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Question

日期转换有问题。如果我在应用程序中使用 ymd，它 returns 一个数字，或者如果我只使用 ymd 它是有效的。有人会知道吗？

library(lubridate)
a <- data.frame(dates1=c("2011-01-01",'2011-02-01','2011-03-01'),
            dates2=c("2013-01-01",'2013-02-01','2013-03-01'))
apply(a, 2, FUN = function(x) ymd(x))

     dates1     dates2
[1,] 1293840000 1356998400
[2,] 1296518400 1359676800
[3,] 1298937600 1362096000

ymd(a$dates1)

[1] "2011-01-01 UTC" "2011-02-01 UTC" "2011-03-01 UTC"

ymd(a$dates2)

[1] "2013-01-01 UTC" "2013-02-01 UTC" "2013-03-01 UTC"

Answer 1

使用lapply():

lapply(a, function(x) ymd(x))

Update/Explaination

apply() 文档指出：

In all cases the result is coerced by as.vector to one of the basic vector types before the dimensions are set, so that (for example) factor results will be coerced to a character array.

正如 J.Ulrich 在对 this answer "dates aren't strictly vectors" 的评论中所述。 lapply() 与 apply() 非常相似，但它 return 是一个列表而不是向量。因此它能够处理日期。

Answer 2

apply 尝试简化它，如果输入 X 是二维的，则使用 as.matrix，请参见此示例：

as.matrix(ymd("2013-01-01"))
      [,1]
[1,] 15706

?apply

Details
If X is not an array but an object of a class with a non-null dim value (such as a data frame), apply attempts to coerce it to an array via as.matrix if it is two-dimensional (e.g., a data frame) or via as.array.

Answer 3

这是一个使用mutate_each

的选项

library(dplyr)
a %>%
    mutate_each(funs(ymd))
#     dates1     dates2
#1 2011-01-01 2013-01-01
#2 2011-02-01 2013-02-01
#3 2011-03-01 2013-03-01

涂抹和润滑

Apply and lubridate

r

date

lubridate