R:填写时间序列中缺少的日期?

问题描述：

我有一个缺少时间的动物园时间序列. 为了填充它并进行连续的序列处理，我要做...

I have a zoo time series with missing days. In order to fill it and have a continuous series I do...

我从头到尾生成一个时间日期-时间序列.

I generate a chron date-time sequence from start to end.

我将我的系列与此系列合并.

I merge my series with this one.

我使用na.locf将las观测值替换为NAs.

I use na.locf to substitute NAs with las obsservation.

我删除了同步时间顺序.

I remove the syntetic chron sequence.

我可以更轻松地这样做吗? 也许有一些与频率相关的索引功能?

Can I do same easier? Maybe with some index function related to the frequency?

答

如果使用带有索引的空" zoo对象，则会稍微容易一些.

It's slightly easier if you use a "empty" zoo object with an index.

> x <- zoo(1:10,Sys.Date()-10:1)[c(1,3,5,7,10)]
> empty <- zoo(order.by=seq.Date(head(index(x),1),tail(index(x),1),by="days"))
> na.locf(merge(x,empty))
2010-08-14 2010-08-15 2010-08-16 2010-08-17 2010-08-18 
         1          1          3          3          5 
2010-08-19 2010-08-20 2010-08-21 2010-08-22 2010-08-23 
         5          7          7          7         10

对于日内数据(使用Gabor出色的xout=建议):

For intra-day data (using Gabor's excellent xout= suggestion):

> index(x) <- as.POSIXct(index(x))
> na.locf(x, xout=seq(head(index(x),1),tail(index(x),1),by="15 min"))

R:填写时间序列中缺少的日期?

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