The perils of benchmarking Q
Suppose we had a list of 10000000 phone numbers and wanted to take just the first 8 digits of each.
We could create the list like this:q)l:10000000 10#99?"0123456789"Here, 99?... creates a vector of 99 random characters from the set [0-9] and 10#... takes 10 characters from each generated "phone number". 10000000 ... repeats it that many times. A simple 8#l will just gives us the first 8 elements of the list which is not what we want.
q)f1:{x@\:til 8}
q)f2:{8#/:x}
q)f3:{8#'x}
The above 3 solutions will give us the first 8 digits of every phone number in the list. Which solution is the fastest one, though? The list is very very long specifically to make benchmarking easier. q)\t f1 l 906\t here gives us execution time for f1 in milliseconds. We proceed to time f2 and f3:
q)\t f2 l 738 q)\t f3 l 738Looks like f1 is much slower but is it really? Let's run the benchmark several times...
q)\t f1 l 624 q)\t f2 l 735See this thread in the non-commercial q/kdb forum for an additional note from Stevan Apter.
