how to cumulatively add values in one vector in R

Question

I have a data set that looks like this

id  name    year    job    job2
1   Jane    1980    Worker  0
1   Jane    1981    Manager 1
1   Jane    1982    Manage         


        

Answer 1

Here is the succinct dplyr solution for the same problem.

NOTE: Make sure that stringsAsFactors = FALSE while reading in the data.

library(dplyr)
dat %>%
  group_by(name, job) %>%
  filter(job != "Boss" | year == min(year)) %>%
  mutate(cumu_job2 = cumsum(job2))

Output:

   id name year     job job2 cumu_job2
1   1 Jane 1980  Worker    0         0
2   1 Jane 1981 Manager    1         1
3   1 Jane 1982 Manager    1         2
4   1 Jane 1983 Manager    1         3
5   1 Jane 1984 Manager    1         4
6   1 Jane 1985 Manager    1         5
7   1 Jane 1986    Boss    0         0
8   2  Bob 1985  Worker    0         0
9   2  Bob 1986  Worker    0         0
10  2  Bob 1987 Manager    1         1
11  2  Bob 1988    Boss    0         0

Explanation

1. Take the dataset
2. Group by name and job
3. Filter each group based on condition
4. Add cumu_job2 column.

Answer 2

Here is a base solution using within and ave. We assume that the input is DF and that the data is sorted as in the question.

DF2 <- within(DF, {
    seq = ave(id, id, job, FUN = seq_along)
    job2 = (job == "Manager") + 0
    cumu_job2 = ave(job2, id, job, FUN = cumsum)
})
subset(DF2, job != 'Boss' | seq == 1, select = - seq)

REVISION: Now uses within.

Answer 3

I think this does what you want, although the data must be sorted as you have presented it.

my.df <- read.table(text = '
id  name    year    job    job2
1   Jane    1980    Worker  0
1   Jane    1981    Manager 1
1   Jane    1982    Manager 1
1   Jane    1983    Manager 1
1   Jane    1984    Manager 1
1   Jane    1985    Manager 1
1   Jane    1986    Boss    0
1   Jane    1987    Boss    0
2   Bob     1985    Worker  0
2   Bob     1986    Worker  0
2   Bob     1987    Manager 1
2   Bob     1988    Boss    0
2   Bob     1989    Boss    0
2   Bob     1990    Boss    0
2   Bob     1991    Boss    0
2   Bob     1992    Boss    0
', header = TRUE, stringsAsFactors = FALSE)

my.seq <- data.frame(rle(my.df$job)$lengths)

my.df$cumu_job2 <- as.vector(unlist(apply(my.seq, 1, function(x) seq(1,x))))

my.df2 <- my.df[!(my.df$job=='Boss' & my.df$cumu_job2 != 1),]
my.df2$cumu_job2[my.df2$job != 'Manager'] <- 0

   id name year     job job2 cumu_job2
1   1 Jane 1980  Worker    0         0
2   1 Jane 1981 Manager    1         1
3   1 Jane 1982 Manager    1         2
4   1 Jane 1983 Manager    1         3
5   1 Jane 1984 Manager    1         4
6   1 Jane 1985 Manager    1         5
7   1 Jane 1986    Boss    0         0
9   2  Bob 1985  Worker    0         0
10  2  Bob 1986  Worker    0         0
11  2  Bob 1987 Manager    1         1
12  2  Bob 1988    Boss    0         0

Answer 4

@BrodieG's is way better:

The Data

dat <- read.table(text="id  name    year    job    job2
1   Jane    1980    Manager 1
1   Jane    1981    Manager 1
1   Jane    1982    Manager 1
1   Jane    1983    Manager 1
1   Jane    1984    Manager 1
1   Jane    1985    Manager 1
1   Jane    1986    Boss    0
1   Jane    1987    Boss    0
2   Bob     1985    Manager 1
2   Bob     1986    Manager 1
2   Bob     1987    Manager 1
2   Bob     1988    Boss    0
2   Bob     1989    Boss    0
2   Bob     1990    Boss    0
2   Bob     1991    Boss    0
2   Bob     1992    Boss    0", header=TRUE)

