在R中更改日期格式

本文介绍了在R中更改日期格式的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 29岁程序员，3月因学历无情被辞！ 我在R中有一些非常简单的数据，需要更改日期格式： 日期中点 1 31/08/2011 0.8378 2 31/07/2011 0.8457 3 30/06/2011 0.8147 4 31/05/2011 0.7970 5 30/04/2011 0.7877 31 31/03/2011 0.7411 7 28/02/2011 0.7624 8 31/01/2011 0.7665 9 31/12/2010 0.7500 10 30/11/2010 0.7734 31 31/10/2010 0.7511 12 30/09/2010 0.7263 13 31/08/2010 0.7158 14 31/07/2010 0.7110 15 30/06/2010 0.6921 16 31/05/2010 0.7005 17 30/04/2010 0.7113 18 31/03/2010 0.7027 19 28 / 02/2010 0.6973 20 31/01/2010 0.7260 21 31/12/2009 0.7154 22 30/11/2009 0.7287 23 31/10/2009 0.7375 而不是％d /％m /％Y ，我想用标准R格式％Y-％m-％d 我做这个改变？我已经尝试过： $ p $ nzd $ date< - format（as.Date（nzd $ date），％Y /％但是，这只是切断了一年，并增加了一天的零： $> b $ b / p> [1]0031/08/200031/07/200030/06/200031 / 05/200030/04/20 [6]0031/03/200028/02/200031/01/200031/12/200030/11 / 20 [11]0031/10/200030/09/200031/08/200031/07/200030/06/20 [ 16]0031/05/200030/04/200031/03/200028/02/200031/01/20 [21]0031/12/20 0030/11/200031/10/200030/09/200031/08/20 [26]0031/07/200030/06/20 0031/05/200030/04/200031/03/20 [31]0028/02/200031/01/200031/12/20 0030/11/200031/10/20 [36]0030/09/200031/08/200031/07/200030/06/200031 / 05/20 感谢！ 解决方案这里有两个步骤： 解析数据。你的例子不是完全可重复的，是文件中的数据，还是文本或因子变量中的变量？让我们假设后者，那么如果你的data.frame被称为X，你可以做 lockquote $ $ p $ （$ X $ date），％d /％m /％Y） newdate 列应该是类型的 日期。 格式化数据。这是调用 format（）或 strftime（）： 更完整的例子： R> nzd + mid = c（0.8378,0.8457， 0.8147）） R> nzd 日期中 1 31/08/2011 0.8378 2 31/07/2011 0.8457 3 30/06/2011 0.8147 R> nzd $ newdate< - strptime（as.character（nzd $ date），％d /％m /％Y） R> nzd $ txtdate< - format（nzd $ newdate，％Y-％m-％d） R> nzd 日期中值新值txtdate 1 31/08/2011 0.8378 2011-08-31 2011-08-31 2 31/07/2011 0.8457 2011-07-31 2011-07- 31 3 30/06/2011 0.8147 2011-06-30 2011-06-30 R> 第三列和第四列的区别在于： newdate 是类日期，而 txtdate 是字符。 I have some very simple data in R that needs to have its date format changed: date midpoint1 31/08/2011 0.83782 31/07/2011 0.84573 30/06/2011 0.81474 31/05/2011 0.79705 30/04/2011 0.78776 31/03/2011 0.74117 28/02/2011 0.76248 31/01/2011 0.76659 31/12/2010 0.750010 30/11/2010 0.773411 31/10/2010 0.751112 30/09/2010 0.726313 31/08/2010 0.715814 31/07/2010 0.711015 30/06/2010 0.692116 31/05/2010 0.700517 30/04/2010 0.711318 31/03/2010 0.702719 28/02/2010 0.697320 31/01/2010 0.726021 31/12/2009 0.715422 30/11/2009 0.728723 31/10/2009 0.7375Rather than %d/%m/%Y, I would like it in the standard R format of %Y-%m-%dHow can I make this change? I have tried:nzd$date <- format(as.Date(nzd$date), "%Y/%m/%d")But that just cut off the year and added zeros to the day: [1] "0031/08/20" "0031/07/20" "0030/06/20" "0031/05/20" "0030/04/20" [6] "0031/03/20" "0028/02/20" "0031/01/20" "0031/12/20" "0030/11/20" [11] "0031/10/20" "0030/09/20" "0031/08/20" "0031/07/20" "0030/06/20" [16] "0031/05/20" "0030/04/20" "0031/03/20" "0028/02/20" "0031/01/20" [21] "0031/12/20" "0030/11/20" "0031/10/20" "0030/09/20" "0031/08/20" [26] "0031/07/20" "0030/06/20" "0031/05/20" "0030/04/20" "0031/03/20" [31] "0028/02/20" "0031/01/20" "0031/12/20" "0030/11/20" "0031/10/20" [36] "0030/09/20" "0031/08/20" "0031/07/20" "0030/06/20" "0031/05/20"Thanks! 解决方案 There are two steps here:Parse the data. Your example is not fully reproducible, is the data in a file, or the variable in a text or factor variable? Let us assume the latter, then if you data.frame is called X, you can doNow the newdate column should be of type Date.Format the data. That is a matter of calling format() or strftime():A more complete example:R> nzd <- data.frame(date=c("31/08/2011", "31/07/2011", "30/06/2011"),+ mid=c(0.8378,0.8457,0.8147))R> nzd date mid1 31/08/2011 0.83782 31/07/2011 0.84573 30/06/2011 0.8147R> nzd$newdate <- strptime(as.character(nzd$date), "%d/%m/%Y")R> nzd$txtdate <- format(nzd$newdate, "%Y-%m-%d")R> nzd date mid newdate txtdate1 31/08/2011 0.8378 2011-08-31 2011-08-312 31/07/2011 0.8457 2011-07-31 2011-07-313 30/06/2011 0.8147 2011-06-30 2011-06-30R>The difference between columns three and four is the type: newdate is of class Date whereas txtdate is character. 这篇关于在R中更改日期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！ 上岸，阿里云！