使用 dplyr 和 rle 计算具有条件的组中的连续值

Question

My question is very similar to the one posed below, however I want to add an additional command to return only cases when a sequence has more than 2 consecutive values.

How do I count the number of consecutive "success" (ie 1 in $consec) when a given sequence run has more than 2 consecutive numbers, within a given Era and a given Year?

Similar question to: Summarize consecutive failures with dplyr and rle . For comparison, I've modified the example used in that question:

library(dplyr)
df <- data.frame(Era=c(1,1,1,1,1,1,1,1,1,1),Year = c(1,2,2,3,3,3,3,3,3,3), consec = c(0,0,1,0,1,1,0,1,1,1))

df %>%
  group_by(Era,Year) %>%
  do({tmp <- with(rle(.$consec==1), lengths[values])
      data.frame(Year= .$Year, Count=(length(tmp)))}) %>% 
  slice(1L)

> Source: local data frame [3 x 3]
> Groups: Era, Year

>   Era Year Count
> 1   1    1     0
> 2   1    2     1
> 3   1    3     2
> 

All I need now is to add a condition to include only cases of consecutive numbers in a sequence of >2. Desired result:

> Source: local data frame [3 x 3]
> Groups: Era, Year

>   Era Year Count
> 1   1    1     0
> 2   1    2     0
> 3   1    3     1

Any advice would be greatly appreciated.

Answer 1

We need to create a logical index with lengths and get the sum of it

df %>%
   group_by(Era, Year) %>% 
   do({ tmp <- with(rle(.$consec), sum(lengths > 2))
   data.frame(Count = tmp)})
#   Era  Year Count
#  <dbl> <dbl> <int>
#1     1     1     0    
#2     1     2     0
#3     1     3     1