Question

这个问题与我以前的帖子有关： Consecutive exceedance above a threshold and additional conditions in R

以下是数据：

 dat <- 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))

我想要的

我想获得满足以下修改条件的第一时间步：

[1] V1 > 0 at the time step

[2] In the succeeding FOUR time steps (including the timestep in [1]), V1 > 0 in AT LEAST THREE timesteps

[3] Accumulated value of the next FOUR timesteps (including the timestep in [1]) should be greater than 1.

这是到目前为止的脚本：

library(dplyr)

newx <- dat %>% as_tibble() %>%
mutate(time = 1: n()) %>%  
filter(V1 > 0, dplyr::lead(V1, 1) > 0, dplyr::lead(V1, 2) > 0, 
(dplyr::lead(V1, 1) + dplyr::lead(V1, 2) + dplyr::lead(V1, 3) + 
dplyr::lead(V1, 4)) > 1)

输出

> newx
# A tibble: 7 x 2
    V1  time
   <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

问题

我不知道如何正确实现第二个条件。它应该检查四个时间步长中是否有三个>0。是否连续是无关紧要的。

预期产量

正确答案应该是28。

我会尽力帮助您。

Answer 1

如果我已正确理解，并且希望第一行符合条件，则可以使用zoo::rollsum：

library(zoo)
library(dplyr)

dat %>%
  rownames_to_column() %>%
  filter(V1 > 0 &
           rollsum(V1 > 0, 4, fill = NA, align = "left") >= 3 &
           rollsum(V1, 4, fill = NA, align = "left") > 1) %>%
  slice(1)

  rowname      V1
1      28 2.32796

Answer 2

使用stats::filter进行总和：

which(
  (dat$V1 > 0) &
  (rev(stats::filter(rev(dat$V1 > 0), rep(1,4), sides=1)) >= 3) &
  rev(stats::filter(rev(dat$V1), rep(1,4), sides=1))
)[1]
#[1] 28

或者如果您必须并入dplyr：

dat %>% 
  slice(
    which(
      (rev(stats::filter(rev(V1 > 0), rep(1,4), sides=1)) >= 3) &
      (V1 > 0) &
      rev(stats::filter(rev(V1), rep(1,4), sides=1))
    )[1]
  )
## A tibble: 1 x 1
#     V1
#  <dbl>
#1  2.33

Answer 3

Wordier：

 library(dplyr)

 dat2 <- dat %>%
   tibble::rowid_to_column() %>%
   mutate(gtz = (V1 > 0) * 1,
          gtz_cuml = cumsum(gtz),
          gtz_next_three = lead(gtz_cuml, 3) - lag(gtz_cuml),
          cuml_V1 = cumsum(V1),
          V1_next_three = lead(cuml_V1, 3) - lag(cuml_V1)) %>%
   filter(gtz > 0,
          gtz_next_three >= 3,
          V1_next_three > 1) %>%
   slice(1)


#>  dat2
#  rowid      V1 gtz gtz_cuml gtz_next_three   cuml_V1 V1_next_three
#1    28 2.32796   1        2              3 -71.22716      9.959473

使用R（dplyr）在过滤器中实现“至少”条件

3 个答案: