从时间上提取小时的最快方式(HH:MM)

Question

希望fastPOSIXct工作 - 但在这种情况下不工作.

这是我的时间数据(没有日期) - 我需要从他们那里得到小时数.

times <- c("9:46","11:06", "14:17", "19:53", "0:03", "3:56")

这是错误的输出fastPOSIXct:

fastPOSIXct(times, "GMT")
[1] "1970-01-01 00:00:00 GMT" "1970-01-01 00:00:00 GMT"
[3] "1970-01-01 00:00:00 GMT" "1970-01-01 00:00:00 GMT"
[5] "1970-01-01 00:00:00 GMT" "1970-01-01 00:00:00 GMT"

没有正确存在日期,它无法识别时间.

在hour从方法data.table与as.ITime解决的目的,但看起来像大时代阵列慢.

library(data.table)
hour(as.ITime(times))
# [1]  9 11 14 19  0  3

想知道是否有更快的方式(就像fastPOSIXct,但无需日期工作).

fastPOSIXct 真的很像快照,但错了.

Answer 1

您也可以尝试substr:as.integer(substr(vals, start = 1, stop = nchar(vals) - 3))

在具有10e6元素的向量的基准测试中,stringi::stri_sub是最快的,并且是substr第二个.

vals <- sample(c("9:46", "11:06", "14:17", "19:53", "0:03", "3:56"), 1e6, replace = TRUE)

fun_substr <- function(vals) as.integer(substr(vals, start = 1, stop = nchar(vals) - 3))

grab.hrs <- function(vals) as.integer(sub(pattern = ":.*", replacement = "", x = vals))

fun_strtrim <- function(vals) as.integer(strtrim(vals, nchar(vals) - 3))

library(chron)
fun_chron <- function(vals) hours(times(paste0(vals, ":00")))

fun_lt <- function(vals) as.POSIXlt(vals, format="%H:%M")$hour

library(stringi)
fun_stri_sub <- function(vals) as.integer(stri_sub(vals, from = 1, to = -4))

library(microbenchmark)
microbenchmark(fun_substr(vals),
               fun_stri_sub(vals),      
               grab.hrs(vals),
               fun_strtrim(vals),
               fun_lt(vals),
               fun_chron(vals),
               unit = "relative", times = 5)
# Unit: relative
#               expr       min        lq      mean    median        uq       max neval
#   fun_substr(vals)  2.186714  1.902074  2.015082  1.968542  1.945007  2.090236     5
# fun_stri_sub(vals)  1.000000  1.000000  1.000000  1.000000  1.000000  1.000000     5
#     grab.hrs(vals)  2.656630  2.397918  2.687133  2.426223  2.446902  3.263962     5
#  fun_strtrim(vals) 31.177869 27.601380 26.009818 27.423562 17.902507 29.426989     5
#       fun_lt(vals) 47.296929 41.122287 42.266556 40.647465 30.539030 52.710992     5
#    fun_chron(vals)  5.594931  5.159192  5.961775  7.746242  5.286944  6.189742     5

Answer 2

您也可以使用包中的times函数执行此操作chron:

library(chron)
vals <- c("9:46","11:06", "14:17", "19:53", "0:03", "3:56")
dat <- times(paste0(vals, ":00"))
hours(dat)
# [1]  9 11 14 19  0  3

如果速度很重要,您可以使用字符串操作更快地提取小时数:

grab.hrs <- function(vals) as.numeric(sub(pattern = ":.*", replacement = "",
                                      x = vals))
grab.hrs(vals)
# [1]  9 11 14 19  0  3

times而且as.POSIXlt(来自@ tonytonov的解决方案)似乎比这更快as.ITime,字符串操作更快:

library(microbenchmark)
library(data.table)
microbenchmark(hours(times(paste0(vals, ":00"))),
               hours(as.ITime(vals)),
               as.POSIXlt(vals, format="%H:%M")$hour,
               grab.hrs(vals))
# Unit: microseconds
#                                     expr     min       lq   median       uq      max neval
#        hours(times(paste0(vals, ":00"))) 174.544 184.9485 193.5630 204.6950 5047.195   100
#                    hours(as.ITime(vals)) 665.833 678.8790 705.6445 735.0525 3030.574   100
#  as.POSIXlt(vals, format = "%H:%M")$hour 158.264 169.8880 171.9670 180.1800  301.840   100
#                           grab.hrs(vals)  10.637  15.4540  20.0995  21.1285   55.985   100

Answer 3

这是一个选择吗？这是一个base解决方案.

as.POSIXlt(times, format="%H:%M")$hour
#[1]  9 11 14 19  0  3

Answer 4

要真正加速,你也可以从字符串中删除lsat 3字符.它比使用更快regex.

as.numeric(strtrim(times, nchar(times) - 3)) 
## [1]  9 11 14 19  0  3

以下是基准测试结果

Unit: microseconds
                                         expr     min       lq   median       uq      max neval
            hours(times(paste0(vals, ":00"))) 200.670 212.9720 218.7960 221.8420  352.370   100
                        hours(as.ITime(vals)) 453.174 478.9680 487.3805 496.7885 1607.321   100
      as.POSIXlt(vals, format = "%H:%M")$hour  41.278  46.4945  49.7310  51.3115   56.453   100
                               grab.hrs(vals)  12.352  15.4295  18.3850  20.3390   31.349   100
  as.numeric(gsub("(.*):.*", "\\\\1", times))  14.528  17.7225  20.6390  23.4530   53.683   100
 as.numeric(strtrim(times, nchar(times) - 3))   9.621  11.6605  12.7435  13.2520  147.446   100