Question

考虑此示例

mydata <- data_frame(ind_1 = c(NA,NA,3,4),
                     ind_2 = c(2,3,4,5),
                     ind_3 = c(5,6,NA,NA),
                     y = c(28,34,25,12),
                     group = c('a','a','b','b'))

> mydata
# A tibble: 4 x 5
  ind_1 ind_2 ind_3     y group
  <dbl> <dbl> <dbl> <dbl> <chr>
1    NA     2     5    28 a    
2    NA     3     6    34 a    
3     3     4    NA    25 b    
4     4     5    NA    12 b

在这里，我想针对每个group，对该组中不缺少的任何变量进行回归y，并将对应的lm对象存储在list-column中。 / p>

也就是说：

对于组a，这些变量对应于ind_2和ind_3
对于组b，它们对应于ind_1和ind_2

我尝试了以下操作，但这不起作用

mydata %>% group_by(group) %>% nest() %>% 
  do(filtered_df <- . %>% select(which(colMeans(is.na(.)) == 0)),
     myreg = lm(y~ names(filtered_df)))

有什么想法吗？谢谢！

Answer 1

我们可以使用map和mutate。如果您要保留过滤后的数据（select），我们可以nestdat1一步一步地map进行建模，也可以使用两个nestdat2分步进行建模{<1} / p>

library(tidyverse)

nestdat1 <- mydata %>%
  group_by(group) %>%
  nest() %>%
  mutate(model = data %>% map(~ select_if(., funs(!any(is.na(.)))) %>%
                                lm(y ~ ., data = .)))

nestdat2 <- mydata %>%
  group_by(group) %>%
  nest() %>%
  mutate(data = data %>% map(~ select_if(., funs(!any(is.na(.))))),
         model = data %>% map(~ lm(y ~ ., data = .)))

输出：

它们产生不同的data列：

> nestdat1 %>% pull(data)
[[1]]
# A tibble: 2 x 4
  ind_1 ind_2 ind_3     y
  <dbl> <dbl> <dbl> <dbl>
1    NA     2     5    28
2    NA     3     6    34

[[2]]
# A tibble: 2 x 4
  ind_1 ind_2 ind_3     y
  <dbl> <dbl> <dbl> <dbl>
1     3     4    NA    25
2     4     5    NA    12

> nestdat2 %>% pull(data)
[[1]]
# A tibble: 2 x 3
  ind_2 ind_3     y
  <dbl> <dbl> <dbl>
1     2     5    28
2     3     6    34

[[2]]
# A tibble: 2 x 3
  ind_1 ind_2     y
  <dbl> <dbl> <dbl>
1     3     4    25
2     4     5    12

但同一列model列：

> nestdat1 %>% pull(model)
[[1]]

Call:
lm(formula = y ~ ., data = .)

Coefficients:
(Intercept)        ind_2        ind_3  
         16            6           NA  

[[2]]

Call:
lm(formula = y ~ ., data = .)

Coefficients:
(Intercept)        ind_1        ind_2  
         64          -13           NA  


> nestdat2 %>% pull(model)
[[1]]

Call:
lm(formula = y ~ ., data = .)

Coefficients:
(Intercept)        ind_2        ind_3  
         16            6           NA  

[[2]]

Call:
lm(formula = y ~ ., data = .)

Coefficients:
(Intercept)        ind_1        ind_2  
         64          -13           NA

Answer 2

这是另一个tidyverse选项，如果您希望将其保留在mydata$model中，请分配给tibble：

library(tidyverse)
mydata %>%
  nest(-group) %>%
  pull(data) %>%
  map(~lm(y ~., discard(.,anyNA)))
# [[1]]
# 
# Call:
# lm(formula = y ~ ., data = discard(., anyNA))
# 
# Coefficients:
# (Intercept)        ind_2        ind_3  
#          16            6           NA  
# 
# 
# [[2]]
# 
# Call:
# lm(formula = y ~ ., data = discard(., anyNA))
# 
# Coefficients:
# (Intercept)        ind_1        ind_2  
#          64          -13           NA  
# 
#

在purrr循环中选择非缺失变量

2 个答案: