请问各位如何选取一个月中最后五天的数据，由于是交易日，所以每个月最后五天都不一样，取完后还需要对这五天的数据进行平均，谢谢各位

barnett874 个人愚见：没有考虑跨年的情况，因此还需要增加 Year 变量，但是 Day 变量感觉没有必要。谨慎起见有必要对数据按照日期升序排列，然后按照 Year、Month分组。既然走 tidyverse 风格，可以考虑将 do(tail(., 5)) 换成 slice_tail(n = 5) 。

[未知用户] [未知用户] 原始数据不一样

[未知用户] 不需要呀。原代码已经排序了。

求助：如何选取一个月中最后五天的数据

chuxinyuan

barnett874 个人愚见：

没有考虑跨年的情况，因此还需要增加 Year 变量，但是 Day 变量感觉没有必要。
谨慎起见有必要对数据按照日期升序排列，然后按照 Year、Month分组。
既然走 tidyverse 风格，可以考虑将do(tail(., 5))换成slice_tail(n = 5)。

chuxinyuan

library(dplyr)
expand.grid(
  year = 2020:2021,
  month = 4:5,
  day = 14:19
  ) %>% 
  arrange(year, month, day) %>%
  mutate(value = 1:nrow(.)) %>%
  group_by(year, month) %>%
  slice_tail(n = 5) %>%
  summarise(
    mean = mean(value), 
    .groups = "keep"
  )
#> # A tibble: 4 x 3
#> # Groups:   year, month [4]
#>    year month  mean
#>   <int> <int> <dbl>
#> 1  2020     4     4
#> 2  2020     5    10
#> 3  2021     4    16
#> 4  2021     5    22

<sup>Created on 2021-05-17 by the reprex package (v0.3.0)</sup>

zzy98

chuxinyuan 谢谢兄弟，会了！

wangbinzjcc

来个base版：

dat <- expand.grid(year = 2015:2021, month = 1:5, day = 10:19)
dat$value <- seq_len(nrow(dat))

do.call(rbind, with(dat, by(
   dat[order(year, month, day), ],  interaction(year, month),
   function(i) with(i, c(year = year[1], month = month[1], value_mean = mean(tail(value, 5))))
)))

chuxinyuan

wangbinzjcc 貌似结果不对。但是我也看不出来哪里不对，我这 base R 功力太弱了。

ryo

wangbinzjcc wangbinzjcc

✓ [main*]> system.time({
     dat <- expand.grid(year = 2015:2021, month = 1:5, day = 10:19) |> 
         {\(i) within(i, value <- seq_len(nrow(i))) }();
     
     dat |> 
         split(~ month + year) |>
         lapply(\(i) with(i, c(year = year[1], month = month[1], value_mean = mean(tail(value[order(day)], 5))))) |>
         {\(i) do.call(rbind, i) }()  
 })
   user  system elapsed 
  0.009   0.000   0.009 
✓ [main*]> 
✓ [main*]> system.time({
     dat <- expand.grid(year = 2015:2021, month = 1:5, day = 10:19);
     dat$value <- seq_len(nrow(dat));
     
     dat <- dat[order(dat$year, dat$month, dat$day), ];
     do.call(rbind, by(dat, interaction(dat$month, dat$year),
                       function(i) c(year = i$year[1], month = i$month[1], value_mean = mean(tail(i$value, 5)))
     ))
 })
   user  system elapsed 
  0.034   0.093   0.125

barnett874

chuxinyuan
增加一行能好，但是baseR功力太弱，不知道怎么改的更好看

dat <- dat[order(dat$year, dat$month, dat$day),]

wangbinzjcc

更正base版：

dat <- expand.grid(year = 2015:2021, month = 1:5, day = 10:19)
dat$value <- seq_len(nrow(dat))
 
dat <- dat[order(dat$year, dat$month, dat$day), ]
do.call(rbind, by(dat, interaction(dat$month, dat$year),
   function(i) c(year = i$year[1], month = i$month[1], value_mean = mean(tail(i$value, 5)))
))

chuxinyuan barnett874 多谢二位的指正。

zovey_lee

dplyr解决方法

library(tidyverse)
#构造数据
data <- data.frame(
  stringsAsFactors = FALSE,
  date = c("2020-4-5","2020-4-6",
           "2020-4-7","2020-4-8","2020-4-9","2020-4-10","2020-5-15",
           "2020-5-16","2020-5-17","2020-5-18","2020-5-19",
           "2020-5-20","2020-5-21"),
  value = c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L)
)
data$date = as.Date(data$date)