#The code:

inds1 <- rle(dat$job2)
inds2 <- cumsum(inds1[[1]])[inds1[[2]] == 1] + 1

ends <- cumsum(inds1[[1]])
starts <- c(1, head(ends + 1, -1))
inds3 <- mapply(":", starts, ends)
dat$id <- rep(1:length(inds3), sapply(inds3, length))
dat <- do.call(rbind, lapply(split(dat[, 1:5], dat$id ), function(x) {
    if(x$job2[1] == 0){ 
        x$cumu_job2 <- rep(0, nrow(x))
    } else { 
        x$cumu_job2 <- 1:nrow(x)
    }
    x
}))


keeps <- dat$job2 > 0
keeps[inds2] <- TRUE
dat2 <- data.frame(dat[keeps, ], row.names = NULL)
dat2

##    id name year     job job2 cumu_job2
## 1   1 Jane 1980 Manager    1         1
## 2   1 Jane 1981 Manager    1         2
## 3   1 Jane 1982 Manager    1         3
## 4   1 Jane 1983 Manager    1         4
## 5   1 Jane 1984 Manager    1         5
## 6   1 Jane 1985 Manager    1         6
## 7   2 Jane 1986    Boss    0         0
## 8   3  Bob 1985 Manager    1         1
## 9   3  Bob 1986 Manager    1         2
## 10  3  Bob 1987 Manager    1         3
## 11  4  Bob 1988    Boss    0         0

Answer 5

Contributed by Matthew Dowle:

dt[, .SD[job != "Boss" | year == min(year)][, cumjob := cumsum(job2)],
     by = list(name, job)]

Explanation

1. Take the dataset
2. Run a filter and add a column within each Subset of Data (.SD)
3. Grouped by name and job

---

Older versions:

You have two different split apply combines here. One to get the cumulative jobs, and the other to get the first row of boss status. Here is an implementation in data.table where we basically do each analysis separately (well, kind of), and then collect everything in one place with rbind. The main thing to note is the by=id piece, which basically means the other expressions are evaluated for each id grouping in the data, which was what you correctly noted was missing from your attempt.

library(data.table)
dt <- as.data.table(df)
dt[, cumujob:=0L]  # add column, set to zero
dt[job2==1, cumujob:=cumsum(job2), by=id]  # cumsum for manager time by person 
rbind(
  dt[job2==1],                     # this is just the manager portion of the data
  dt[job2==0, head(.SD, 1), by=id] # get first bossdom row
)[order(id, year)]                 # order by id, year
#       id name year     job job2 cumujob
#   1:  1 Jane 1980 Manager    1       1
#   2:  1 Jane 1981 Manager    1       2
#   3:  1 Jane 1982 Manager    1       3
#   4:  1 Jane 1983 Manager    1       4
#   5:  1 Jane 1984 Manager    1       5
#   6:  1 Jane 1985 Manager    1       6
#   7:  1 Jane 1986    Boss    0       0
#   8:  2  Bob 1985 Manager    1       1
#   9:  2  Bob 1986 Manager    1       2
#  10:  2  Bob 1987 Manager    1       3
#  11:  2  Bob 1988    Boss    0       0

Note this assumes table is sorted by year within each id, but if it isn't that's easy enough to fix.

---

Alternatively you could also achieve the same with:

ans <- dt[, .I[job != "Boss" | year == min(year)], by=list(name, job)]
ans <- dt[ans$V1]
ans[, cumujob := cumsum(job2), by=list(name,job)] 

The idea is to basically get the row numbers where the condition matches (with .I - internal variable) and then subset dt on those row numbers (the $v1 part), then just perform the cumulative sum.