连续超过R的阈值和附加条件

时间:2019-04-04 09:20:19

标签: r time-series multiple-conditions

我想使用R获得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时间步长):

[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))

我到目前为止有什么

我能够组合条件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,只需输入:

first_exceed_seq(dat[,1])

这给出28。这应该是正确的答案,但我想知道以下问题。

问题

1)我想在上述函数中添加第三个条件,以使从29到32的总和大于1。 通过上面的函数,我将最小长度设置为3。我将这个长度应用于多个时间序列,并且我可能会遇到一个具有四个连续正值或更多正值的时间序列,并且从此开始的第一时间步不满足[3]而是第二或第三时间步等。

关于如何执行此R的任何建议?我将不胜感激。

更新:我尝试了以下解决方案,但dplyr发出警告消息。

  

1:在filter_impl(.data,quo)中:     强制进行lead的混合评估。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。

1 个答案:

答案 0 :(得分:1)

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

如果只希望第一次出现,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:像我一样包含dplyr软件包,或将lead替换为filter::lead