R returns the minimum column index for each row

Question

I have a data.frame that contains 4 columns (given below). I want to find the minimum column index (NOT VALUE) for each row. Any idea should achieve this?

> d V1 V2 V3 V4 1 0.388116155 0.98999967 0.41548536 0.76093748 2 0.495971331 0.47173142 0.51582728 0.06789924 3 0.436495321 0.48699268 0.21187838 0.54139290 4 0.313514389 0.50265539 0.08054103 0.46019601 5 0.277275961 0.39055360 0.29594162 0.70622532 6 0.264804739 0.86996266 0.85708635 0.61136741 7 0.627344463 0.54277873 0.96769568 0.80399490 8 0.814420492 0.35362949 0.39023446 0.39246250 9 0.517459983 0.65895805 0.93662382 0.06762166 10 0.498319937 0.67081260 0.43225997 0.42139151 11 0.046862110 0.97304915 0.06542971 0.09779383 12 0.619009734 0.82363618 0.14514799 0.52858058 13 0.007262782 0.82203403 0.08573499 0.61094206 14 0.001602586 0.33241230 0.57762669 0.45285004 15 0.698388370 0.83541257 0.21051568 0.84431347 16 0.296088411 0.34363164 0.02179999 0.70551493 17 0.897869571 0.50625928 0.92861583 0.61249019 18 0.372497428 0.29025182 0.23201891 0.55737699 19 0.172931860 0.03604668 0.50291560 0.10850847 20 0.988827604 0.15800337 0.87999839 0.09899663 

So, I want the following output:

 1 1 2 4 3 3 4 3 

which continues for all rows. Thanks

Answer 1

In your English description you are invited to:

  apply( df, 1, which.min) 

But the answer you give is not formatted as a vector and is not the correct answer if the above interpretation is correct. Oh wait, you were expecting usurers.

  as.matrix(apply( d, 1, which.min)) [,1] 1 1 2 4 3 3 4 3 5 1 6 1 7 2 8 2 9 4 10 4 11 1 12 3 13 1 14 1 15 3 16 3 17 2 18 3 19 2 20 4 

Answer 2

Another option is max.col of d times -1

 max.col(-d) # [1] 1 4 3 3 1 1 2 2 4 4 1 3 1 1 3 3 2 3 2 4 

If you need a matrix as output, use

 cbind(1:nrow(d), # row max.col(-d)) # column position of minimum 

---

Here is a benchmark for two approaches

 set.seed(42) dd <- as.data.frame(matrix(runif(1e5 * 100), nrow = 1e5, ncol = 100)) library(microbenchmark) library(ggplot2) b <- microbenchmark( apply = apply(dd, 1, which.min), max_col = max.col(-dd), times = 25 ) autoplot(b) 

 b #Unit: milliseconds # expr min lq mean median uq max neval cld # apply 705.7478 855.7112 906.2340 892.3214 933.4655 1211.5016 25 b # max_col 162.8273 175.6363 227.1156 206.0213 225.2973 406.9124 25 a