在R中一次重新格式化多列中的日期答案

【问题标题】：Reformatting dates in multiple columns at once in R在R中一次重新格式化多列中的日期
【发布时间】：2021-09-14 17:08:40
【问题描述】：

我正在尝试将多个列从“字符”转换为日期，但还要重新格式化日期。我可以逐列进行，但我希望编写某种循环来迭代所有变量。

例如，我可以做

dates_test$date.d.m.y <- format(as.Date(dates_test$date.d.m.y, "%d/%m/%Y), "%m/%d/%Y")

我将如何编写代码，以便将 date.d.m.y 和 d.m.y.test 的格式同时更改为 %m/%d/%Y 格式？

这里的数据集：

dput(head(dates_test,20)

structure(list(date.m.d.y = c("5/13/2013", "5/14/2013", "5/15/2013", 
"5/16/2013", "5/17/2013", "5/18/2013", "5/19/2013", "5/20/2013", 
"5/21/2013", "5/22/2013", "5/23/2013", "5/24/2013", "5/25/2013", 
"5/26/2013", "5/27/2013", "5/28/2013", "5/29/2013", "5/30/2013", 
"5/31/2013", "6/1/2013"), date.d.m.y = c("2/2/2012", "2/2/2012", 
"2/2/2012", "2/2/2012", "2/2/2012", "9/2/2012", "9/2/2012", "9/2/2012", 
"9/2/2012", "9/2/2012", "16/2/2012", "16/2/2012", "16/2/2012", 
"16/2/2012", "16/2/2012", "23/2/2012", "23/2/2012", "23/2/2012", 
"23/2/2012", "23/2/2012"), date.y.m.d = c("2010-12-11", "2010-12-12", 
"2010-12-13", "2010-12-14", "2010-12-15", "2010-12-16", "2010-12-17", 
"2010-12-18", "2010-12-19", "2010-12-20", "2010-12-21", "2010-12-22", 
"2010-12-23", "2010-12-24", "2010-12-25", "2010-12-26", "2010-12-27", 
"2010-12-28", "2010-12-29", "2010-12-30"), date.d.m.y.2 = c("13.5.2013", 
"14.5.2013", "15.5.2013", "16.5.2013", "17.5.2013", "18.5.2013", 
"19.5.2013", "20.5.2013", "21.5.2013", "22.5.2013", "23.5.2013", 
"24.5.2013", "25.5.2013", "26.5.2013", "27.5.2013", "28.5.2013", 
"29.5.2013", "30.5.2013", "31.5.2013", "1.6.2013"), date.m.d.y.2 = c("13-May-2013", 
"14-May-2013", "15-May-2013", "16-May-2013", "17-May-2013", "18-May-2013", 
"19-May-2013", "20-May-2013", "21-May-2013", "22-May-2013", "23-May-2013", 
"24-May-2013", "25-May-2013", "26-May-2013", "27-May-2013", "28-May-2013", 
"29-May-2013", "30-May-2013", "31-May-2013", "1-Jun-2013"), d.m.y.test = c("2/2/2012", 
"2/2/2012", "2/2/2012", "2/2/2012", "2/2/2012", "9/2/2012", "9/2/2012", 
"9/2/2012", "9/2/2012", "9/2/2012", "16/2/2012", "16/2/2012", 
"16/2/2012", "16/2/2012", "16/2/2012", "23/2/2012", "23/2/2012", 
"23/2/2012", "23/2/2012", "23/2/2012"), new = c("13/05/2013", 
"14/05/2013", "15/05/2013", "16/05/2013", "17/05/2013", "18/05/2013", 
"19/05/2013", "20/05/2013", "21/05/2013", "22/05/2013", "23/05/2013", 
"24/05/2013", "25/05/2013", "26/05/2013", "27/05/2013", "28/05/2013", 
"29/05/2013", "30/05/2013", "31/05/2013", "01/06/2013")), row.names = c(NA, 
20L), class = "data.frame")

【问题讨论】：

图片不是发布数据（或代码）的好方法。请参阅 this Meta 和 relevant xkcd。您可以以dput 格式发布示例数据吗？请使用您尝试过的代码和dput(dates_test) 的输出来编辑问题。或者，如果 dput(head(dates_test, 20)) 的输出太大。
请注意，当您 format() 时，您会将日期转换回字符串。这是你想要的吗？
@RuiBarradas - 刚刚添加了 dput 输出。感谢您的专业提示和您对我的耐心！
@MrFlick - 不，所以我想这是第 2 个问题。本质上，我正在尝试将所有变量转换为 %m/%d/%Y 日期格式。
到目前为止，第一列和最后一列似乎完全相同。这是否意味着最后一列已被编辑？

