

As with the other question, one workaround is to define a wrapper for lm() that composes a custom call directly on the dataset, which is then captured by lm() as an unevaluated expression.

# Composes a custom lm() expression and evaluates it
lm2 <- function(data, ...)
    eval( rlang::expr(lm(data=!!rlang::enexpr(data), !!!list(...))) )

# Now mapping using lm2, instead of lm
map_test <- nest(df, data=everything()) %>% 
    mutate(map_lm       = map(data, lm2, y ~ x),
           param_map_lm = map(map_lm, broom::tidy))

# The data is stored directly inside the lm object
# segmented() now has no problems accessing it
map_lm_model <- map_test[[2]][[1]]
map_seg_model <- segmented(obj = map_lm_model, seg.Z = ~ x)
# Call: segmented.lm(obj = map_lm_model, seg.Z = ~x)
# 
# Meaningful coefficients of the linear terms:
# (Intercept)            x         U1.x  
#   1.000e+00    6.344e-15    1.607e+00  
# 
# Estimated Break-Point(s):
# psi1.x  
#  3.889

or as a single mutate() chain:

map_test <- nest(df, data=everything()) %>% 
    mutate(map_lm       = map(data, lm2, y ~ x),
           param_map_lm = map(map_lm, broom::tidy),
           seg_lm       = map(map_lm, segmented, seg.Z=~x))
# # A tibble: 1 x 4
#             data map_lm param_map_lm     seg_lm    
#   <list<df[,2]>> <list> <list>           <list>    
# 1       [10 × 2] <lm>   <tibble [2 × 5]> <segmentd>

来源：https://stackoverflow.com/questions/58525603/error-using-segmented-with-lm-extracted-from-output-of-tidyverse-map-in-r

标签

purrr