滚动滞后差异 [英] Rolling lagged differences
问题描述
好，所以我希望在R中创建滚动的滞后差异.
Ok so I am looking to create rolling lagged differences in R.
vec <- c(43.79979, 44.04865, 44.17308, 44.54638, 44.79524, 44.79524, 44.79524, 44.42195, 44.54638, 44.79524, 44.42195, 43.30206, 43.30206, 43.17764, 43.30206)
> length(vec)
[1] 15
这是我到目前为止尝试过的:
This is what I have tried so far:
vec1 <- rollapply(vec, width = 2, fill = NA, FUN = diff)
这将给出以下输出:
[1] 0.24886 0.12443 0.37330 0.24886 0.00000 0.00000 -0.37329 0.12443 0.24886 -0.37329 -1.11989 0.00000 -0.12442 0.12442 NA
> length(vec1)
[1] 15
请注意，元素15中有一个NA值.
Note we have an NA value in element 15.
所以我想对延迟1,2和3进行延迟比较...所以上面的代码无法满足此要求，所以我尝试以下操作:
So I want to do this diff in lags for say lags 1,2 and 3... So the above code doesn't cater for this, so I try the below:
lag1 <- diff(vec, lag = 1, differences = 1, arithmetic = TRUE, na.pad = TRUE)
lag2 <- diff(vec, lag = 2, differences = 1, arithmetic = TRUE, na.pad = TRUE)
lag3 <- diff(vec, lag = 3, differences = 1, arithmetic = TRUE, na.pad = TRUE)
length(lag1)
length(lag2)
length(lag3)
结果:
> lag1
[1] 0.24886 0.12443 0.37330 0.24886 0.00000 0.00000 -0.37329 0.12443 0.24886 -0.37329 -1.11989 0.00000 -0.12442 0.12442
> lag2
[1] 0.37329 0.49773 0.62216 0.24886 0.00000 -0.37329 -0.24886 0.37329 -0.12443 -1.49318 -1.11989 -0.12442 0.00000
> lag3
[1] 0.74659 0.74659 0.62216 0.24886 -0.37329 -0.24886 0.00000 0.00000 -1.24432 -1.49318 -1.24431 0.00000
> length(lag1)
[1] 14
> length(lag2)
[1] 13
> length(lag3)
[1] 12
请注意，当执行上述滞后差值时...将diff结果放在减去了该值的那一行上，因此采用了我们的当前值-滞后值.它将差异结果放置在滞后值位置.然后我们失去向量的长度.我想实际将diff-滞后结果放置在起始编号(diff)上，并放置前导NA以解决数据集起始处的缺失值.
Notice that when do the lagged difference above... it places the diff result on the line that it subtracted the value on... so it took our current value - lagged value. It places the diff result on the lagged value position. We then lose the length of the vector. I want to actually place the diff - lagged result on the start number (diff) and place leading NA's to account for the missing values at the start of the data set.
以滞后2为例，这是我想要的结果:
Using lag 2 as en example, this is my desired result:
> lag2
[1] NA NA 0.37329 0.49773 0.62216 0.24886 0.00000 -0.37329 -0.24886 0.37329 -0.12443 -1.49318 -1.11989 -0.12442 0.00000
有人知道如何纠正此问题吗?
Does anyone know a way on how to correct this??
也许还要解释一下:
这是向量的开始:
vec <- c(43.79979, 44.04865, 44.17308.....
所以，如果我们做一个滞后的2个差异...
So if we do a lagged 2 difference...
我们取第三个元素... 44.17308-43.79979 = 0.37329的结果.
We take the 3rd element... 44.17308 - 43.79979 = the result of 0.37329.
所以我想拥有NA NA 0.37329
So I want to have NA NA 0.37329
而不是将0.37329放在新的lag2向量的第一个位置上.
Instead of placing 0.37329 on the first position in the new lag2 vector.
推荐答案
就像
2018年5月10日 @thistleknot对我说(谢谢！)，
On May 10th 2018 it was pointed to me by @thistleknot (thanks!) that 我想我找到了罪魁祸首:github.com/tidyverse/dplyr/issues/1586
答:这是拥有大量R包的自然结果.
只是明确地使用stats :: lag或dplyr :: lag I think I found the culprit: github.com/tidyverse/dplyr/issues/1586
answer: This is a natural consequence of having lots of R packages.
Just be explicit and use stats::lag or dplyr::lag 这篇关于滚动滞后差异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！
dplyr掩盖了stats自己的lag泛型.因此，请确保没有附加dplyr，或者明确地运行stats::lag，否则我的代码将无法运行.
dplyr masks stats's own lag generic. Therefore make sure you don't have dplyr attached, or instead run stats::lag explicitly, otherwise my code won't run.