标签： r date format as.date reformatting

【解决方案1】：

我们可以使用anydate将across不同日期格式的列转换为Date类，然后应用format返回想要的格式——%m/%d/%Y

library(anytime)
library(dplyr)
addFormats("%d.%m.%Y")
dates_test1 <- dates_test %>% 
     mutate(across(everything(), ~ format(anydate(.), "%m/%d/%Y")))

-输出

dates_test1
 date.m.d.y date.d.m.y date.y.m.d date.d.m.y.2 date.m.d.y.2 d.m.y.test        new
1  05/13/2013 02/02/2012 12/11/2010   05/13/2013   05/13/2013 02/02/2012 05/13/2013
2  05/14/2013 02/02/2012 12/12/2010   05/14/2013   05/14/2013 02/02/2012 05/14/2013
3  05/15/2013 02/02/2012 12/13/2010   05/15/2013   05/15/2013 02/02/2012 05/15/2013
4  05/16/2013 02/02/2012 12/14/2010   05/16/2013   05/16/2013 02/02/2012 05/16/2013
5  05/17/2013 02/02/2012 12/15/2010   05/17/2013   05/17/2013 02/02/2012 05/17/2013
6  05/18/2013 09/02/2012 12/16/2010   05/18/2013   05/18/2013 09/02/2012 05/18/2013
7  05/19/2013 09/02/2012 12/17/2010   05/19/2013   05/19/2013 09/02/2012 05/19/2013
8  05/20/2013 09/02/2012 12/18/2010   05/20/2013   05/20/2013 09/02/2012 05/20/2013
9  05/21/2013 09/02/2012 12/19/2010   05/21/2013   05/21/2013 09/02/2012 05/21/2013
10 05/22/2013 09/02/2012 12/20/2010   05/22/2013   05/22/2013 09/02/2012 05/22/2013
11 05/23/2013 02/16/2012 12/21/2010   05/23/2013   05/23/2013 02/16/2012 05/23/2013
12 05/24/2013 02/16/2012 12/22/2010   05/24/2013   05/24/2013 02/16/2012 05/24/2013
13 05/25/2013 02/16/2012 12/23/2010   05/25/2013   05/25/2013 02/16/2012 05/25/2013
14 05/26/2013 02/16/2012 12/24/2010   05/26/2013   05/26/2013 02/16/2012 05/26/2013
15 05/27/2013 02/16/2012 12/25/2010   05/27/2013   05/27/2013 02/16/2012 05/27/2013
16 05/28/2013 02/23/2012 12/26/2010   05/28/2013   05/28/2013 02/23/2012 05/28/2013
17 05/29/2013 02/23/2012 12/27/2010   05/29/2013   05/29/2013 02/23/2012 05/29/2013
18 05/30/2013 02/23/2012 12/28/2010   05/30/2013   05/30/2013 02/23/2012 05/30/2013
19 05/31/2013 02/23/2012 12/29/2010   05/31/2013   05/31/2013 02/23/2012 05/31/2013
20 06/01/2013 02/23/2012 12/30/2010   01/06/2013   06/01/2013 02/23/2012 06/01/2013

【讨论】：

这个几乎成功了，但不幸的是，第二列（date.d.m.y）中的月/日发生了一些变化

【解决方案2】：

这是一种使用包lubridate 函数parse_date_time 的方法。主要技巧是首先从列名中获取格式。

library(lubridate)

fmt <- names(dates_test)
fmt <- sub("date", "", fmt)
fmt <- unique(gsub("[^dmy]", "", fmt))
fmt <- fmt[nchar(fmt) == 3]

dates_new <- lapply(dates_test, parse_date_time, orders = fmt)
dates_new <- do.call(cbind.data.frame, dates_new)

str(dates_new)
#'data.frame':  20 obs. of  7 variables:
# $ date.m.d.y  : POSIXct, format: "2013-05-13" ...
# $ date.d.m.y  : POSIXct, format: "2012-02-02" ...
# $ date.y.m.d  : POSIXct, format: "2010-12-11" ...
# $ date.d.m.y.2: POSIXct, format: "2013-05-13" ...
# $ date.m.d.y.2: POSIXct, format: "2013-05-13" ...
# $ d.m.y.test  : POSIXct, format: "2012-02-02" ...
# $ new         : POSIXct, format: "2013-05-13" ...

