Lubridate mdy函数

Question

我正在尝试转换以下内容，但其中一个日期1不成功。"4/2/10“变成了"0010-04-02”。

有没有办法纠正这个问题？

谢谢，Vivek

data <- data.frame(initialDiagnose = c("4/2/10","14.01.2009", "9/22/2005", 
        "4/21/2010", "28.01.2010", "09.01.2009", "3/28/2005", 
        "04.01.2005", "04.01.2005", "Created on 9/17/2010", "03 01 2010"))

mdy <- mdy(data$initialDiagnose) 
dmy <- dmy(data$initialDiagnose) 
mdy[is.na(mdy)] <- dmy[is.na(mdy)] # some dates are ambiguous, here we give 
data$initialDiagnose <- mdy        # mdy precedence over dmy
data

   initialDiagnose
1       0010-04-02
2       2009-01-14
3       2005-09-22
4       2010-04-21
5       2010-01-28
6       2009-09-01
7       2005-03-28
8       2005-04-01
9       2005-04-01
10      2010-09-17
11      2010-03-01

Answer 1

我认为这是因为mdy()函数更喜欢使用%Y (实际年份)来匹配年份，而不是%y (两位数的年份缩写，默认为19XX或20XX)。

不过，还是有一个变通办法的。我查看了lubridate::parse_date_time (?parse_date_time)的帮助文件，在帮助文件的底部，有一个添加参数的示例，该参数优先于今年的%y格式，而不是%Y格式。帮助文件中的相关代码：

## ** how to use `select_formats` argument **
## By default %Y has precedence:
parse_date_time(c("27-09-13", "27-09-2013"), "dmy")
## [1] "13-09-27 UTC"   "2013-09-27 UTC"

## to give priority to %y format, define your own select_format function:

my_select <-   function(trained){
   n_fmts <- nchar(gsub("[^%]", "", names(trained))) + grepl("%y", names(trained))*1.5
   names(trained[ which.max(n_fmts) ])
}

parse_date_time(c("27-09-13", "27-09-2013"), "dmy", select_formats = my_select)
## '[1] "2013-09-27 UTC" "2013-09-27 UTC"

因此，对于您的示例，您可以调整此代码，并将mdy <- mdy(data$initialDiagnose)行替换为：

# Define a select function that prefers %y over %Y. This is copied 
# directly from the help files
my_select <-   function(trained){
  n_fmts <- nchar(gsub("[^%]", "", names(trained))) + grepl("%y", names(trained))*1.5
  names(trained[ which.max(n_fmts) ])
}

# Parse as mdy dates
mdy <- parse_date_time(data$initialDiagnose, "mdy", select_formats = my_select)
# [1] "2010-04-02 UTC" NA               "2005-09-22 UTC" "2010-04-21 UTC" NA              
# [6] "2009-09-01 UTC" "2005-03-28 UTC" "2005-04-01 UTC" "2005-04-01 UTC" "2010-09-17 UTC"
#[11] "2010-03-01 UTC"

然后运行你问题中的其余代码行，它给出了这个数据框作为结果：

   initialDiagnose
1       2010-04-02
2       2009-01-14
3       2005-09-22
4       2010-04-21
5       2010-01-28
6       2009-09-01
7       2005-03-28
8       2005-04-01
9       2005-04-01
10      2010-09-17
11      2010-03-01