Probably is a good idea. On this topic I have a question: is there a rule of thumb where certain operations or built in functions are vectorized and the others are not ?
Will it make any sense for the developers to make all functions and operations vectorized by default? Or is this too difficult at this point to do? On Friday, 6 June 2014 22:58:10 UTC-4, Stefan Karpinski wrote: > > We should probably vectorize the ifelse function. > > On Jun 6, 2014, at 10:51 PM, David Einstein <[email protected] > <javascript:>> 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. >>>>> >>>> >>>
