Performance of grepl versus more crude method using substr() and ==

I wanted test if the name of parameters begins with max. First I used a rather crude substr(var, begin, length) and then I had a better idea: using grepl. But was the more crude version really the worst ?

Here is a little test that permits to use the function microbenchmark in the package of the same name.

> times <- microbenchmark( grepl("^max", "max_152"), substr("max_152", 1, 3) == "max", times=1e3)
> times
Unit: nanoseconds
                             expr  min   lq     mean median     uq   max neval
         grepl("^max", "max_152") 3980 4255 4863.691 4539.5 4941.0 30161  1000
 substr("max_152", 1, 3) == "max"  995 1205 1492.998 1340.5 1579.5 16659  1000


The returned data frame has the following informations:
expr: the tested expressions
neval: how many time they have been evaluated
min, lq, mean, median, uq, max are respectively the minimum, lower quartile, mean, median, upper quartile and maximum time for the evaluation of 1 iteration. Note that Max value is wrong.


And finally... grepl is not efficient at all !

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