[julia-users] Re: Indexing problem

[Patrik Waldmann]

OK, that works fine. Thanks. I think it would be a good idea to drop the 
matlab-ism in future versions.

Patrik

On Thursday, November 17, 2016 at 11:51:09 AM UTC+1, Simon Byrne wrote:

> 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))]
>
> 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))] = 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))] = 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))] = 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))] = 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))] = 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

>>>

[julia-users] Re: Indexing problem

[Simon Byrne]

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))]

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))] = 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))] = 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))] = 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))] = 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))] = 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
>>>
>>

[julia-users] Re: Indexing problem

[Patrik Waldmann]

I guess I should have explained my problem clearer. If I run the code 
without w[1,(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))] = 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))] = 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))] = 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))] = 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
>>
>

[julia-users] Re: Indexing problem

[Jeffrey Sarnoff]

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))] = 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))] = 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))] = 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
>