Counting continuous sequences of months

Question

If I have a vector of year and month coded like this:

ym <- c(
  201401,
  201403:201412,
  201501:201502,
  201505:201510,
  201403
)

And I'd like to end up with a vector that looks like this:

 [1]  1  1  2  3  4  5  6  7  8  9 10  11  12  1  2  3  4  5  6  1

That is, I want to count continuous sequences of month records. Can anyone recommend an approach? I've spinning my wheels with something like this:

ym_date <- as.Date(paste0(ym, 01), format = "%Y%m%d")

diff(ym_date)

but haven't been able to get any farther because I'm not sure how to flag that start of a sequence when we are dealing with months. Any base R, tidyverse, data.frame centric or not solution would be welcomed.

Maybe of some help: [How to partition a vector into groups of regular, consecutive sequences?](https://stackoverflow.com/questions/5222061/how-to-partition-a-vector-into-groups-of-regular-consecutive-sequences) — markus, Feb 14 '20 at 22:02
Or [Number of months between two dates](https://stackoverflow.com/questions/1995933/number-of-months-between-two-dates) — markus, Feb 14 '20 at 22:11
I might have poorly asked my question but I don't quite see how that one answers this question. — boshek, Feb 14 '20 at 22:12
@akrun it isn't clear to me why my question was closed. The answer sort of answer the original post with something like this `tibble(ym, group = cumsum(c(1, diff(ym) != 1))) %>% group_by(group) %>% mutate(months = length(group))` but I don't consider it is trivial leap. — boshek, Feb 14 '20 at 22:21
Your expected output based on the input `201501 201502 201505 201506` is `1 2 3 4` is that right — akrun, Feb 14 '20 at 22:32
`as.numeric(substr(ym, 5, 6))`? It's sort of unclear what's being asked. — Rich Pauloo, Feb 14 '20 at 22:39
Yes and that was terrible SO manners by me. I can't say which one is objectively better. — boshek, Feb 14 '20 at 23:41
@boshek It's ok. Just that I was going in a different route with the logic — akrun, Feb 14 '20 at 23:42

akrun · Answer 1 · 2020-02-14T23:41:50.947

1

We can use

library(lubridate)
mth <- month(ym_date)
new <- mth + cumsum(c(0, (mth %/% 12)[-length(mth)])) * 12
ave(mth, cumsum(c(TRUE, diff(new) != 1)), FUN = seq_along)
#[1]  1  1  2  3  4  5  6  7  8  9 10 11 12  1  2  3  4  5  6  1

It can be also written in a more compact way

ave(mth, cumsum(c(TRUE, diff(c(0, head(cumsum(mth == 12), -1)) * 12 + mth) != 1)), FUN = seq_along)
#[1]  1  1  2  3  4  5  6  7  8  9 10 11 12  1  2  3  4  5  6  1

edited Feb 14 '20 at 23:41

answered Feb 14 '20 at 21:56

akrun

874,273
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Oops - I screwed up my expected output. I need the count to span the end of the year and into the next. So the output _should_ be `1 1 2 3 4 5 6 7 8 9 10 11 12 1 2 3 4 5 6 1` – boshek Feb 14 '20 at 22:57

ThomasIsCoding · Accepted Answer · 2020-02-14T23:03:18.660

1

Maybe you can try the following base R code with rle

r <- unlist(sapply(rle(cumsum(c(1,round(as.numeric(diff(ym_date))/30.24)!=1)))$lengths,seq_along))

or with ave

r <- ave(ym,cumsum(c(1,round(as.numeric(diff(ym_date))/30.24)!=1)),FUN = seq_along)

such that

> r
 [1]  1  1  2  3  4  5  6  7  8  9 10  11  12  1  2  3  4  5  6  1

edited Feb 14 '20 at 23:03

answered Feb 14 '20 at 22:52

ThomasIsCoding

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Thank you. Question: can you explain why 30.24? It works but I'm not totally sure why. – boshek Feb 14 '20 at 23:20
1

@boshek this is used to calculate convert days to month, you can take it as a constant for this kind of calculation – ThomasIsCoding Feb 14 '20 at 23:22

Counting continuous sequences of months

2 Answers2