在 R 中更改日期格式 [英] Changing date format in R

问题描述

我在 R 中有一些非常简单的数据需要更改其日期格式:

 日期中点1 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 0.7375

而不是%d/%m/%Y，我更喜欢%Y-%m-%d

的标准R格式

我该如何进行此更改?我试过了:

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

但这只是截断了年份并在当天添加了零:

 [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"

谢谢！

解决方案

这里有两个步骤:

• 解析数据.您的示例不能完全重现，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，你可以这样做

<块引用>

 X$newdate <- strptime(as.character(X$date), "%d/%m/%Y")

现在 newdate 列的类型应该是 Date.

• 格式化数据.这是调用format() 或strftime() 的问题:

<块引用>

 格式(X$newdate, "%Y-%m-%d")

一个更完整的例子:

R>nzd <- data.frame(date=c("31/08/2011", "31/07/2011", "30/06/2011"),+ 中=c(0.8378,0.8457,0.8147))R＞新西兰日期中间1 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$txtdate <- 格式(nzd$newdate, "%Y-%m-%d")R＞新西兰日期中间新日期 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＞

第三列和第四列的区别在于类型:newdate 属于 Date 类，而 txtdate 是字符.

I have some very simple data in R that needs to have its date format changed:

 date midpoint
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
6   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
11  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

Rather than %d/%m/%Y, I would like it in the standard R format of %Y-%m-%d

How 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 do

 X$newdate <- strptime(as.character(X$date), "%d/%m/%Y")

Now the newdate column should be of type Date.

• Format the data. That is a matter of calling format() or strftime():

 format(X$newdate, "%Y-%m-%d")

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    mid
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
        date    mid    newdate    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> 

The difference between columns three and four is the type: newdate is of class Date whereas txtdate is character.

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