连续超过R的阈值和附加条件

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))

到目前为止我有什么

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，只需键入：

first_exceed_seq(dat[,1])

得出28。这应该是正确的答案，但是我想知道以下问题。

问题

1）我想在上面的函数中添加第三个条件，以使29到32的总和大于1。从上面的函数中，我将最小长度设置为3。我将把它应用于多个时间序列，可能会遇到具有四个或四个以上连续正值的时间序列，并且由此产生的第一个时间步不满足[3]，而是第二个或第三个时间步等。

关于如何执行此R有任何建议吗？我将不胜感激。

更新：我尝试了下面的解决方案，但dplyr发出警告消息。

1：在filter_impl（.data，quo）中：强制执行的混合评估lead。请使用dplyr :: lead（）或library（dplyr）删除此警告。

同样正确的答案应该是28，因为它首先满足所有三个条件。