4 ms·

I would suggest something more like: teachers[age > 30].count (Because I'm working on something exactly like this right now)

• moonchild 6y ago
  In apl (incl. apl-alikes—kdb, etc.), you can essentially say this directly. 'age > 30' produces an array with the same shape as 'age' with a 1 for instances that are >30 and 0 else. So for instance: 27 42 44 > 30 0 1 1 If all you want is a count of teachers older than 30, you can simply sum up the resulting array, so: +/0 1 1 ⍝note: '/' is reduction, so +/ is sum reduction 2 If you want the indices, you can use 'where', which will produce an array of indices corresponding to 1s in its argument, so ⍸0 1 1 1 2 Indicates that '0 1 1' has 1s at indices 1 and 2. (We could also take the length of this array to get the count of teachers older than 30, although that would be a bit silly compared to just summing the original array.) We can then use these indices to index a different array: ages←27 42 44 names←'anthony' 'barbara' 'bert' names[1 2] barbara bert names[⍸ ages>30] barbara bert However, we don't even need to use 'where' in this case; the reduction operator (/) has a second use which lets us say: 0 1 1 / names barbara bert (ages > 30) / names barbara bert Good luck with your database, and hope this helps!