Replace values in a column unless there's already a "1" there-CodePudding

I have data like this:

df<-structure(list(a = c(1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0), b = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 1, 0, 0, 0)), row.names = c(NA, -19L), class = c("tbl_df", 
"tbl", "data.frame"))

I would like to replace values in column A based on column B. If column B has a "1" in it, I want to replace the row in column A with a 1.

I know this can do that:

df<-df %>%mutate(a=ifelse(str_detect(b,"1"),1,0))

The problem is, this replaces everything in column A based on those rules, overwriting what was already there. I only want to replace A if it didn't already have a "1". So my expected output would be:

CodePudding user response：

We may need just | on the binary column to replace the values in 'a' where 'b' is also 1

library(dplyr)
df %>%
    mutate(a =  (a|b))

-output

# A tibble: 19 × 2
       a     b
   <int> <dbl>
 1     1     0
 2     0     0
 3     0     0
 4     0     0
 5     0     0
 6     0     0
 7     0     0
 8     0     0
 9     0     0
10     0     0
11     0     0
12     0     0
13     0     0
14     0     0
15     0     0
16     1     1
17     0     0
18     0     0
19     0     0

Or in base R

df$a[df$b == 1] <- 1