数据整理，根据时间阈值保留行

Question

I have a data frame where I have a time variable (seconds) and a site variable, but basically the issue is just regarding the time. I want to select which rows are "good" or "bad" based on a specific time threshold.

Hypothetically, let's assume that my threshold is time_threshold >= 250

Here's an example dataset:

data.frame(site = c("site1",
                    "site2",
                    "site2",
                    "site2",
                    "site2",
                    "site2",
                    "site3",
                    "site3",
                    "site1",
                    "site3",
                    "site3"),
           time_difference = c(250,
                               300,
                               277,
                               137,
                               75,
                               85,
                               108,
                               91,
                               0,
                               118,
                               113))

Ultimately I want to have something like this, where each row is assigned either a "good" or "bad":

data.frame(site = c("site1",
                    "site2",
                    "site2",
                    "site2",
                    "site2",
                    "site2",
                    "site3",
                    "site3",
                    "site1",
                    "site3",
                    "site3"),
           time_difference = c(250,
                               300,
                               277,
                               137,
                               75,
                               85,
                               108,
                               91,
                               0,
                               118,
                               113),
status = c("good",
                      "good",
                      "good",
                      "good",
                      "bad",
                      "good",
                      "bad",
                      "good",
                      "bad",
                      "bad",
                      "good"))

The way that each row is assigned a status is based on time streaks. I'll try to explain: starting from the first row we see that time_difference is equal to 250 which is equal to our threshold thus a good is assigned to the status column, the next two rows are also "good" as they are above the threshold.

Once we get to row four we see that the time difference is 137, in this case we need to cumulatively add all following rows until our threshold is reached. In this case 137+75+85 = 297 . Once this is established, the first row is given a good and the last row becomes the start of the new streak, whilst the row that has 75 is given a "bad" (anything between the starting row and start of the next streak is given "bad").

This process continues until the end of the dataset. (ie 85+108+91 = 284 keep 85 and 91 and gives "bad" to 108; 91+118+0+113 = 332 keep 91 and 113 gives "bad" to 118 and 0).

I hope this is relatively clear, basically I want to keep the first row of each 250 streak and make the last row of the steak the next starting row.

Answer 1

I think this loop captures your logic.

df$status   <- 'bad'
running_sum <- 0
mem         <- 1

for(i in 1:length(df$time_difference)) {
  
  if(running_sum == 0 & df$time_difference[i] != 0) mem <- i
  
  running_sum <- running_sum + df$time_difference[i]
  
  if(running_sum >= 250) {
    running_sum    <- df$time_difference[i]
    df$status[mem] <- 'good'
    df$status[i]   <- 'good'
  }
}

df
#>     site time_difference status
#> 1  site1             250   good
#> 2  site2             300   good
#> 3  site2             277   good
#> 4  site2             137   good
#> 5  site2              75    bad
#> 6  site2              85   good
#> 7  site3             108    bad
#> 8  site3              91   good
#> 9  site1               0    bad
#> 10 site3             118    bad
#> 11 site3             113   good

^{Created on 2022-03-17 by the reprex package (v2.0.1)}