现在这些列都属于"POSIXct"、lapply 类，相应的format 方法。 R 的 S3 类机制会自动调用它。

dates_new[] <- lapply(dates_new, format, format = "%m/%d/%Y")
str(dates_new)
#'data.frame':  20 obs. of  7 variables:
# $ date.m.d.y  : chr  "05/13/2013" "05/14/2013" "05/15/2013" "05/16/2013" ...
# $ date.d.m.y  : chr  "02/02/2012" "02/02/2012" "02/02/2012" "02/02/2012" ...
# $ date.y.m.d  : chr  "12/11/2010" "12/12/2010" "12/13/2010" "12/14/2010" ...
# $ date.d.m.y.2: chr  "05/13/2013" "05/14/2013" "05/15/2013" "05/16/2013" ...
# $ date.m.d.y.2: chr  "05/13/2013" "05/14/2013" "05/15/2013" "05/16/2013" ...
# $ d.m.y.test  : chr  "02/02/2012" "02/02/2012" "02/02/2012" "02/02/2012" ...
# $ new         : chr  "05/13/2013" "05/14/2013" "05/15/2013" "05/16/2013" ...

【讨论】：

作为一个学习练习，如果我只想更改某些列的输出，是否可以将fmt <- names(dates_test) 替换为fmt <- c("date.m.d.y","date.d.m.y")？
@user16368087 你会提前知道列名或编号吗？
是的！虽然我认为这个解决方案只会忽略未格式化为日期的列？

【解决方案3】：

您可以使用 parsedate 包并根据需要格式化输出。

library(parsedate)

for (i in colnames(dates_test)) {
  dates_test[,i] <- format(parse_date(dates_test[,i]),"%m/%d/%Y")
}

dates_test
   date.m.d.y date.d.m.y date.y.m.d date.d.m.y.2 date.m.d.y.2 d.m.y.test        new
1  05/13/2013 02/02/2012 12/11/2010   05/13/2013   05/13/2013 02/02/2012 05/13/2013
2  05/14/2013 02/02/2012 12/12/2010   05/14/2013   05/14/2013 02/02/2012 05/14/2013
3  05/15/2013 02/02/2012 12/13/2010   05/15/2013   05/15/2013 02/02/2012 05/15/2013
4  05/16/2013 02/02/2012 12/14/2010   05/16/2013   05/16/2013 02/02/2012 05/16/2013
5  05/17/2013 02/02/2012 12/15/2010   05/17/2013   05/17/2013 02/02/2012 05/17/2013
6  05/18/2013 09/02/2012 12/16/2010   05/18/2013   05/18/2013 09/02/2012 05/18/2013
7  05/19/2013 09/02/2012 12/17/2010   05/19/2013   05/19/2013 09/02/2012 05/19/2013
8  05/20/2013 09/02/2012 12/18/2010   05/20/2013   05/20/2013 09/02/2012 05/20/2013
9  05/21/2013 09/02/2012 12/19/2010   05/21/2013   05/21/2013 09/02/2012 05/21/2013
10 05/22/2013 09/02/2012 12/20/2010   05/22/2013   05/22/2013 09/02/2012 05/22/2013
11 05/23/2013 02/16/2012 12/21/2010   05/23/2013   05/23/2013 02/16/2012 05/23/2013
12 05/24/2013 02/16/2012 12/22/2010   05/24/2013   05/24/2013 02/16/2012 05/24/2013
13 05/25/2013 02/16/2012 12/23/2010   05/25/2013   05/25/2013 02/16/2012 05/25/2013
14 05/26/2013 02/16/2012 12/24/2010   05/26/2013   05/26/2013 02/16/2012 05/26/2013
15 05/27/2013 02/16/2012 12/25/2010   05/27/2013   05/27/2013 02/16/2012 05/27/2013
16 05/28/2013 02/23/2012 12/26/2010   05/28/2013   05/28/2013 02/23/2012 05/28/2013
17 05/29/2013 02/23/2012 12/27/2010   05/29/2013   05/29/2013 02/23/2012 05/29/2013
18 05/30/2013 02/23/2012 12/28/2010   05/30/2013   05/30/2013 02/23/2012 05/30/2013
19 05/31/2013 02/23/2012 12/29/2010   05/31/2013   05/31/2013 02/23/2012 05/31/2013
20 06/01/2013 02/23/2012 12/30/2010   06/01/2013   06/01/2013 02/23/2012 01/06/2013

【讨论】：