dplyr 通过比较变量和不同大小的向量来改变变量

Question

我有以下类型的数据框

df <- tibble::tribble(~x,
                      c("A", "B"),
                      c("A", "B", "C"),
                      c("A", "B", "C", "D"),
                      c("A", "B"))

和像这样的载体

vec1 <- c("A", "B")
vec2 <- c("A", "B", "C")
vec3 <- c("A", "B", "C", "D")

我想改变一个变量 y 来显示哪一行有哪个向量。我尝试了以下操作，但得到了带有警告的空 y 变量："longer object length is not a multiple of shorter object length"

df_new <- df %>%
  mutate(y = case_when(x == vec1 ~ "vec1",
                       x == vec2 ~ "vec2",
                       x == vec2 ~ "vec3"))

期望的输出是

df_new <- tibble::tribble(~x,                      ~y,
                          c("A", "B"),             "vec1",
                          c("A", "B", "C"),        "vec2",
                          c("A", "B", "C", "D"),   "vec3",
                          c("A", "B"),             "vec1")

Answer 1

使用 map2_lgl 和 identical 评估向量是否相同的解决方案。

library(tidyverse)

df_new <- df %>%
  mutate(y = case_when(
    map2_lgl(x, list(vec1), ~identical(.x, .y))  ~"vec1",
    map2_lgl(x, list(vec2), ~identical(.x, .y))  ~"vec2",
    map2_lgl(x, list(vec3), ~identical(.x, .y))  ~"vec3"
  ))
df_new
# # A tibble: 4 x 2
#   x         y    
#   <list>    <chr>
# 1 <chr [2]> vec1 
# 2 <chr [3]> vec2 
# 3 <chr [4]> vec3 
# 4 <chr [2]> vec1

Answer 2

这是一个更具编程性的替代方案 - 您不需要明确指定每个向量

数据

df <- tibble::tribble(~x,
                      c("A", "B"),
                      c("A", "B", "C"),
                      c("A", "B", "C", "D"),
                      c("A", "B"))

vec1 <- c("A", "B")
vec2 <- c("A", "B", "C")
vec3 <- c("A", "B", "C", "D")

解决方案 - 利用 ls(...) 到 return 相关矢量名称使用模式

vecs <- ls(pattern="vec")
L <- lapply(vecs, get)
names(L) <- vecs
df %>%
  mutate(y = names(L)[match(x, L)])

# A tibble: 4 x 2
  # x         y    
  # <list>    <chr>
# 1 <chr [2]> vec1 
# 2 <chr [3]> vec2 
# 3 <chr [4]> vec3 
# 4 <chr [2]> vec1

Answer 3

这也行得通：

comp <- list(vec1, vec2, vec3)

df %>% 
  mutate(y = map_chr(df$x, ~ paste0("vec", which(comp %in% list(.)))))

# A tibble: 4 x 2
  x         y    
  <list>    <chr>
1 <chr [2]> vec1 
2 <chr [3]> vec2 
3 <chr [4]> vec3 
4 <chr [2]> vec1

Answer 4

一个选项是使用 compare::compareEqual 和 apply 作为：

library(dplyr)
library(compare)

df$y <- apply(df,1, function(x){
  dplyr::case_when(
    isTRUE(compareEqual(x[[1]], vec1)) ~ "vec1",
    isTRUE(compareEqual(x[[1]], vec2)) ~ "vec2",
    isTRUE(compareEqual(x[[1]],vec3)) ~ "vec3"
  )})



df
# # A tibble: 4 x 2
# x         y    
# <list>    <chr>
# 1 <chr [2]> vec1 
# 2 <chr [3]> vec2 
# 3 <chr [4]> vec3 
# 4 <chr [2]> vec1

Answer 5

很遗憾，您不能对异国情调的格式进行 left_join，但我们可以欺骗并加入 dput 字符串：

library(tidyverse)
lkp <- enframe(map_chr(lst(vec1,vec2,vec3),~capture.output(dput(.x))))
df %>%
  mutate(value = map_chr(x,~capture.output(dput(.x)))) %>%
  left_join(lkp) %>%
  select(-value)

# # A tibble: 4 x 2
#           x  name
#      <list> <chr>
# 1 <chr [2]>  vec1
# 2 <chr [3]>  vec2
# 3 <chr [4]>  vec3
# 4 <chr [2]>  vec1

dplyr 通过比较变量和不同大小的向量来改变变量

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