How to replace every nth element in a column if the element is a specific value in R

Question

I want to transform the following dataframe without using a loop, changing the value of 'a' in the 1st and every 3rd element after to a value of 5 if the value of 'b' at that row is 1.

df <- data.frame(a =c(2,2,2,2,2,2,2,2,2,2),b =  c(1,1,0,1,1,0,0,1,1,1))


df
   a b
1  2 1
2  2 1
3  2 0
4  2 1
5  2 1
6  2 0
7  2 0
8  2 1
9  2 1
10 2 1

transforming df such that the value of 'a' at index 1,4,10 of column a is replaced by 5 since the value of 'b' is 1 at index of 1,4,10 in column b

 df
   a b
1  5 1
2  2 1
3  2 0
4  5 1
5  2 1
6  2 0
7  2 0
8  2 1
9  2 1
10 5 1

I have tried using df[(seq(1,to=nrow(df),by=3))]==1 to check for value in 'b' at those specific indexes, but I am not sure how to incorporate it to change the value in 'a' without the use of a loop

Answer 1

You may try

library(dplyr)

df %>%
  tibble::rownames_to_column() %>%
  rowwise %>%
  mutate(rowname = as.numeric(rowname),
         a = ifelse((rowname %% 3 == 1) && b == 1 , 5, a)) %>%
  select(-rowname)

       a     b
   <dbl> <dbl>
 1     5     1
 2     2     1
 3     2     0
 4     5     1
 5     2     1
 6     2     0
 7     2     0
 8     2     1
 9     2     1
10     5     1

Answer 2

For a base R solution,

df[intersect(seq(1,nrow(df),by=3), which(df$b == 1)),'a'] <- 5

Returns,

> df
   a b
1  5 1
2  2 1
3  2 0
4  5 1
5  2 1
6  2 0
7  2 0
8  2 1
9  2 1
10 5 1

Answer 3

You can keep the dplyr solution to a single line using @Park's calculation.

library(dplyr)

df <- data.frame(a =c(2,2,2,2,2,2,2,2,2,2),b =  c(1,1,0,1,1,0,0,1,1,1))

mutate(df, a = ifelse(row_number() %% 3 == 1 & b == 1, 5, a))
#>    a b
#> 1  5 1
#> 2  2 1
#> 3  2 0
#> 4  5 1
#> 5  2 1
#> 6  2 0
#> 7  2 0
#> 8  2 1
#> 9  2 1
#> 10 5 1

Answer 4

In base R, you can use

df$a[seq_len(nrow(df)) %% 3 == 1 & df$b == 1] <- 5
df
#   a b
#1  5 1
#2  2 1
#3  2 0
#4  5 1
#5  2 1
#6  2 0
#7  2 0
#8  2 1
#9  2 1
#10 5 1