# How to add new column in R data frame showing sum of a value in a current row and a prior row, if certain conditions are met in the 2 rows?

## Issue

Suppose you have a data frame of columns "a" and "b" with the values shown below, generated with `df <- data.frame(a=c(0, 1, 2, 2, 3), b=c(1, 3, 8, 9, 4))`. Suppose you want to add a column "c", whereby if a value in "a" equals the value in the immediately preceding row in col "a", then the corresponding row values in col "b" are summed; otherwise a 0 value is shown. A column "c" is added to the below to illustrate what I’m trying to do:

``````   a  b   add col c
1  0  1       0
2  1  3       0
3  2  8       0
4  2  9       17 (since the values in col "a" rows 3 and 4 are equal, add the values in col b rows 3 and 4)
5  3  4       0
``````

Or in this scenario, whereby cols "a" and "b" are generated by `df <- data.frame(a=c(0,1,2,2,2,3), b=c(1,2,3,4,5,6))`:

``````   a  b    add col c
1  0  1        0
2  1  2        0
3  2  3        0
4  2  4        7 (3+4 from col "b")
5  2  5        9 (4+5 from col "b")
6  3  6        0 (since 2 from prior row <> 3 from current row)
``````

What is the easiest way to do this in native R?

## Solution

As we are interested in the adjacent values to be equal, use `rleid` (from `data.table`) to create a grouping index, then create the ‘c’, by adding the ‘b’ with `lag` of ‘b’ and replace the default first value of `lag` (`NA`) to 0

``````library(dplyr)
library(data.table)
library(tidyr)
df %>%
group_by(grp = rleid(a)) %>%
mutate(c = replace_na(b + lag(b), 0)) %>%
ungroup %>%
select(-grp)
``````

-output

``````# A tibble: 6 × 3
a     b     c
<dbl> <dbl> <dbl>
1     0     1     0
2     1     2     0
3     2     3     0
4     2     4     7
5     2     5     9
6     3     6     0
``````

Or using `base R` – a similar approach is with the `rle` to create the ‘grp’, then use `ave` to do the addition of previous with current value (by removing the first and last) and then append 0 at the beginning

``````grp <- with(rle(df\$a), rep(seq_along(values), lengths))
df\$c <- with(df, ave(b, grp, FUN = function(x) c(0, x[-1] + x[-length(x)])))
``````