# 解决方案
data %>% 
  slice_max(date, n = 5) %>% # 选中后5天
  summarise(mean = mean(value)) # 计算value的平均值

wangbinzjcc

R 4.1.0 版

dat <- expand.grid(year = 2015:2021, month = 1:5, day = 10:19) |> 
  {\(i) within(i, value <- seq_len(nrow(i))) }() 

dat |> 
  split(~ month + year) |>
  lapply(\(i) with(i, c(year = year[1], month = month[1], value_mean = mean(tail(value[order(day)], 5))))) |>
  {\(i) do.call(rbind, i) }()

chuxinyuan

wangbinzjcc ryo 多测算几次求平均这个思路没错，但是数据量太少了，说明不了问题。

重新开个帖子讨论这个问题吧！楼主的问题早就干净利索的解决了。

wangbinzjcc

test_4.1.0 <- function(){
  # R 4.1.0
  system.time({
    
    dat <- expand.grid(year = 2015:2021, month = 1:12, day = 1:31) |> 
      {\(i) within(i, value <- seq_len(nrow(i))) }()
    
    dat |> 
      split(~ month + year) |>
      lapply(\(i) with(i, c(year = year[1], month = month[1], value_mean = mean(tail(value[order(day)], 5))))) |>
      {\(i) do.call(rbind, i) }()  
    
    })
}

times_4.1.0 <- replicate(10, test_4.1.0()[1:3])
times_4.1.0
#>           [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
#> user.self    0    0    0    0    0 0.02 0.01 0.02    0  0.01
#> sys.self     0    0    0    0    0 0.00 0.00 0.00    0  0.00
#> elapsed      0    0    0    0    0 0.01 0.01 0.02    0  0.01





test_4.0.5 <- function(){
  # R 4.0.5
  system.time({
    
    dat <- expand.grid(year = 2015:2021, month = 1:12, day = 1:31)
    dat <- within(dat, value <- seq_len(nrow(dat)))
    
    do.call(rbind, lapply(
      split(dat, dat[, c("month", "year")]), 
      function(i) with(i, c(year = year[1], month = month[1], value_mean = mean(tail(value[order(day)], 5))))
      ))
    
  })
}

times_4.0.5 <- replicate(10, test_4.0.5()[1:3])
times_4.0.5
#>           [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
#> user.self    0    0    0 0.01    0 0.01    0    0    0     0
#> sys.self     0    0    0 0.00    0 0.00    0    0    0     0
#> elapsed      0    0    0 0.01    0 0.01    0    0    0     0

<sup>Created on 2021-05-21 by the reprex package (v2.0.0)</sup>

运行速度上没有区别。

yihui

chuxinyuan 是的，正规测速不是这么测的。system.time() 可以用来粗测，但对运行时间短的代码仅仅跑 10 遍求平均是不太靠谱的。可以考虑 microbenchmark::microbenchmark() 等工具。

ryo

✓ 1.9 GiB [main*]> microbenchmark(
     test_4.1.0 = {
         dat1 <- expand.grid(year = 2015:2021, month = 1:12, day = 1:31) |> 
             {\(i) within(i, value <- seq_len(nrow(i))) }(); 
         dat1 |> 
             split(~ month + year) |> 
             lapply(\(i) with(i, c(year = year[1], month = month[1], value_mean = mean(tail(value[order(day)], 5))))) |> 
             {\(i) do.call(rbind, i) }()}, 
     test_4.0.5 = {
         dat2 <- expand.grid(year = 2015:2021, month = 1:12, day = 1:31)
         dat2 <- within(dat2, value <- seq_len(nrow(dat2)))
         
         do.call(rbind, lapply(
             split(dat2, dat2[, c("month", "year")]), 
             function(i) with(i, c(year = year[1], month = month[1], value_mean = mean(tail(value[order(day)], 5))))
         ))}, 
     ## https://github.com/nmatzke/BioGeoBEARS
     #test_BioGeoBEARS.alt = {
     #  dat <- expand.grid.alt(year = 2015:2021, month = 1:12, day = 1:31) |> 
     #    {\(i) within(i, value <- seq_len(nrow(i))) }(); 
     #  dat |> 
     #    split(~ month + year) |> 
     #    lapply(\(i) with(i, c(year = year[1], month = month[1], value_mean = mean(tail(value[order(day)], 5))))) |> 
     #    {\(i) do.call(rbind, i) }()}, 
     ## https://github.com/nmatzke/BioGeoBEARS
     #test_BioGeoBEARS.jc = {
     #  dat <- expand.grid.jc(year = 2015:2021, month = 1:12, day = 1:31) |> 
     #    {\(i) within(i, value <- seq_len(nrow(i))) }(); 
     #  dat |> 
     #    split(~ month + year) |> 
     #    lapply(\(i) with(i, c(year = year[1], month = month[1], value_mean = mean(tail(value[order(day)], 5))))) |> 
     #    {\(i) do.call(rbind, i) }()}, 
     
     test_dplyr = {
         dat3 <- expand.grid(year = 2015:2021, month = 1:12, day = 1:31) %>% 
             mutate(value = 1:nrow(.))
         dat3 %>% ddply(., .(year, month), summarise, mean_value = mean(tail(value, 5)))
         #}, 
         ## https://rdrr.io/cran/noncompliance/man/expand.grid.DT.html
         ## https://atrebas.github.io/post/2019-03-03-datatable-dplyr
         #test_exg.DT = {
         #dat4 <- noncompliance::expand.grid.DT(2015:2021, 1:12, 1:31)
         #setnames(dat4, old = c('seq1', 'seq2', 'seq3'), new = c('year', 'month', 'day'))
         #dat4[, value := 1:.N][]
        
