我有以下数据集：
dat = structure(list(C86_1981 = c("Outer London", "Buckinghamshire",
NA, "Ross and Cromarty", "Cornwall and Isles of Scilly", NA,
"Kirkcaldy", "Devon", "Kent", "Renfrew"), C96_1981 = c("Outer London",
"Buckinghamshire", NA, "Ross and Cromarty", "Not known/missing",
NA, "Kirkcaldy", NA, NA, NA), C00_1981 = c("Outer London", "Inner London",
"Lancashire", "Ross and Cromarty", NA, "Humberside", "Kirkcaldy",
NA, NA, NA), C04_1981 = c("Kent", NA, NA, "Ross and Cromarty",
NA, "Humberside", "Not known/missing", NA, NA, "Renfrew"), C08_1981 = c("Kent",
"Oxfordshire", NA, "Ross and Cromarty", "Cornwall and Isles of Scilly",
"Humberside", "Dunfermline", NA, NA, "Renfrew"), C12_1981 = c("Kent",
NA, NA, "Ross and Cromarty", "Cornwall and Isles of Scilly",
"Humberside", "Dunfermline", NA, NA, "Renfrew")), row.names = c(NA,
-10L), class = c("tbl_df", "tbl", "data.frame"), .Names = c("C86_1981",
"C96_1981", "C00_1981", "C04_1981", "C08_1981", "C12_1981"))
我希望每列dplyr::count()。预期产出：
# A tibble: 10 x 3
C86_1981 dat86_n dat96_n ...
<chr> <int> <int>
1 Buckinghamshire 1 1
2 Cornwall and Isles of Scilly 1 NA
3 Devon 1 NA
4 Kent 1 NA
5 Kirkcaldy 1 1
6 Outer London 1 1
7 Renfrew 1 NA
8 Ross and Cromarty 1 1
9 <NA> 2 5
10 Not known/missing NA 1
目前我手动执行此操作，然后dplyr::full_join()结果：
library("tidyverse")
dat86_n = dat %>%
count(C86_1981) %>%
rename(dat86_n = n)
dat96_n = dat %>%
count(C96_1981) %>%
rename(dat96_n = n)
# ...
dat_counts = dat86_n %>%
full_join(dat96_n, by = c("C86_1981" = "C96_1981"))
# ...
哪个有效，但如果我以后的任何数据发生变化，则不完全健壮。我曾希望以编程方式执行此操作。
我试过一个循环：
lapply(dat, count)
# Error in UseMethod("groups") :
# no applicable method for 'groups' applied to an object of class "character"
（purrr::map()给出了同样的错误）。我认为这个错误是因为count()期望tbl和变量作为单独的参数，所以我也尝试过：
lapply(dat, function(x) {
count(dat, x)
})
# Error in grouped_df_impl(data, unname(vars), drop) :
# Column `x` is unknown
同样，purrr::map()给出了同样的错误。我还尝试了summarise_all()的变体：
dat %>%
summarise_all(count)
