It makes sense now. The ternary operator only evaluates one branch, while a function would force both to be evaluated. So the answer to the original question should be
inside_disc(x,y,radius) = ifelse((sqrt(x.^2 .+ y.^2) .< radius), 1 , 0) This should be faster than the map version. Amazingly (at least to me), the ifelse version is only 10% faster than the map version when processing million item vectors of floats, and generates just a slight amount more garbage. On Saturday, June 7, 2014 1:05:20 PM UTC-4, Stefan Karpinski wrote: > > That's because ifelse is just a function, not an operator with special > syntax. The ternary operator is just a one-line form of the if-else and > requires the condition to be a single boolean value. It is a genuine > control flow construct, not a function. > > > On Fri, Jun 6, 2014 at 11:54 PM, David Einstein <[email protected] > <javascript:>> wrote: > >> It looks like you have vectorized ifelse, the operator version just >> doesn't work. This code is on line 379 of operators.jl >> >> function ifelse(c::AbstractArray{Bool}, x, y) >> reshape([ifelse(ci, x, y) for ci in c], size(c)) >> end >> >> If I do >> >> ifelse((1:10 .< 5) , 1 , 0) >> >> then everything works. >> >> however >> >> (1:10 .< 5) ? 1 : 0 >> >> does not. I'm not sure why. >> >> Also it appears that ifelse is not automatically imported like the rest >> of the operators from base are. >> >> >> On Friday, June 6, 2014 10:58:10 PM UTC-4, Stefan Karpinski wrote: >> >>> We should probably vectorize the ifelse function. >>> >>> On Jun 6, 2014, at 10:51 PM, David Einstein <[email protected]> wrote: >>> >>> sqrt(x.^2 .+ y.^2) .< radius >>> is a vector of booleans >>> >>> ?: >>> expects a boolean >>> >>> you probably want something like: >>> >>> inside_disc(x,y,radius) = map(good -> good ? 1 : 0, >>> sqrt(x.^2+y.^2).<radius) >>> >>> >>> >>> On Friday, June 6, 2014 10:18:36 PM UTC-4, Zahirul ALAM wrote: >>>> >>>> When I pass two arrays the function returns: >>>> >>>> type: non-boolean (BitArray{1}) used in boolean context while loading >>>> In[10], in expression starting on line 1 >>>> >>>> >>>> I have modified the function to address the element wise operation as >>>> follows: >>>> >>>> >>>> inside_disc(x,y,radius) = sqrt(x.^2 .+ y.^2) .< radius ? 1 : 0 >>>> >>>> seems to me that the error is being thrown by .< operation. What am I >>>> missing? >>>> >>>> >>>> On Friday, 6 June 2014 20:34:22 UTC-4, Miguel Bazdresch wrote: >>>>> >>>>> I'm not sure I understand the question. Do you mean something like >>>>> this? >>>>> >>>>> inside_disc(x,y,radius) = sqrt(x^2+y^2)<radius ? 1 : 0 >>>>> >>>>> -- mb >>>>> >>>>> >>>>> On Fri, Jun 6, 2014 at 8:28 PM, Zahirul ALAM <[email protected]> >>>>> wrote: >>>>> >>>>>> I guess one can do a for loop. But how do I vectorize the code? >>>>>> >>>>>> >>>>>> On Friday, 6 June 2014 20:27:46 UTC-4, Zahirul ALAM wrote: >>>>>>> >>>>>>> How would one implement a step function like behaviour in julia? In >>>>>>> mathematica one can write the following to create a circle with value >>>>>>> of 1 >>>>>>> within the radius and 0 outside >>>>>>> >>>>>>> UnitBox[Sqrt[X^2 + Y^2]*0.5/radius]; >>>>>>> >>>>>>> X and Y are the coordinates. >>>>>>> >>>>>> >>>>> >
