I'm not familiar with the package in question, but this line:
w = Any[ 0.1*randn(1,13), 0 ]
may be what is causing the problem. It is creating a 2-element Vector, the
first element of which is a 1x13 Matrix, and the second element is a scalar
0. The analogous object in R would be:
W = list(matrix(0.1*rnorm(13),nrow=1), 0)
In Julia, extraneous dimensions have an implicit index of 1 (this is a
matlab-ism, and may disappear in future), so w[1], w[1,1], w[1,1,1] are all
identical (and equivalent to W[[1]] in R). w[1,:] is a bit of an odd case
in that it returns a 1-element Vector containing a Matrix, but would be
equivalent to W[1] in R.
I think what you may want is actually
w[1][(w[1].<z) & (w[1].>-(z))]
which can be written more clearly as
w[1][-z .< w[1] .< z]
-Simon
On Thursday, 17 November 2016 09:39:14 UTC, Patrik Waldmann wrote:
>
> I guess I should have explained my problem clearer. If I run the code
> without w[1,(w[1].<z)&(w[1].>-(z))] = 0, and do:
> dump(w)
> Array{Any}((2,))
> 1:
> Array{Float64}((1,13))
> [-0.681392 0.595298 … 0.893845 -3.5044]
> 2: Float64 22.447679788630705
>
> and
> println(w[1])
> [-0.681392 0.595298 -0.394906 0.776983 -1.11178 3.11679 -0.0984956
> -2.18501 0.928204 -0.484802 -1.86844 0.893845 -3.5044]
>
> println(w[1,1])
> [-0.681392 0.595298 -0.394906 0.776983 -1.11178 3.11679 -0.0984956
> -2.18501 0.928204 -0.484802 -1.86844 0.893845 -3.5044]
>
> println(w[1,:])
> Any[
> [-0.681392 0.595298 -0.394906 0.776983 -1.11178 3.11679 -0.0984956
> -2.18501 0.928204 -0.484802 -1.86844 0.893845 -3.5044]]
>
>
> This is very confusing for an R user like me. How do I access the column
> indexes of w[1] and apply the logical expression
> w[1,(w[1].<z)&(w[1].>-(z))] = 0 ?
>
> Patrik
>
>
>
>
> On Thursday, November 17, 2016 at 6:58:14 AM UTC+1, Jeffrey Sarnoff wrote:
>
>> good things to know about how indexing works
>>
>>
>> The indices for a Vector, or a column or row of a Matrix start at *1*
>>
>> ```
>> length(avector) # gets the number of elements in avector
>>
>> avector[1] # gets the first item in avector
>> avector[end] # gets the final item in avector
>> avector[1:end] # gets all elements of avector
>>
>> int_column_vector = [10, 20, 30]
>> 10
>> 20
>> 30
>>
>> int_column_vector[1]
>> 10
>> # do not use zero as an index
>> int_column_vector[ 0 ]
>> ERROR: BoundsError:
>> # do not use false, true as indices because avec[ false ] means avec[ 0 ]
>>
>> ```
>>
>> in ` w[1,(w[1].<z)&(w[1].>-(z))] = 0 `, the second index can simplify to
>> `false` (consider this)
>> ```
>> avec = [ 10, 20, 30 ]
>> avec1 = avec[ 1 ]
>> avec1 == avec[ 1 + false ]
>> avec2 = avec[ 2 ]
>> avec2 == avec[ 1 + true ]
>> ```
>>
>> As a start, recheck indexing expressions, be more sure they do what you
>> want them to do.
>>
>>
>> On Wednesday, November 16, 2016 at 1:36:57 PM UTC-5, Patrik Waldmann
>> wrote:
>>>
>>> Hi,
>>>
>>> I'm an R user trying to learn Julia. I got hold of some code from the
>>> Knet package that I was playing around with. My goal is to set values to
>>> zero in a loop based on a logical expression, but I cannot figure out how
>>> the indexing works. Any help would be appreciated (the problem lies in
>>> w[1,(w[1].<z)&(w[1].>-(z))] = 0):
>>>
>>> using Knet
>>> predict(w,x) = w[1]*x .+ w[2]
>>> lambda = 2
>>> z = Array{Float64}(1,13)
>>> loss(w,x,y) = sumabs2(y - predict(w,x)) / size(y,2)
>>> lossgradient = grad(loss)
>>> function train(w, data; lr=.1)
>>> for (x,y) in data
>>> dw = lossgradient(w, x, y)
>>> z[:] = lr * lambda
>>> w[1] -= lr * dw[1]
>>> w[2] -= lr * dw[2]
>>> w[1,(w[1].<z)&(w[1].>-(z))] = 0
>>> end
>>> return w
>>> end
>>> url = "
>>> https://archive.ics.uci.edu/ml/machine-learning-databases/housing/housing.data
>>> "
>>> rawdata = readdlm(download(url))
>>> x = rawdata[:,1:13]'
>>> x = (x .- mean(x,2)) ./ std(x,2)
>>> y = rawdata[:,14:14]'
>>> w = Any[ 0.1*randn(1,13), 0 ]
>>> niter = 25
>>> lossest = zeros(niter)
>>> for i=1:niter; train(w, [(x,y)]); lossest[i]=loss(w,x,y); end
>>>
>>>
>>> Best regards,
>>>
>>> Patrik
>>>
>>
