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如何返回不同组中最后n行的平均值(由变量表示)

tidyverse dplyr r

jakes · 技术社区 · 6 年前

data <- structure(list(seq = c(1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 
4L, 5L, 5L, 5L, 5L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 
6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 7L, 7L, 7L, 
7L, 7L, 8L, 8L, 9L, 9L, 9L, 10L, 10L, 10L), new_seq = c(2, 2, 
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 
2, 2, 2, 2, NA, NA, NA, NA, NA, 4, 4, 4, 4, NA, NA, NA, NA, NA, 
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 
NA, NA, NA, 6, 6, 6, 6, 6, NA, NA, 8, 8, 8, NA, NA, NA), value = c(2L, 
0L, 0L, 1L, 0L, 5L, 5L, 3L, 0L, 3L, 2L, 3L, 2L, 3L, 4L, 1L, 0L, 
0L, 0L, 1L, 1L, 0L, 2L, 5L, 3L, 0L, 1L, 0L, 0L, 0L, 1L, 1L, 3L, 
5L, 3L, 1L, 1L, 1L, 0L, 1L, 0L, 4L, 3L, 0L, 3L, 1L, 3L, 0L, 0L, 
1L, 0L, 0L, 3L, 4L, 5L, 3L, 5L, 3L, 5L, 0L, 1L, 1L, 3L, 2L, 1L, 
0L, 0L, 0L, 0L, 5L, 1L, 1L, 0L, 4L, 1L, 5L, 0L, 3L, 1L, 2L, 1L, 
0L, 3L, 0L, 1L, 1L, 3L, 0L, 1L, 1L, 2L, 2L, 1L, 0L, 4L, 0L, 0L, 
3L, 0L, 0L)), row.names = c(NA, -100L), class = c("tbl_df", "tbl", 
"data.frame"))

列 new_seq 指的是 seq . 对于中的每个值 新建\u seq 哪一个不是 NA 2 行 value 序号 . 例如,行 1:2 新列的值应为 0.5 (行的平均值) 49:50 ),行 51:54 0.5 49:50 也一样),但是 60:63 4 (行的平均值) 58:59 ). 我怎么能用它呢 tidyverse

2 回复 | 直到 6 年前

Z.Lin 6 年前

像这样的?

# calculate the mean value based on the last two rows of each seq
lookup <- data %>%
  group_by(seq) %>%
  mutate(rank = seq(n(), 1)) %>% 
  filter(rank <= 2) %>%
  summarise(new_column = mean(value)) %>%
  ungroup()

# match back to original dataset (only non-NA values of new_seq can be matched)
left_join(data, lookup, by = c("new_seq" = "seq"))

# A tibble: 100 x 4
     seq new_seq value new.column
   <int>   <dbl> <int>      <dbl>
 1     1       2     2        0.5
 2     1       2     0        0.5
 3     2      NA     0       NA  
 4     2      NA     1       NA  
...

divibisan 6 年前

嗯,只有一半 tidyverse 我相信有人可以做得更好,但这里有一个尝试。

group_by mutate 使计算组中最后两行的平均值变得容易,但我不知道如何获得组之间的连接 seq 和 new_seq

dat2 <- dat %>%
    group_by(seq) %>%
    mutate(end_val = (nth(value, -1L) + nth(value, -2L))/2)

dat3$result <- apply(dat2, 1, function(x) {
    dat2[dat2$seq == x['new_seq'], 'end_val'][[1]][1]
})

rowid

dat3 %>% tibble::rowid_to_column() %>% .[c(1:3,50:55,59:64),] 

# A tibble: 15 x 6
# Groups:   seq [6]
   rowid   seq new_seq value end_val result
   <int> <int>   <dbl> <int>   <dbl>  <dbl>
 1     1     1       2     2     1      0.5
 2     2     1       2     0     1      0.5
 3     3     2      NA     0     0.5   NA  
 4    50     2      NA     1     0.5   NA  
 5    51     3       2     0     3.5    0.5
 6    52     3       2     0     3.5    0.5
 7    53     3       2     3     3.5    0.5
 8    54     3       2     4     3.5    0.5
 9    55     4      NA     5     4     NA  
10    59     4      NA     5     4     NA  
11    60     5       4     0     2      4  
12    61     5       4     1     2      4  
13    62     5       4     1     2      4  
14    63     5       4     3     2      4  
15    64     6      NA     2     2     NA