R - greatest common divisor dplyr routine

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 of the schoolmath library. I am looping through my data and store the values in the vector v. Finally, I use the min of v to find the gcd of my data.

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 for loop).

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

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

Any clue ?

Looks like the `gcd` is not vectorized. Perhaps `dt %>% mutate(dur1 = lag(dur, default = dur[1])) %>% rowwise() %>% mutate(new1 = gcd(dur, dur1))` — akrun, Aug 14 '16 at 14:09
Here's a vectorized version of gcd that could be helpful http://stackoverflow.com/a/21504113/3001626 — David Arenburg, Aug 14 '16 at 14:30
Using this function, you could do `dt %>% mutate(res = gcd(dur, lag(dur))) %>% summarise(Min = min(res, na.rm = TRUE))` which will be probably much faster than the `schoolmath::gcd` one — David Arenburg, Aug 14 '16 at 14:42

akrun · Accepted Answer · 2016-08-14T14:25:37.433

2

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 Vectorized 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.

edited Aug 14 '16 at 14:25

answered Aug 14 '16 at 14:15

akrun

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thank you - great answer. I am surprised actually that the code has to be so long. – giac Aug 14 '16 at 17:12
1

@giacomoV I was extracting the `min` as a single value. If you want as a data.frame, it could be `dt %>% summarise(Min = min(gcdV(dur, lag(dur, default = dur[1]))[-1]))` – akrun Aug 14 '16 at 17:20

R - greatest common divisor dplyr routine

1 Answers1