我定期进行一组测量,但有些测量结果丢失了:
measurement_date value
1 2011-01-17 13:00:00 5
2 2011-01-17 13:04:00 5
3 2011-01-17 13:08:00 7
4 2011-01-17 13:12:00 8
5 2011-01-17 13:16:00 4
6 2011-01-17 13:24:00 6
7 2011-01-17 13:28:00 5
8 2011-01-17 13:32:00 6
9 2011-01-17 13:36:00 9
10 2011-01-17 13:40:00 8
11 2011-01-17 13:44:00 6
12 2011-01-17 13:48:00 6
13 2011-01-17 13:52:00 4
14 2011-01-17 13:56:00 6
Run Code Online (Sandbox Code Playgroud)
我有一个函数,它将处理值并可以处理缺少的值,但行必须在那里,所以我生成一个每分钟有一行的数组,如下所示:
times <- timeSequence(from=.., length=60, by="min")
Run Code Online (Sandbox Code Playgroud)
现在我每小时都有一行,但我需要合并数据.我试过这样的事情,但不能完全正确:
lapply(times, function(time) {
n <- as.numeric(time)
v <- Position(function(candidate) {
y <- as.numeric(candiated)
n == y
}
.. insert the value into the row here ..
}
Run Code Online (Sandbox Code Playgroud)
但我只是得到错误和警告.我是否以正确的方式解决问题?我真的想要一个具有每分钟值的"完整"数组,因为将有许多不同的函数将运行读数,如果它们可以假设它就在那里,它就更容易实现它们.
DF <- data.frame(measurement_date = seq(as.POSIXct("2011-01-17 13:00:00"),
as.POSIXct("2011-01-17 13:56:00"),
by = "mins")[seq(1, 57, by = 4)][-6],
value = c(5,5,7,8,4,6,5,6,9,8,6,6,4,6))
full <- data.frame(measurement_date = seq(as.POSIXct("2011-01-17 13:00:00"),
by = "mins", length = 60),
value = rep(NA, 60))
Run Code Online (Sandbox Code Playgroud)
可以使用两种方法,第一种方法merge:
> v1 <- merge(full, DF, by.x = 1, by.y = 1, all = TRUE)[, c(1,3)]
> names(v1)[2] <- "value" ## I only reset this to pass all.equal later
> head(v1)
measurement_date value
1 2011-01-17 13:00:00 5
2 2011-01-17 13:01:00 NA
3 2011-01-17 13:02:00 NA
4 2011-01-17 13:03:00 NA
5 2011-01-17 13:04:00 5
6 2011-01-17 13:05:00 NA
Run Code Online (Sandbox Code Playgroud)
第二个是通过使用%in%以下方式导出的指标变量:
> want <- full$measurement_date %in% DF$measurement_date
> full[want, "value"] <- DF[, "value"]
> head(full)
measurement_date value
1 2011-01-17 13:00:00 5
2 2011-01-17 13:01:00 NA
3 2011-01-17 13:02:00 NA
4 2011-01-17 13:03:00 NA
5 2011-01-17 13:04:00 5
6 2011-01-17 13:05:00 NA
> all.equal(v1, full)
[1] TRUE
Run Code Online (Sandbox Code Playgroud)
合并版本是强烈的首选,但需要一点点的工作.该%in%解决方案只因为数据在时间顺序都在这里工作DF和full,因此我先前的"首选".然而,很容易按时间顺序获得/确保这两个对象,因此这两种方法都需要一点精细才能工作.我们可以修改%in%方法以按顺序获取两个变量(重新开始full):
full2 <- data.frame(measurement_date = seq(as.POSIXct("2011-01-17 13:00:00"),
by = "mins", length = 60),
value = rep(NA, 60))
full2 <- full2[order(full2[,1]), ] ## get full2 in order
DF2 <- DF[order(DF[,1]), ] ## get DF in order
want <- full$measurement_date %in% DF$measurement_date
full2[want, "value"] <- DF2[, "value"]
> all.equal(full, full2)
[1] TRUE
> all.equal(full2, v1)
[1] TRUE
>
Run Code Online (Sandbox Code Playgroud)
在你的函数中,as.numeric(candiated)应该是as.numeric(候选者).还有一个支架丢失.我不知道你在你的功能中究竟想要实现什么,但它对我来说看起来非常复杂.
尝试
merge(Data,times,by.x=1,by.y=1,all.y=T)
Run Code Online (Sandbox Code Playgroud)
这应该给你一些工作.