关于 r：如何将指标列转换为连接列(列名)

How to convert indicator columns to a concatenated column (of column names)

我有 3 列由指标 (0/1)

组成

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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 \\’qualifiers\\’，这样带有指示符 == 1 的列名连接在一个字符串中，如下所示：

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*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。

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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。

你可以试试：

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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：

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

相关讨论

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]))

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。

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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 包 (v0.2.1) 创建于 2019-02-27

这是一个疯狂的方式：

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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

来源：https://www.codenong.com/54917703/

How to convert indicator columns to a concatenated column (of column names)

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