Question

我有一个类似于以下内容的数据框：

library(tidyverse)
set.seed(4214)

df <- data.frame(value = sample(x = 1:50, 70, replace = TRUE),
                 group = sample(x = letters, 70, replace = TRUE),
                 stringsAsFactors = FALSE) %>% 
  as_tibble() %>% 
  arrange(group)

其中group是我的分组变量，并且每个值以不同的频率出现（例如group == "a"出现5次，group == "b"出现6次，依此类推）。

我需要将此数据尽可能平均地分成n = 9个子数据帧。但是，要注意的是，我无法在子集之间拆分相同的分组变量。例如，group == "b"不能同时出现在子集1和子集2中。

n <- 9
df %>% 
  mutate(divider = rep(x = 1:n, 
                       each = ceiling(nrow(.)/n), 
                       length.out = nrow(.))) %>%
  split(.$divider)

在这里，我创建了一个divider列，希望将数据分成子集。但是group的给定值可能具有divider的两个不同值。因此，此处将分组变量划分为子集。我一直在尝试使用nest和lag来改善这一点，但到目前为止没有成功。

我知道子集的行号将不相等，但是我希望有以下类似的东西：

$`1`
# A tibble: 11 x 3
  value group divider
  <int> <chr>   <int>
1    43 a           1
2    22 a           1
3     1 a           1
4     5 a           1
5     4 a           1
6    18 b           1
7    32 b           1
8    33 b           1
9    47 b           1
10   43 b           1
11   35 b           1

$`2`
# A tibble: 6 x 3
  value group divider
  <int> <chr>   <int>
1    24 c           2
2     3 d           2
3    12 d           2
4    13 e           2
5     6 e           2
6    45 f           2

$`3`
...

Answer 1

一种实现方法，但这取决于数据的顺序，是按组对实例进行计数，并用与所需组数最接近的整数将它们分开。

如果需要9组，则将累积频率相加并除以9。取整数并将其用作数据集的新拆分变量

dftab <- as.data.frame(table(df$group)) %>%
  mutate(nobs = cumsum(Freq),
         newgrouping = ceiling(nobs/9)) %>%
  group_by(newgrouping ) %>%
  summarise(number_obs = sum(Freq))

dftab

# A tibble: 8 x 2
  newgrouping number_obs
        <dbl>      <int>
1           1          5
2           2         12
3           3          9
4           4         10
5           5          9
6           6          7
7           7         11
8           8          7

对于“尽可能均匀”，我们可以对各组观测值的标准差进行愚蠢的优化。在这里，依靠组变量的顺序可以帮助完成此过程。

set.seed(4214)

df <- data.frame(value = sample(x = 1:50, 70, replace = TRUE),
                 group = sample(x = letters, 70, replace = TRUE),
                 stringsAsFactors = FALSE) %>% 
  as_tibble() %>% 
  arrange(group)


store_group <- list()
store_sd <- NA_integer_

for(i in 1:1000){

  dftab <- table(df$group) %>%
    as.data.frame() %>% 

    # important step is to shuffle the group variable every iteration
    mutate(group = factor(Var1, levels = df$group %>%
                            unique %>%
                            sample)) %>%
    arrange(group) %>%

    mutate(nobs = cumsum(Freq),
           newgrouping = ceiling(nobs/9)) %>%

    select(newgrouping, group, Freq)

  store_group[[i]] <- dftab

  df_sd <- dftab %>%
    group_by(newgrouping) %>%
    summarise(number_obs = sum(Freq))

  store_sd[i] <- sd(df_sd$number_obs)
}

这将导致

store_group[[which.min(store_sd)]] %>%
       group_by(newgrouping) %>%
       summarise(number_obs = sum(Freq))

  newgrouping number_obs
        <dbl>      <int>
1           1          9
2           2          9
3           3          9
4           4          8
5           5          9
6           6          9
7           7          8
8           8          9

其中store_group[[which.min(store_sd)]]拥有原始数据，并具有“最佳”分组的可能（给定循环中的迭代次数），而当您按{{1}拆分数据集时，整个数据集没有相同的group }变量

Answer 2

假设您想要按字母顺序排列的解决方案，如预期输出所示；您可以将cumsum除以所需的分割数（即9），以改变上限和下限，并更均匀地分配组。这将导致向量x，其中分裂指示符已分配给group变量的每个类别。 x单独拆分，然后给出一个列表，可使用lapply拆分数据帧。

x <- round(cumsum(table(dat$group)) / (nrow(dat) / 9))
result <- lapply(lapply(split(x, x), names), function(i) dat[dat$group %in% i, ])

行在结果列表中的分布

t(Map(nrow, result))
#      1  2 3 4 5 6 7 8 9
# [1,] 11 6 9 8 7 7 8 7 7

> sapply(result, "[", 2)
$`1.group`
 [1] "a" "a" "a" "a" "a" "b" "b" "b" "b" "b" "b"

$`2.group`
[1] "c" "d" "d" "e" "e" "f"

$`3.group`
[1] "g" "g" "g" "g" "i" "j" "j" "j" "j"

$`4.group`
[1] "k" "k" "l" "l" "l" "l" "l" "l"

$`5.group`
[1] "n" "n" "o" "p" "p" "p" "p"

$`6.group`
[1] "q" "q" "q" "q" "r" "r" "r"

$`7.group`
[1] "s" "s" "s" "t" "u" "u" "u" "v"

$`8.group`
[1] "w" "w" "w" "x" "x" "x" "x"

$`9.group`
[1] "y" "y" "y" "y" "z" "z" "z"

数据

dat <- structure(list(value = c(43L, 22L, 1L, 5L, 4L, 18L, 32L, 33L, 
47L, 43L, 35L, 24L, 3L, 12L, 13L, 6L, 45L, 12L, 5L, 22L, 47L, 
35L, 20L, 36L, 34L, 15L, 22L, 9L, 41L, 1L, 7L, 2L, 21L, 3L, 8L, 
33L, 12L, 39L, 19L, 2L, 34L, 45L, 7L, 22L, 24L, 25L, 20L, 19L, 
45L, 36L, 25L, 23L, 47L, 13L, 45L, 36L, 23L, 14L, 12L, 15L, 12L, 
11L, 25L, 31L, 41L, 14L, 38L, 15L, 13L, 6L), group = c("a", "a", 
"a", "a", "a", "b", "b", "b", "b", "b", "b", "c", "d", "d", "e", 
"e", "f", "g", "g", "g", "g", "i", "j", "j", "j", "j", "k", "k", 
"l", "l", "l", "l", "l", "l", "n", "n", "o", "p", "p", "p", "p", 
"q", "q", "q", "q", "r", "r", "r", "s", "s", "s", "t", "u", "u", 
"u", "v", "w", "w", "w", "x", "x", "x", "x", "y", "y", "y", "y", 
"z", "z", "z")), row.names = c(6L, 21L, 50L, 66L, 69L, 15L, 36L, 
46L, 48L, 62L, 67L, 34L, 18L, 54L, 31L, 51L, 3L, 7L, 9L, 24L, 
39L, 55L, 8L, 11L, 27L, 29L, 59L, 70L, 19L, 23L, 40L, 45L, 52L, 
68L, 26L, 43L, 44L, 16L, 38L, 63L, 65L, 10L, 49L, 56L, 61L, 1L, 
13L, 64L, 22L, 35L, 47L, 4L, 25L, 33L, 53L, 37L, 14L, 17L, 60L, 
2L, 5L, 12L, 57L, 28L, 32L, 41L, 42L, 20L, 30L, 58L), class = "data.frame")

均匀分割数据框而不分离分组变量

2 个答案: