如何将指标列转换为连续列（列名称）

Question

我有3列由指标（0/1）组成

icols <- 
structure(list(delivery_group = c(0, 1, 1, 0, 0), culturally_tailored = c(0, 
0, 1, 0, 1), integrated_intervention = c(1, 0, 0, 0, 0)), class = c("tbl_df", 
"tbl", "data.frame"), row.names = c(NA, -5L))

我想返回单个字符列'限定符'，这样带有指示符== 1的列名在字符串中连接如下：

*qualifiers*
    integrated_intervention
    delivery_group
    delivery_group, culturally_tailored

    culturally_tailored 

我尝试了extdplyr :: ind（有各种选项）但没有成功。下面的一个撞毁了我的R会话。

icols <- extdplyr::ind_to_char(col = qualifiers, ret_factor = FALSE, remove = TRUE,
              from = c("delivery_group", "culturally_tailored", "integrated_intervention"),
              mutually_exclusive = FALSE, collectively_exhaustive = FALSE)

我找到了Convert Boolean indicator columns to a single factor column，但认为可能有一个更简单的解决方案。

Answer 1

你可以试试：

icols$collapsed <- apply(icols, 1, function(x) paste0(names(icols)[x == 1], collapse = ", "))

icols

  delivery_group culturally_tailored integrated_intervention                           collapsed
1              0                   0                       1             integrated_intervention
2              1                   0                       0                      delivery_group
3              1                   1                       0 delivery_group, culturally_tailored
4              0                   0                       0                                    
5              0                   1                       0                 culturally_tailored

或者，正如Maurits建议的那样：

apply(icols, 1, function(x) toString(names(icols)[x == 1]))

Answer 2

我不确定这是一个“简单”的解决方案，但这是一个使用tidyverse的解决方案。

library(tidyverse)

icols <- tibble(
  delivery_group = c(0, 1, 1, 0, 0),
  culturally_tailored = c(0, 0, 1, 0, 1),
  integrated_intervention = c(1, 0, 0, 0, 0)
)

icols %>%
  rowid_to_column(var = "rowid") %>%
  gather(key = "qualifiers", value = "indicator", -rowid) %>%
  filter(indicator == 1) %>%
  group_by(rowid) %>%
  summarize(qualifiers = paste(qualifiers, collapse = ", ")) %>%
  ungroup() %>%
  complete(rowid = 1:nrow(icols)) %>%
  select(qualifiers)
#> # A tibble: 5 x 1
#>   qualifiers                         
#>   <chr>                              
#> 1 integrated_intervention            
#> 2 delivery_group                     
#> 3 delivery_group, culturally_tailored
#> 4 <NA>                               
#> 5 culturally_tailored

由reprex package创建于2019-02-27（v0.2.1）

Answer 3

这是一个疯狂的方式：

library(tidyverse)

icols %>%
  mutate(qualifiers = case_when(
    delivery_group & culturally_tailored == 1 ~ "delivery_group, culturally_tailored",
    delivery_group & integrated_intervention == 1 ~ "delivery_group, integrated_intervation",
    culturally_tailored & integrated_intervention == 1 ~ "culturally_tailored, integrated_intervation",
    culturally_tailored == 1 ~ "culturally_tailored",
    integrated_intervention == 1 ~ "integrated_intervention",
    delivery_group == 1 ~ "delivery_group"))
# A tibble: 5 x 4
  delivery_group culturally_tailored integrated_intervention qualifiers                         
           <dbl>               <dbl>                   <dbl> <chr>                              
1              0                   0                       1 integrated_intervention            
2              1                   0                       0 delivery_group                     
3              1                   1                       0 delivery_group, culturally_tailored
4              0                   0                       0 NA                                 
5              0                   1                       0 culturally_tailored