如何将具有混合日期格式的变量转换为一种格式? [英] How to convert variable with mixed date formats to one format?

问题描述

我的数据框示例:

                date
1   25 February 1987
2     20 August 1974
3     9 October 1984
4     18 August 1992
5  19 September 1995
6          16-Oct-63
7          30-Sep-65
8        22 Jan 2008
9         13-11-1961
10    18 August 1987
11         15-Sep-70
12    5 October 1994
13   5 December 1984
14          03/23/87
15    30 August 1988
16        26-10-1993
17    22 August 1989
18         13-Sep-97

我有一个包含多种日期格式的日期变量的大型数据框.变量中的大多数格式如上所示 - 还有一些非常罕见的其他格式.之所以有多种格式，是因为数据是从各个使用不同格式的网站收集而来的.

I have a large dataframe with a date variable that has multiple formats for dates. Most of the formats in the variable are shown above- there are a couple of very rare others too. The reason why there are multiple formats is that the data were pulled together from various websites that each used different formats.

我尝试过使用简单的转换，例如

I have tried using straightforward conversions e.g.

strftime(mydf$date,"%d/%m/%Y")

但是如果有多种格式，这些类型的转换将不起作用.我不想诉诸多个 gsub 类型编辑.我想知道我是否缺少更简单的解决方案?

but these sorts of conversion will not work if there are multiple formats. I don't want to resort to multiple gsub type editing. I was wondering if I am missing a more simple solution?

代码示例:

    structure(list(date = structure(c(12L, 8L, 18L, 6L, 7L, 4L, 14L, 
10L, 1L, 5L, 3L, 17L, 16L, 11L, 15L, 13L, 9L, 2L), .Label = c("13-11-1961", 
"13-Sep-97", "15-Sep-70", "16-Oct-63", "18 August 1987", "18 August 1992", 
"19 September 1995", "20 August 1974", "22 August 1989", "22 Jan 2008", 
"03/23/87", "25 February 1987", "26-10-1993", "30-Sep-65", "30 August 1988", 
"5 December 1984", "5 October 1994", "9 October 1984"), class = "factor")), .Names = "date", row.names = c(NA, 
-18L), class = "data.frame")

推荐答案

您可以尝试 parse_date_time 在包 lubridate 中允许用户指定几个格式顺序来使用 orders 参数处理异构日期时间字符表示".像……

You may try parse_date_time in package lubridate which "allows the user to specify several format-orders to handle heterogeneous date-time character representations" using the orders argument. Something like...

library(lubridate)
parse_date_time(x = df$date,
                orders = c("d m y", "d B Y", "m/d/y"),
                locale = "eng")

...应该能够处理您的大部分格式.请注意 b/B 格式是 locale 敏感的.

...should be able to handle most of your formats. Please note that b/B formats are locale sensitive.

可以在orders 中使用的其他日期时间格式列在?strptime 的详细信息 部分.

Other date-time formats which can be used in orders are listed in the Details section in ?strptime.

这篇关于如何将具有混合日期格式的变量转换为一种格式?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！