giacomoV - 1 year ago 71

R Question

I need to find the greatest common divisor (gcd) for a set of durations:

`dur`

My data look like this

`actrec dur`

1 c Personal Care 120

2 c Free Time 10

3 c Free Time 70

4 c Free Time 40

5 b Unpaid 10

6 c Free Time 20

7 c Personal Care 30

8 c Free Time 40

9 c Free Time 40

10 c Free Time 10

I am using the function

`gcd`

`schoolmath`

I am looping through my data and store the values in the vector

`v`

Finally, I use the

`min`

`v`

`library(schoolmath)`

l = length(dt$dur)

v = array(0, l)

for(i in 2:l){

v[i] = gcd(dt$dur[i], dt$dur[i-1])

}

minV = min(v[-1])

minV

Which gives

`10`

However, I have trouble translating this routine into

`dplyr`

I thought of something like (

`lag`

`dt %>% mutate(gcd(dur, lag(dur, 0)))`

But it isn't working. And I am unsure how to insert

`min`

Any clue ?

Answer Source

We can use `rowwise`

to apply the `gcd`

function on each row after taking the `lag`

of 'dur, extract the 'new1' and get the `min`

```
dt %>%
mutate(dur1 = lag(dur, default = dur[1])) %>%
rowwise() %>%
mutate(new1 = gcd(dur, dur1)) %>%
.$new1 %>%
tail(.,-1) %>%
min
#[1] 10
```

Or we create a `Vectorize`

d function of 'gcd' and apply on the 'dur' column

```
gcdV <- Vectorize(function(x,y) gcd(x, y))
dt %>%
mutate(new1 = gcdV(dur, lag(dur, default = dur[1])))
```

and get the `min`

as in the above solution.