如何通过从列表向量列粘贴来制作单个字符串列？

Question

我正在尝试清理表以用作索引。由于一些条目对应于多个列，因此我想在必要时连接成一个字符串。我正在尝试做的例子是：

tibble(
 day = c("Mo", "Tu","Tu", "We","Th","Fr","Fr","Fr"),
 see = c("cat", "cat", "dog", NA, "cat","cat","dog","bird" )
) %>% 
 nest(data = see) %>% 
 mutate(see = paste0(data)) %>%  
 select(-data)

这几乎可以工作，但是 paste0 不只是粘贴列表中的数据，而是：

# A tibble: 5 × 2
  day   see                                        
  <chr> <chr>                                      
1 Mo    "list(see = \"cat\")"                      
2 Tu    "list(see = c(\"cat\", \"dog\"))"          
3 We    "list(see = NA)"                           
4 Th    "list(see = \"cat\")"                      
5 Fr    "list(see = c(\"cat\", \"dog\", \"bird\"))"

我需要改变什么？我希望

see

列仅包含逗号或空格分隔的字符串，或者在 NA.

处保持为空

Answer 1

我认为我们不需要

nest()

，但可以用

summarise()

代替

paste(..., collapse = ", ")

。

library(dplyr)

tibble(
  day = c("Mo", "Tu","Tu", "We","Th","Fr","Fr","Fr"),
  see = c("cat", "cat", "dog", NA, "cat","cat","dog","bird" )
) %>% 
  summarise(data = paste(see, collapse = ", "), .by = "day")

#> # A tibble: 5 × 2
#>   day   data          
#>   <chr> <chr>         
#> 1 Mo    cat           
#> 2 Tu    cat, dog      
#> 3 We    NA            
#> 4 Th    cat           
#> 5 Fr    cat, dog, bird

^{创建于 2023-03-16 与 reprex v2.0.2}

Answer 2

另一种方法是使用

toString

：

tibble(
  day = c("Mo", "Tu","Tu", "We","Th","Fr","Fr","Fr"),
  see = c("cat", "cat", "dog", NA, "cat","cat","dog","bird" )
) %>% 
  group_by(day) %>% 
  summarise(see = toString(see))
# A tibble: 5 × 2
  day   see           
  <chr> <chr>         
1 Fr    cat, dog, bird
2 Mo    cat           
3 Th    cat           
4 Tu    cat, dog      
5 We    NA   

Answer 3

您可以尝试

tidyverse

解决方案

stringr::str_flatten_comma

：

library(dplyr) #1.1.0+
library(stringr)
tib <- tibble(
 day = c("Mo", "Tu","Tu", "We","Th","Fr","Fr","Fr"),
 see = c("cat", "cat", "dog", NA, "cat","cat","dog","bird" )
)

tib %>% 
  summarise(see = stringr::str_flatten_comma(see), .by = day)

# A tibble: 5 × 2
  day   see           
  <chr> <chr>         
1 Mo    cat           
2 Tu    cat, dog      
3 We    NA            
4 Th    cat           
5 Fr    cat, dog, bird

Answer 4

为了完整性，

data.table

方式

library(data.table)
setDT(df)

df[, lapply(.SD, toString), day]