关于unnest

library(dplyr)
library(tidyr)

df %>% 
  mutate(species = strsplit(species, ",")) %>% 
  unnest(species)
# A tibble: 12 × 3
   species time    loc
   <chr>   <chr> <int>
 1 Barbar  9:30      1
 2 Barbar  9:30      1
 3 Barbar  10:37     4
 4 Barbar  12:03     2
 5 Pippip  12:03     2
 6 Barbar  09:52     5
 7 Pippip  09:52     5
 8 Hypsav  09:52     5
 9 Pippip  07:45     5
10 Barbar  07:45     5
11 Barbar  00:00     3
12 Pippip  00:00     3

数据

df <- structure(list(species = c("Barbar,Barbar", "Barbar", "Barbar,Pippip", 
"Barbar,Pippip,Hypsav", "Pippip,Barbar", "Barbar,Pippip"), time = c("9:30", 
"10:37", "12:03", "09:52", "07:45", "00:00"), loc = c(1L, 4L, 
2L, 5L, 5L, 3L)), class = "data.frame", row.names = c(NA, -6L))

尝试使用以下语句，使用,作为当前分隔符

library(tidyverse)

data_split <- data %>%
  separate(species, into = c("species1", "species2", "species3"), sep = ",") %>%
  pivot_longer(cols = starts_with("species"), values_to = "species") %>%
  filter(!is.na(species)) %>%
  select(-name)
  print(data_split)

或者，可以使用data.table方法：

library(data.table)

# convert df to a data.table
setDT(df)

# at first split the species and reassign it to the same column
# then unlist to distribute the species for every "loc" and "time"
df[,species:=strsplit(x = species, split = ","),][
  ,.(species = unlist(species)), by=.(loc,time)]

#    loc  time species
# 1:   1  9:30  Barbar
# 2:   1  9:30  Barbar
# 3:   4 10:37  Barbar
# 4:   2 12:03  Barbar
# 5:   2 12:03  Pippip
# 6:   5 09:52  Barbar
# 7:   5 09:52  Pippip
# 8:   5 09:52  Hypsav
# 9:   5 07:45  Pippip
#10:   5 07:45  Barbar
#11:   3 00:00  Barbar
#12:   3 00:00  Pippip

根据您的一般工作流程或数据大小，您可以评估最适合您的工作。