我想用R满足以下条件的时间序列中的时间步长(应该是满足以下条件的第一步):
[1] V1 > 0 at the time step
[2] V1 > 0 in at least 3 consecutive time steps from the timestep obtained in [1]
[3] Accumulated value of the next four timesteps following [1] should be greater than 1.
这是数据
structure(list(V1 = c(-3.85326, -2.88262, -4.1405, -3.95193,
-6.68925, -2.04202, -2.47597, -4.91161, -2.5946, -2.82873, 2.68839,
-4.1287, -4.50296, -0.143476, -1.12174, -0.756168, -1.67556,
-1.92704, -1.89279, -2.37569, -5.71746, -2.7247, -4.12986, -2.29769,
-1.52835, -2.63623, -2.31461, 2.32796, 4.14354, 4.47055, -0.557311,
-0.425266, -2.37455, -5.97684, -5.22391, 0.374004, -0.986549,
2.36419, 0.218283, 2.66014, -3.44225, 3.46593, 1.3309, 0.679601,
5.42195, 10.6555, 8.34144, 1.64939, -1.64558, -0.754001, -4.77503,
-6.66197, -4.07188, -1.72996, -1.15338, -8.05588, -6.58208, 1.32375,
-3.69241, -5.23582, -4.33509, -7.43028, -3.57103, -10.4991, -8.68752,
-8.98304, -8.96825, -7.99087, -8.25109, -6.48483, -6.09004, -7.05249,
-4.78267)), class = "data.frame", row.names = c(NA, -73L))
到目前为止我有什么
我能够结合条件1和2.这是脚本。
first_exceed_seq <- function(x, thresh = 0, len = 3)
{
# Logical vector, does x exceed the threshold
exceed_thresh <- x > thresh
# Indices of transition points; where exceed_thresh[i - 1] !=
exceed_thresh[i]
transition <- which(diff(c(0, exceed_thresh)) != 0)
# Reference index, grouping observations after each transition
index <- vector("numeric", length(x))
index[transition] <- 1
index <- cumsum(index)
# Break x into groups following the transitions
exceed_list <- split(exceed_thresh, index)
# Get the number of values exceeded in each index period
num_exceed <- vapply(exceed_list, sum, numeric(1))
# Get the starting index of the first sequence where more then len
exceed thresh
transition[as.numeric(names(which(num_exceed >= len))[1])]
}
然后,使用上面的函数,只需键入:
first_exceed_seq(dat[,1])
这给出了28.这应该是正确的答案,但我想知道以下问题。
问题
1)我想在上面的函数中添加第三个条件,使得从29到32之和大于1.从上面的函数中,我将最小长度设置为3.I将这个应用于多个时间序列而我可能会遇到一个具有四个连续正值或更多的时间序列,并且第一次从此步骤不满足[3]而不是第二或第三个时间步等。
关于如何做这个R的任何建议?我会感激任何帮助。
更新:我尝试了下面的解决方案,但dplyr给出了警告消息。
1:在filter_impl(.data,quo)中:强制为
lead
进行混合评估。请使用dplyr :: lead()或库(dplyr)删除此警告。
正确的答案应该是28,因为它首先满足所有三个条件。
这是使用dplyr
包和lead
函数的解决方案。在以下代码中,x
是您提供的数据:
library(dplyr)
newx <- x %>% as_tibble() %>%
mutate(time = 1: n()) %>%
filter(V1 > 0, lead(V1, 1) > 0, lead(V1, 2) > 0,
lead(V1, 1) + lead(V1, 2) + lead(V1, 3) + lead(V1, 4) > 1)
# A tibble: 7 x 2
V1 idx
<dbl> <int>
1 2.33 28
2 2.36 38
3 3.47 42
4 1.33 43
5 0.680 44
6 5.42 45
7 10.7 46
如果您只想要第一次出现,可以使用slice
:
slice(newx, 1)
# A tibble: 1 x 2
V1 idx
<dbl> <int>
1 2.33 28
关于错误:要么像我一样包括dplyr
包,要么用lead
替换filter::lead
。