[q-lang-users] lambda

[Eddie R. <ed...@bm...>]

Me again. When I run the following:

// assume X is sorted
medianS X = if (#X) mod 2 = 0 then
	      (X!(#X div 2 - 1) +  X!(#X div 2))/2
	    else
	      X!(#X div 2);

outliers X = filter (\X. (X < LF) or (X > RF)) X
	       where (LF, RF) = (LQ - IRQ, UQ + IRQ)
	       where IRQ = 1.5 * (UQ - LQ)
	       where UQ = medianS (drop (trunc (#X/2 + 0.5)) L)
	       where LQ = medianS (take (trunc (#X/2)) L)
	       where L = sort (<) X;

> outliers $ [1] ++ [20..30]
! Error in conditional

the debugger's last line before the "! Error in conditional"
  0>  stdlib.q, line 134: filter (\[...] . (...) or (...)) [1,20,21,...] 
==>  [1|filter (...) [...]] if (\[...] . (...) or (...)) 1


When I change the (\X. (X < LF) or (X > RF)) to (\Y. (Y < LF) or (Y > RF))
it works

> outliers $ [1] ++ [20..30]
=> [1]

Why should it matter that I'm using X inside of the lambda function? I
know that X is a parameter to the outliers equation, but shouldn't
lambda's X be private to (\X. (X < LF) or (X > RF))?

Eddie

P.S. I really don't mean to be such an aggrivation. I really am enjoying Q!