应用lubridate返回数字

我有一个看起来像这样的数据集

birds[,1:3]
Source: local data frame [15 x 3]

   year month day
1  2015     5  13
2  2015     5  14
3  2015     5  15
4  2015     5  16
5  2015     5  17
6  2014     5  28
7  2014     5  29
8  2014     5  30
9  2014     5  31
10 2014     6   1
11 2013     5   8
12 2013     5   9
13 2013     5  10
14 2013     5  11
15 2013     5  12

我想要做的是将这些列组合成一个“日期”列,所以我想我可以将它们粘贴在一起并将它们传递给lubridate.

这有效:

ymd(paste(birds[1,1], birds[1,2], birds[1,3], sep="-"))
[1] "2015-05-13 UTC"

但是当我尝试使用apply来执行每行时,我得到了这个:

apply(birds[,c("year","month","day")], 1, 
function(x) ymd(paste(x[1], x[2], x[3], sep="-")))
 [1] 1431475200 1431561600 1431648000 1431734400 1431820800 1401235200 1401321600 1401408000 1401494400
 [10] 1401580800 1367971200 1368057600 1368144000 1368230400 1368316800

为什么会发生这种情况,我该如何解决?

最佳答案 我们不需要申请MARGIN = 1.相反,我们可以用(粘贴(年,月,日,sep =“ – ”))粘贴列,并用as.Date包装以转换为’Date’类. ymd的输出是POSIXct类,在apply中,它将被强制为’numeric’形式.

library(lubridate)
library(dplyr)
mutate(birds, date=ymd(paste(year, month, day)))

或者我们可以使用tidyr中的unite然后转换为POSIXct类

library(tidyr)
unite(birds, date, year:day, sep="-") %>% 
           mutate(date=ymd(date))

或者使用基数为R的do.call和ymd

birds$date <- ymd(do.call(paste, birds))

或者你可以从基数R使用as.Date

as.Date(do.call(paste, c(birds,sep="-")))

要修复使用apply获得的输出

res <- apply(birds[,c("year","month","day")], 1, 
      function(x) ymd(paste(x[1], x[2], x[3], sep="-")))
unname(as.POSIXct(res, origin='1970-01-01',tz='UTC'))
#[1] "2015-05-13 UTC" "2015-05-14 UTC" "2015-05-15 UTC" "2015-05-16 UTC"
#[5] "2015-05-17 UTC" "2014-05-28 UTC" "2014-05-29 UTC" "2014-05-30 UTC"
#[9] "2014-05-31 UTC" "2014-06-01 UTC" "2013-05-08 UTC" "2013-05-09 UTC"
#[13] "2013-05-10 UTC" "2013-05-11 UTC" "2013-05-12 UTC"

数据

birds <- structure(list(year = c(2015L, 2015L, 2015L, 2015L, 2015L, 
2014L, 
2014L, 2014L, 2014L, 2014L, 2013L, 2013L, 2013L, 2013L, 2013L
), month = c(5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 6L, 5L, 5L, 
5L, 5L, 5L), day = c(13L, 14L, 15L, 16L, 17L, 28L, 29L, 30L, 
31L, 1L, 8L, 9L, 10L, 11L, 12L)), .Names = c("year", "month", 
"day"), class = "data.frame", row.names = c("1", "2", "3", "4", 
"5", "6", "7", "8", "9", "10", "11", "12", "13", "14", "15"))
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