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
>>
>