# Error in summarise_impl(.data, dots) :
# Evaluation error: no applicable method for 'groups' applied to an object of class "character".
我觉得我错过了一些明显的东西，解决方案应该是直截了当的。 dplyr解决方案特别受欢迎，因为这是我最常使用的。
答案 0 :(得分：4)
使用tidyr包，以下代码可以解决问题：
dat %>% tidyr::gather(name, city) %>% dplyr::group_by(name, city) %>% dplyr::count() %>% dplyr::ungroup %>% tidyr::spread(name, n)
结果：
# A tibble: 15 x 7
city C00_1981 C04_1981 C08_1981 C12_1981 C86_1981 C96_1981
* <chr> <int> <int> <int> <int> <int> <int>
1 Buckinghamshire NA NA NA NA 1 1
2 Cornwall and Isles of Scilly NA NA 1 1 1 NA
3 Devon NA NA NA NA 1 NA
4 Dunfermline NA NA 1 1 NA NA
5 Humberside 1 1 1 1 NA NA
6 Inner London 1 NA NA NA NA NA
7 Kent NA 1 1 1 1 NA
8 Kirkcaldy 1 NA NA NA 1 1
9 Lancashire 1 NA NA NA NA NA
10 Not known/missing NA 1 NA NA NA 1
11 Outer London 1 NA NA NA 1 1
12 Oxfordshire NA NA 1 NA NA NA
13 Renfrew NA 1 1 1 1 NA
14 Ross and Cromarty 1 1 1 1 1 1
15 <NA> 4 5 3 4 2 5
答案 1 :(得分：3)
@ You-leee只是打败了我;）
使用tidyverse;
library(tidyverse)
df <-
dat %>%
gather (year, county) %>%
group_by(year, county) %>%
summarise(no = n()) %>%
spread (year, no)
# A tibble: 15 x 7
county C00_1981 C04_1981 C08_1981 C12_1981 C86_1981 C96_1981
* <chr> <int> <int> <int> <int> <int> <int>
1 Buckinghamshire NA NA NA NA 1 1
2 Cornwall and Isles of Scilly NA NA 1 1 1 NA
3 Devon NA NA NA NA 1 NA
4 Dunfermline NA NA 1 1 NA NA
5 Humberside 1 1 1 1 NA NA
6 Inner London 1 NA NA NA NA NA
7 Kent NA 1 1 1 1 NA
8 Kirkcaldy 1 NA NA NA 1 1
9 Lancashire 1 NA NA NA NA NA
10 Not known/missing NA 1 NA NA NA 1
11 Outer London 1 NA NA NA 1 1
12 Oxfordshire NA NA 1 NA NA NA
13 Renfrew NA 1 1 1 1 NA
14 Ross and Cromarty 1 1 1 1 1 1
15 <NA> 4 5 3 4 2 5
答案 2 :(得分：1)
先前带有gather +count+spread的答案很好用，但不适用于非常大的数据集（大型组或许多变量）。这是一种替代方法，使用map-count + join可以处理非常大的数据，速度似乎快了2倍：
library(tidyverse)
N <- 1000000
df <- tibble(x1=sample(letters, N, replace = TRUE),
x2=sample(letters, N, replace = TRUE),
x3=sample(letters, N, replace = TRUE),
x4=sample(letters, N, replace = TRUE),
x5=sample(letters, N, replace = TRUE))
res1 <- map(c("x1", "x2", "x3", "x4", "x5"), function(x) select_at(df, x) %>% count(!!rlang::sym(x)) %>%
rename(value=!!rlang::sym(x),
!!rlang::sym(x):=n)) %>%
reduce(full_join, by = "value")
res2 <- df %>%
tidyr::gather(variable, value) %>%
dplyr::group_by(variable, value) %>%
dplyr::count() %>% dplyr::ungroup()%>%
tidyr::spread(variable, n)
all.equal(res1, res2)
#> [1] TRUE
library(microbenchmark)
microbenchmark(s1=map(c("x1", "x2", "x3", "x4", "x5"), function(x) select_at(df, x) %>% count(!!rlang::sym(x)) %>%
rename(value=!!rlang::sym(x),
!!rlang::sym(x):=n)) %>%
reduce(full_join, by = "value"),
s2= df %>%
tidyr::gather(variable, value) %>%
dplyr::group_by(variable, value) %>%
dplyr::count() %>% dplyr::ungroup()%>%
tidyr::spread(variable, n),
times = 50, check = "equal")
#> Unit: milliseconds
#> expr min lq mean median uq max neval
#> s1 214.9027 220.2292 241.8811 229.0913 242.2507 368.5147 50
#> s2 412.8934 447.5347 515.2612 528.0221 561.7649 692.5999 50
由reprex package（v0.3.0）于2020-05-19创建