How to loop ifelse function through a grouped variable with dplyr

Question

I'm trying to apply a rule for a group of IDs that, upon the first instance where the value for a variable in one row equals 1, all values for another variable in all subsequent rows in that group equal 1.

Essentially, here is what I am trying to do:

I have:

I want:

I'm trying to use dplyr to iterate through a series of grouped rows, in each one applying an ifelse function. My code looks like this:

data$prev = 0
data <-   
data %>%
group_by(id)%>%
mutate(prev = if_else(lag(prev) == 1 | lag(d) == 1, 1, 0))

But for some reason, this is not applying the ifelse function over the whole group, resulting in data that looks something like this:

Can anyone help me with this?

Edo · Accepted Answer · 2020-11-27 19:04:18Z

1

What about this:

library(dplyr)
df %>% 
 group_by(ID) %>% 
 mutate(prev = +(cumsum(c(0, D[-length(D)])) > 0)) %>% 
 ungroup()

#> # A tibble: 14 x 3
#>       ID     D  prev
#>    <int> <int> <int>
#>  1     1     1     0
#>  2     1     0     1
#>  3     1     0     1
#>  4     2     0     0
#>  5     2     0     0
#>  6     3     1     0
#>  7     3     0     1
#>  8     3     0     1
#>  9     4     1     0
#> 10     4     0     1
#> 11     4     1     1
#> 12     4     1     1
#> 13     4     1     1
#> 14     4     0     1

To explain what it does, let's just take a simple vector. The calc will be the same for each group.

Be x our vector

x <- c(0,0,0,1,1,0,0,2,3,4)

Do the cumulative sum over x

cumsum(x)
#>  [1]  0  0  0  1  2  2  2  4  7 11

You are interested only on value above zeros, therefore:

cumsum(x)>0
#>  [1] FALSE FALSE FALSE  TRUE  TRUE  TRUE  TRUE  TRUE  TRUE  TRUE

You don't want logical, but numeric. Just a + makes the trick

+(cumsum(x)>0)
#>  [1] 0 0 0 1 1 1 1 1 1 1

However, you want the 1s delayed by 1. Thus, we had a zero on top of x

+(cumsum(c(0,x))>0)
#>  [1] 0 0 0 0 1 1 1 1 1 1 1

We need to keep the same length, so we remove the last value of x.

+(cumsum(c(0, x[-length(x)])) > 0)
#>  [1] 0 0 0 0 1 1 1 1 1 1

And that makes the trick.

edited Nov 27, 2020 at 19:04

answered Nov 27, 2020 at 18:35

Edo

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3 Comments

Izzy Over a year ago

This worked! How did this do what I wanted?

Edo Over a year ago

I've added an edit to explain what it does

Edo Over a year ago

However, @akrun solution is better. You should use that code! :-)

akrun · Accepted Answer · 2020-11-27 19:04:30Z

1

We can use lag

library(dplyr)
df %>% 
   group_by(ID) %>%
   mutate(prev = lag(cumsum(D) > 0, default = 0))

-output

# A tibble: 14 x 3
# Groups:   ID [4]
#      ID     D  prev
#   <dbl> <dbl> <dbl>
# 1     1     1     0
# 2     1     0     1
# 3     1     0     1
# 4     2     0     0
# 5     2     0     0
# 6     3     1     0
# 7     3     0     1
# 8     3     0     1
# 9     4     1     0
#10     4     0     1
#11     4     1     1
#12     4     1     1
#13     4     1     1
#14     4     0     1

data

df <- data.frame(
    ID = c(1, 1, 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 4, 4),
    D = c(1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 1, 1, 0)
)

answered Nov 27, 2020 at 19:04

akrun

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Comments

geom_na · Accepted Answer · 2020-11-27 18:39:55Z

You can use a new function from dplyr dplyr::group_modify to apply function over groups

df <- data.frame(
    ID = c(1, 1, 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 4, 4),
    D = c(1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 1, 1, 0)
)

df %>% group_by(ID) %>% group_modify(
    function(x, y){
        boo <- x[1, ]$D == 1
        ifelse(boo, 
               {x$prev = 1
                x$prev[1] = 0
               },
               {x$prev = 0})
        
        x
    }
)

# A tibble: 14 x 3
# Groups:   ID [4]
      ID     D  prev
   <dbl> <dbl> <dbl>
 1     1     1     0
 2     1     0     1
 3     1     0     1
 4     2     0     0
 5     2     0     0
 6     3     1     0
 7     3     0     1
 8     3     0     1
 9     4     1     0
10     4     0     1
11     4     1     1
12     4     1     1
13     4     1     1
14     4     0     1

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