一种方法，首先在列表的每个数据框中添加年份，然后按 * 国籍 * 分列。

library(dplyr)

years <- c(2022, 2023) # one year per data frame

df_list <- list(df_1, df_2)

lapply(1:length(df_list), function(x) cbind(df_list[[x]], year=years[x])) %>% 
  bind_rows() %>% 
  group_split(nationality)

输出

<list_of<
  tbl_df<
    nationality: character
    total      : double
    male       : double
    year       : integer
  >
>[2]>
[[1]]
# A tibble: 2 × 4
  nationality total  male  year
  <chr>       <dbl> <dbl> <int>
1 german        500   400  2022
2 german        800   300  2023

[[2]]
# A tibble: 2 × 4
  nationality total  male  year
  <chr>       <dbl> <dbl> <int>
1 italian       300   100  2022
2 italian       600    80  2023

正如Jon Spring所要求的，您必须在加入之前使用年份注解数据框，例如，如果您的数据框位于命名列表中，其中名称表示年份：

all_years <- list(year1 = df_1, year2 = df_2)

你可以在tidyverse中做一些事情，比如：

library(tidyverse)
all_years %>% 
  names() %>% 
  map(\(year)
      {all_years[[year]] %>% rename_with(~paste(., year, sep="_"),
                                         -nationality)}) %>% 
  purrr::reduce(left_join, by="nationality") %>% 
  pivot_longer(-nationality, names_to="population_year") %>% 
  separate("population_year", c("population", "year"), sep="_") %>% 
  pivot_wider(id_cols = c(population,nationality), 
              names_from = year, 
              values_from = value) %>% 
  group_split(nationality) %>% 
  set_names(map(., ~.$nationality[1]))

生成的tibble列表包含

$german
# A tibble: 2 × 4
  population nationality year1 year2
  <chr>      <chr>       <dbl> <dbl>
1 total      german        500   800
2 male       german        400   300

$italian
# A tibble: 2 × 4
  population nationality year1 year2
  <chr>      <chr>       <dbl> <dbl>
1 total      italian       300   600
2 male       italian       100    80

这个解决方案使用了一个连接（而不是绑定）操作，这样在 Dataframe 格式不相同的情况下会更安全，例如ncol不同。