在 R xts 中显示时间索引 [英] Display time index in R xts

查看:33
本文介绍了在 R xts 中显示时间索引的处理方法,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

我想用微秒定时索引解析 csv.所以,我写了这样的代码:

I want to parse csv with micro-second timed index. So, I wrote code like this:

t<-read.zoo("test", index.column = 1, sep=",",header=TRUE, format="%Y-%m-%d %H:%M:%OS")
t.xts<-as.xts(t)

此后,我尝试显示此内容,但在索引上看不到时间信息.

after then, I tried to display this but I couldn't see time information on the index.

> t.xts[1:10,4]
           drate  
2010-09-28 " -149"
2010-09-28 " -269"
2010-09-28 " -358"
2010-09-28 " -358"
2010-09-28 " -239"
2010-09-28 " -149"
2010-09-28 " -149"
2010-09-28 " -149"
2010-09-28 " -119"
2010-09-28 " -149"

我尝试了 options(digits.secs=6) 但没有奏效.

I tried options(digits.secs=6) but didn't work.

推荐答案

如果您能提供 CSV 文件中的几行内容,将会有所帮助.设置 options(digits.secs=6) 对我有用.您也可以尝试使用 indexFormat 手动设置格式.

It would help if you could provide a few lines from your CSV file. Setting options(digits.secs=6) works for me. You could also try manually setting the format with indexFormat.

> x <- .xts(1:5, 1:5+runif(5))
> x
                    [,1]
1969-12-31 18:00:01    1
1969-12-31 18:00:02    2
1969-12-31 18:00:03    3
1969-12-31 18:00:04    4
1969-12-31 18:00:05    5
> indexFormat(x) <- "%Y-%m-%d %H:%M:%OS3"
> x
                        [,1]
1969-12-31 18:00:01.915    1
1969-12-31 18:00:02.002    2
1969-12-31 18:00:03.134    3
1969-12-31 18:00:04.981    4
1969-12-31 18:00:05.204    5
> indexFormat(x) <- "%Y-%m-%d %H:%M:%OS"
> options(digits.secs=6)
> x
                           [,1]
1969-12-31 18:00:01.914681    1
1969-12-31 18:00:02.001752    2
1969-12-31 18:00:03.134311    3
1969-12-31 18:00:04.981147    4
1969-12-31 18:00:05.204021    5

这篇关于在 R xts 中显示时间索引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!

查看全文
登录 关闭
扫码关注1秒登录
发送“验证码”获取 | 15天全站免登陆