         ## https://s3.amazonaws.com/assets.datacamp.com/blog_assets/datatable_Cheat_Sheet_R.pdf
         #dat4[order(year, month, day), .(mean_value = mean((value[.N-5:.N]))), by = c('year', 'month')]
         
         ## https://stackoverflow.com/questions/36526141/summarize-data-table-by-group
         ## https://stackoverflow.com/a/16325932/3806250
         ## https://stackoverflow.com/a/34753260/3806250
         ## https://stackoverflow.com/a/53994503/3806250
         ## https://stackoverflow.com/a/50093928/3806250
         ## https://s3.amazonaws.com/assets.datacamp.com/blog_assets/datatable_Cheat_Sheet_R.pdf
         #dat4[dat4[, .I[(.N-4):.N], by = c('year', 'month')]$V1]
         #dat4[dat4[, .I[tail(seq_len(.N),5)], by = c('year', 'month')]$V1]
             
         #dat4[, {
         #value = value
         #value = value[order(day)]
         #.SD[, .(mean_value = mean(tail(value, 5))), 
         #    by=.(year, month)]}]
         
         #dat4[(.N-5):.N, 
         #   .(mean_value = mean(value)), 
         #   by = c('year', 'month')]
         
         #dat4[, mean(value[(.N-5):.N]), by = c('year', 'month')]
         
         #dat4[, mean(tail(value, 5)), by = c('year', 'month')]
         
         #dat4[, .(mean_value = mean(tail(value, 5))), by = c('year', 'month')]
         
         #dat4[, .(mean_value = mean(tail(value[order(day)], 5))), by = c('year', 'month')]

         #dat4[, by = c('year', 'month'),
         #   lapply(tail(.SD, 5), mean),
        #    .SDcols = c('value')]

         #dat4[, keyby = .(year, month),
         #     .(mean_value = mean(tail(value, 5)))]
         
         ## https://stackoverflow.com/questions/33702470/aggregate-sum-and-mean-in-r-with-ddply
         #dat4[, lapply(.SD[c(.N-5, .N)], mean), by = c('year', 'month')]
         #},
         ## https://stackoverflow.com/questions/12948686/why-is-expand-grid-faster-than-data-table-s-cj
         #test_CJ = {
         #dat5 <- CJ(2015:2021, 1:12, 1:31)
         #setnames(dat4, old = c('seq1', 'seq2', 'seq3'), new = c('year', 'month', 'day'))
         #dat4[, value := 1:.N][]
     }
 )
Unit: milliseconds
       expr    min     lq   mean median     uq   max neval
 test_4.1.0  7.129  7.647  8.293  7.915  8.145 19.88   100
 test_4.0.5  7.087  7.609  8.492  7.866  8.254 21.05   100
 test_dplyr 15.888 16.695 17.533 17.089 17.528 28.48   100

这data.table怎么使用？搞了一整天都错，晕 😅 ～

✓ 1.9 GiB [main*]> microbenchmark(
     eg = expand.grid(1800:2050, 1:12, 1:31),
     egDT = noncompliance::expand.grid.DT(expand.grid.DT(1800:2050, 1:12, 1:31)),
     CJ = data.table::CJ(1800:2050, 1:12, 1:31),
     eg.L = expand.grid(1800:2050L, 1:12, 1:31),
     egDT.L = noncompliance::expand.grid.DT(expand.grid.DT(1800:2050L, 1:12, 1:31)),
     CJ.L = data.table::CJ(1800:2050L, 1:12, 1:31),
     eg.LL = expand.grid(1800L:2050L, 1:12, 1:31),
     egDT.LL = noncompliance::expand.grid.DT(expand.grid.DT(1800L:2050L, 1:12, 1:31)),
     CJ.LL = data.table::CJ(1800L:2050L, 1:12, 1:31))
Unit: microseconds
    expr    min     lq   mean median     uq    max neval
      eg 2682.4 2939.9 3171.6 3002.4 3096.1  13861   100
    egDT 4455.3 4652.5 5500.6 4785.3 5015.9  20647   100
      CJ  459.9  531.6 4260.9  594.2  681.8 357043   100
    eg.L 2737.2 2938.1 3345.6 3020.6 3191.0  14063   100
  egDT.L 4378.7 4662.0 5497.7 4836.1 5238.5  16041   100
    CJ.L  459.8  538.6  731.5  586.2  670.0   9731   100
   eg.LL 2725.1 2949.2 3288.2 3022.1 3176.0  13908   100
 egDT.LL 4463.2 4681.4 5278.5 4841.8 5090.6  16971   100
   CJ.LL  464.8  554.1  695.0  612.4  679.5   5130   100