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:如何在不获取 NA 值的情况下将 factor 转换为 date 对象.

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

这是一个类似的帖子:将因子转换为 R 中的日期/时间

在那篇文章中，用户在 date 之前转换为 character 对象.在 as.Date 函数中使用 as.character 转换为 character 对象时，我得到了 NA 值.

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，其日期采用因子格式，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 对象会产生 NA 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 函数是否有效.

Checking if the as.character function worked.

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

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 而不是 %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

现在的问题是你的格式字符串声明日期包括年份和世纪，而你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"

注意，当您使用正确的占位符时，R 会正确显示；小写y.

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

%Y 如果没有世纪的年份，会发生什么似乎取决于操作系统.正如您在 Linux (Fedora 22) 上看到的那样，我没有得到年份部分的填充，而您看到的是零填充.

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.

这篇关于将因子转换为不带 NA 的日期对象 R的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！