如果我没理解错的话：purrr::map_df()函数有一个很好的特性，可以识别final data.frames中的输入列表（理想的命名）：

library(dplyr)
library(tidyr)
# dropping V5 as it will be "automatically" computed
result = function(input, teams) {
  data = teams %>%
  separate_rows(V2) %>%
  left_join(input, by = c("V2" = "V2")) %>%
  replace_na(list(V4 = 0)) %>%
  group_by(V1.x) %>% 
  fill(V1.y, V3) %>%
  summarize(V1.y = first(V1.y),
            V2 = paste(V2, collapse = " + "),
            V3 = first(V3),
            V4 = sum(V4)
            # we do not need V5 here anymore
            )
  return(data)
}

l_objects <- ls()
input_objects <- grep("^input_\\d", all_objects, value = T)
input_test <- lapply(input_objects, get)

# name the object list to use the name for identification
names(input_test) <- input_objects

# use purrr map to data.frame with the .id feature
purrr::map_df(input_test, ~result(.x, teams), .id = "V5")

  V5      V1.x  V1.y      V2           V3        V4
   <chr>   <chr> <chr>     <chr>        <chr>  <dbl>
 1 input_1 team1 Team_2022 Mary + Frank Sydney   131
 2 input_1 team2 Team_2022 Mary + John  Sydney    90
 3 input_1 team3 Team_2022 Mary + Bill  Sydney    76
 4 input_2 team1 Team_2023 Mary + Frank Sydney    23
 5 input_2 team2 Team_2023 Mary + John  Sydney    33
 6 input_2 team3 Team_2023 Mary + Bill  Sydney   136
 7 input_3 team1 Team_2024 Mary + Frank Sydney    26
 8 input_3 team2 Team_2024 Mary + John  Sydney    19
 9 input_3 team3 Team_2024 Mary + Bill  Sydney    71
10 input_4 team1 Team_2025 Mary + Frank Sydney    90
11 input_4 team2 Team_2025 Mary + John  Sydney   132
12 input_4 team3 Team_2025 Mary + Bill  Sydney    44

请注意，这在不命名列表的情况下也可以工作，尽管您将只获得列表项编号，这可能是不够的。
plyr::ldply函数也可以用于绑定命名列表，在结果data.frame中生成一个包含列表名称的新列。

与deparse/substitute不同，在result中也为名称创建一个参数，然后使用该参数

library(dplyr)
library(purrr)
library(tidyr)
result <- function(input, teams, inputnm) {
  data = teams %>%
  separate_rows(V2) %>%
  left_join(input, by = c("V2" = "V2")) %>%
  replace_na(list(V4 = 0)) %>%
  group_by(V1.x) %>% fill(V1.y, V3) %>%
  summarize(V1.y = first(V1.y),
            V2 = paste(V2, collapse = " + "),
            V3 = first(V3),
            V4 = sum(V4),
            V5 = inputnm)
  return(data)
}

• 测试

input_test %>% 
   pull(input_objects) %>%
   mget(inherits = TRUE) %>%
   imap_dfr(~ result(.x, teams, .y))

• 输出

# A tibble: 12 × 6
   V1.x  V1.y      V2           V3        V4 V5     
   <chr> <chr>     <chr>        <chr>  <dbl> <chr>  
 1 team1 Team_2022 Mary + Frank Sydney   131 input_1
 2 team2 Team_2022 Mary + John  Sydney    90 input_1
 3 team3 Team_2022 Mary + Bill  Sydney    76 input_1
 4 team1 Team_2023 Mary + Frank Sydney    23 input_2
 5 team2 Team_2023 Mary + John  Sydney    33 input_2
 6 team3 Team_2023 Mary + Bill  Sydney   136 input_2
 7 team1 Team_2024 Mary + Frank Sydney    26 input_3
 8 team2 Team_2024 Mary + John  Sydney    19 input_3
 9 team3 Team_2024 Mary + Bill  Sydney    71 input_3
10 team1 Team_2025 Mary + Frank Sydney    90 input_4
11 team2 Team_2025 Mary + John  Sydney   132 input_4
12 team3 Team_2025 Mary + Bill  Sydney    44 input_4

如果一次只有一个输入，只需将指定V5的行移出summarize和dplyr管道即可：

result = function(input, teams) {
  data = teams %>%
    separate_rows(V2) %>%
    left_join(input, by = c("V2" = "V2")) %>%
    replace_na(list(V4 = 0)) %>%
    group_by(V1.x) %>% fill(V1.y, V3) %>%
    summarize(V1.y = first(V1.y),
              V2 = paste(V2, collapse = " + "),
              V3 = first(V3),
              V4 = sum(V4),
              #V5 = print(deparse(substitute(input)))
              )
  data$V5 <- deparse(substitute(input))
  return(data)
}

result(input_1, teams)

#  V1.x   V1.y       V2           V3        V4 V5     
#  <chr> <chr>     <chr>        <chr>  <dbl> <chr>  
#  1 team1 Team_2022 Mary + Frank Sydney   131 input_1
#  2 team2 Team_2022 Mary + John  Sydney    90 input_1
#  3 team3 Team_2022 Mary + Bill  Sydney    76 input_1

但是列表的使用使得这有点困难（对于该代码，它将输出X[[i]]）。
为了解决使用lapply时的这个问题，我建议简单地添加一个接受名称的附加输入，并简单地将该名称赋给它，同时对lapply函数进行额外的调整以适应它：

result = function(input, teams, nme) {
  data = teams %>%
    separate_rows(V2) %>%
    left_join(input, by = c("V2" = "V2")) %>%
    replace_na(list(V4 = 0)) %>%
    group_by(V1.x) %>% fill(V1.y, V3) %>%
    summarize(V1.y = first(V1.y),
              V2 = paste(V2, collapse = " + "),
              V3 = first(V3),
              V4 = sum(V4),
              #V5 = print(deparse(substitute(input)))
              )
  data$V5 <- nme 
  return(data)
}

all_objects <- ls()
input_objects <- grep("^input_\\d", all_objects, value = T)
input_test <- lapply(input_objects, get)

# add in assigning names to the list
names(input_test) <- input_objects 

output = lapply(input_objects, function(x) result(input_test[[x]], teams, nme = x)) %>%
  bind_rows()

#     V1.x  V1.y      V2           V3        V4 V5     
#     <chr> <chr>     <chr>        <chr>  <dbl> <chr>  
#   1 team1 Team_2022 Mary + Frank Sydney   131 input_1
#   2 team2 Team_2022 Mary + John  Sydney    90 input_1
#   3 team3 Team_2022 Mary + Bill  Sydney    76 input_1
#   4 team1 Team_2023 Mary + Frank Sydney    23 input_2
#   5 team2 Team_2023 Mary + John  Sydney    33 input_2
#   6 team3 Team_2023 Mary + Bill  Sydney   136 input_2
#   7 team1 Team_2024 Mary + Frank Sydney    26 input_3
#   8 team2 Team_2024 Mary + John  Sydney    19 input_3
#   9 team3 Team_2024 Mary + Bill  Sydney    71 input_3
#  10 team1 Team_2025 Mary + Frank Sydney    90 input_4
#  11 team2 Team_2025 Mary + John  Sydney   132 input_4
#  12 team3 Team_2025 Mary + Bill  Sydney    44 input_4