将因子转换为不带NA的日期对象R [英] Convert factor to date object R without NA

问题描述

问题：如何将因子转换为 date 对象 NA 值。

Question: how can I convert a factor to a date object without getting NA values.

这是类似的帖子：将系数转换为R中的日期/时间

在该帖子中，用户转换为字符 日期之前的对象。使用 as.character character 对象时，我得到 NA 值 as.Date 函数中的$ c>。

In that post, the user converted to a character object before a date. I am getting NA values when converting to character object using as.character inside the as.Date function.

我在数据框中有一列，其日期为因子格式，出现次数不同。这是data.frame中包含的信息。

I have a column in the dataframe with the date in factor format with different numbers of occurrences. Here's the information contained in the data.frame.

> head(fraud, 5)
  TRANSACTION.DATE TRANSACTION.AMOUNT AIR.TRAVEL.DATE POSTING.DATE
1 2/27/14                  25.00                 <NA>          2/28/14
2 2/28/14                  25.00                 <NA>          2/28/14
3 2/27/14                  25.00                 <NA>          2/28/14
4 2/27/14                  20.00              2/27/14          2/28/14
5 2/27/14                  12.13                 <NA>          2/28/14

> str(fraud$TRANSACTION.DATE)
 Factor w/ 519 levels "1/1/14","1/1/15",..: 228 230 228 228 228 230 226 228 230 228 ...

> summary(fraud$TRANSACTION.DATE, 5)
9/30/14 9/17/14 11/4/14 9/23/14 (Other) 
    197     187     171     160   19221 

将因子转换为 date 对象会导致 NA 个值。

Converting the factor to a date object resulted in NA values.

> fraud$TRANSACTION.DATE <- as.Date(as.character(fraud$TRANSACTION.DATE), 
+                                       format = "%m/%d/%Y")
> head(fraud$TRANSACTION.DATE, 5)
[1] NA NA NA NA NA

检查 as.character 函数是否起作用。

> fraud$TRANSACTION.DATE <- as.character(fraud$TRANSACTION.DATE)
> head(fraud$TRANSACTION.DATE)
[1] NA NA NA NA NA NA

编辑：我使用as.Date函数，但格式错误

> fraud$TRANSACTION.DATE <- as.Date(fraud$TRANSACTION.DATE, format = "%m/%d/%Y")
> str(fraud$TRANSACTION.DATE)
 Date[1:19936], format: "0014-02-27" "0014-02-28" "0014-02-27" "0014-02-27" "0014-02-27" ...
> head(fraud$TRANSACTION.DATE, 5)
[1] "0014-02-27" "0014-02-28" "0014-02-27" "0014-02-27" "0014-02-27"

编辑2：这是dput值

> dput(droplevels(head(fraud$TRANSACTION.DATE)))
structure(c(1L, 2L, 1L, 1L, 1L, 2L), .Label = c("2/27/14", "2/28/14"
), class = "factor")

解决方案：使用％y代替％Y

> fraud$TRANSACTION.DATE <- as.Date(fraud$TRANSACTION.DATE, "%m/%d/%y")
> head(fraud$TRANSACTION.DATE, 5)
[1] "2014-02-27" "2014-02-28" "2014-02-27" "2014-02-27" "2014-02-27"

推荐答案

现在的问题是格式字符串指出日期包括年份 with Century ，其中您的日期只包含年份没有世纪。您需要使用％y 占位符，而不是％Y 占位符。

The problem now is that your format string states the dates include the year with century where your dates only contain the year without century. You need to use the %y placeholder, not the %Y one.

dates <- factor(c("2/27/14","2/28/14","2/27/14","2/27/14","2/27/14"))
as.Date(dates, format = "%m/%d/%y") # correct lowercase y
as.Date(dates, format = "%m/%d/%Y") # incorrect uppercase y

> as.Date(dates, format = "%m/%d/%y")
[1] "2014-02-27" "2014-02-28" "2014-02-27" "2014-02-27" "2014-02-27"
> as.Date(dates, format = "%m/%d/%Y")
[1] "14-02-27" "14-02-28" "14-02-27" "14-02-27" "14-02-27"

当您使用正确的格式时，注意R会正确显示占位符小写的 y 。

Notice R gets it right when you use the correct placeholder; lowercase y.

在没有$ ％Y 的情况下会发生什么百年似乎与操作系统有关。正如您在Linux（Fedora 22）上所看到的那样，我没有获得年度填充部分，而您却看到了零填充。

What happens with %Y when you don't have a year with century seems OS dependent. As you can see on Linux (Fedora 22) I get no padding of the year part whereas you are seeing zero-padding.

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