Question

我有3列，其中包括指标（0/1）

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))

我想返回一个字符列“ qualifiers”，以使指示符== 1的列名称串联在一个字符串中，如下所示：

*qualifiers*
    integrated_intervention
    delivery_group
    delivery_group, culturally_tailored

    culturally_tailored

我尝试了extdplyr :: ind（具有各种选项），但没有成功。下面的一个崩溃了我的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，但认为可能会有一个更简单的解决方案。

Answer 1

您可以尝试：

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，如莫里斯所建议的那样紧凑：

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

Answer 2

我不确定这是否是“简单”的解决方案，但这是使用 tidyverse 的解决方案。

third_party

^{由reprex package（v0.2.1）于2019-02-27创建}

Answer 3

这是一种疯狂的方式：

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

如何将指标列转换为（列名的）串联列

3 个答案: