如何将指示符列转换为连接列（列名）答案

【问题标题】：How to convert indicator columns to a concatenated column (of column names)如何将指示符列转换为连接列（列名）
【发布时间】：2019-02-28 02:55:15
【问题描述】：

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))

我想返回一个单一的字符列“限定符”，这样带有指示符 == 1 的列名连接在一个字符串中，如下所示：

*qualifiers*
    integrated_intervention
    delivery_group
    delivery_group, culturally_tailored

    culturally_tailored

我尝试了 extdplyr::ind（使用各种选项）但没有成功。下面的一个使我的 R 会话崩溃。

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，但认为可能有更简单的解决方案。

【问题讨论】：

标签： r dplyr tidyr

【解决方案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，按照 Maurits 的建议，更简洁：

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

【讨论】：

不错的一个（+1）。为了让它更短，你可以使用toString 而不是paste0(..., collapse = ", ")：apply(icols, 1, function(x) toString(names(icols)[x == 1]))

【解决方案2】：

我不确定这是一个“简单”的解决方案，但这里有一个使用 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 (v0.2.1) 于 2019 年 2 月 27 日创建}

【讨论】：

【解决方案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

【讨论】：