如何将指标列转换为串联列(列名) [英] How to convert indicator columns to a concatenated column (of column names)

问题描述

我有 3 列由指标 (0/1) 组成

I have 3 columns consisting of indicator (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))

我想返回一个单字符列限定符"，这样带有指示符 == 1 的列名将连接在一个字符串中，如下所示:

I would like to return a single character column 'qualifiers', such that column names with indicator == 1 are concatenated in a string as below:

*qualifiers*
    integrated_intervention
    delivery_group
    delivery_group, culturally_tailored

    culturally_tailored 

我尝试了 extdplyr::ind(带有各种选项)但没有成功.下面的一个崩溃了我的 R 会话.

I tried extdplyr::ind (with various options) without success. The one below crashed my R session.

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)

我发现 将布尔指标列转换为单因子列，但认为可能有更简单的解决方案.

I found Convert Boolean indicator columns to a single factor column, but thought there might be a simpler solution.

推荐答案

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，更像 Maurits 建议的那样:

Or, even more compactly as Maurits suggested:

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

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