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Question

蘋果	香蕉	橙	水果名稱
1	0	1	蘋果橙
1	0	0	蘋果
1	0	1	蘋果橙
1	1	1	蘋果香蕉橙

如果列的 rest 為 1，我想創建並改變列“frutis_name”。例如，蘋果和橙子得到 1，那么“frutis_name 將是蘋果空間香蕉。

Answer 1

使用purrr::map_chr() ：

df <- data.frame(apple = c(1,1),    banana = c(0,0),    orange = c(1,0))

df$fruit_name <- purrr::pmap_chr(df, ~ paste(names(df)[as.logical(c(...))], collapse = " "))
df
#>   apple banana orange   fruit_name
#> 1     1      0      1 apple orange
#> 2     1      0      0        apple

Answer 2

也許不是最簡單的解決方案，而是一個可行的解決方案，基於將列名重鑄為值的想法，這是由pivot_longer完成的：

library(tidyverse)
df %>%
  # create row ID:
  mutate(row = row_number()) %>%
  # cast longer so that fruit names become values:
  pivot_longer(-row, names_to = "fruits") %>%
  # for each combination of `row` and `value`...
  group_by(row, value) %>%
  # ... combine fruit names if `value == 1`:
  mutate(fruits = ifelse(value == 1, str_c(fruits, collapse = " "), fruits)) %>%
  # remove obsolete rows:
  filter(value == 1 & !duplicated(fruits)) %>%
  # deactivate grouping:
  ungroup() %>%
  # remove unnecessary columns:
  select(-c(value, row)) %>%
  # bind original `df` together with new `fruit` column:
  bind_cols(df, .)
  apple banana orange              fruits
1     1      0      1        apple orange
2     1      0      0               apple
3     1      0      1        apple orange
4     1      1      1 apple banana orange

Answer 3

df$fruits_name <- 
  apply(df, 1, \(x) paste(names(df)[as.logical(x)], collapse = " "))

#   apple banana orange         fruits_name
# 1     1      0      1        apple orange
# 2     1      0      0               apple
# 3     1      0      1        apple orange
# 4     1      1      1 apple banana orange

相同的邏輯但更高效

library(data.table)
df$fruits_name <- 
  vapply(transpose(df), \(x) paste(names(df)[as.logical(x)], collapse = " "), character(1L))

可重現的數據

df <- data.frame(
  apple  = 1,    
  banana = c(0,0,0,1),   
  orange = c(1,0,1,1)
)

Answer 4

您可以使用apply來做到這一點：

 x <- data.frame(apple = c(1,0,1), banana = c(0,0,1), orange = c(1,0,0))
 x$fruits_name <- apply(x, 1, function(y) paste(colnames(x)[y==1], collapse=" "), simplify = TRUE)
 x
 apple banana orange         fruits_name
 1     1      0      1 apple orange
 2     0      0      0              
 3     1      1      0 apple banana

編輯：我用collapse替換了sep參數，現在它產生了一個向量，而不是一個列表