R：按组创建数字序列，并按特定条件启动序列

Question

I would like to create a new variable, Number, which sequentially generate numbers within a group ID, starting at a particular condition (in this case, when Percent > 5).

groupID <- c(1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3)
Percent <- c( 3, 4, 5, 10, 2, 1, 6, 8, 4, 8, 10, 11)

Number <- ifelse (Percent < 5, 0, 1:4)

I get:

> Number
[1] 0 0 3 4 0 0 3 4 0 2 3 4

But I'd like:

    0 0 1 2 0 0 1 2 0 1 2 3

I did not include groupID variable within the ifelse statement and used 1:4 instead, as there are always 4 rows within each groupID.

Any suggestions or clues? Thank you!

Answer 1

 ave(Percent, groupID, FUN=function(x) cumsum(x>=5))
[1] 0 0 1 2 0 0 1 2 0 1 2 3

To the example in the comments below, this is my alternate logical test to be cumsum() -ed:

ave(Percent, groupID, FUN=function(x) cumsum(seq_along(x)>= which(x >=5)[1]) )

Answer 2

It's ugly and throws warnings, but it gets you what you want:

ave(Percent,groupID,FUN=function(x) {x[x<5] <- 0; x[x>=5] <- 1:4; x} )
#[1] 0 0 1 2 0 0 1 2 0 1 2 3

@BondedDust's answer below using cumsum is almost certainly more appropriate though.

If your data was not always in ascending order in each group, you could also replace all the >=5 values like:

Percent <- c( 3, 5, 4, 10, 2, 1, 6, 8, 4, 8, 10, 11)
ave(Percent, list(groupID,Percent>=5), FUN=function(x) cumsum(x>=5))
#[1] 0 1 0 2 0 0 1 2 0 1 2 3

Answer 3

Try this:

ID <- c(1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3)
Percent <- c( 3, 4, 5, 10, 2, 1, 6, 8, 4, 8, 10, 11)


Number <- Percent >= 5

result = lapply(seq_along(Number), function(i){
    if( length(which(! Number[1:i]) ) == 0){start = 1}
    else {start =max(which(! Number[1:i]) )}

    sum( Number[start : i])

  })

> unlist(result)
[1] 0 0 1 2 0 0 1 2 0 1 2 3