Got an exception of type ErrorException outside of a @test: type DataType has no field FactorMargin
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Kevin Liu
Nov 16, 2016, 6:50:28 PM11/16/16
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From this issue https://github.com/JuliaPy/PyPlot.jl/issues/157 I understand
`Factor[:FactorMargin](A, Remove_var, Remain_var, Remove_dims, Remain_dims)` (line 85 of FactorOperations.jl) should pass, as it does on Atom, but not on the REPL, which throws
Got an exception of type ErrorException outside of a @test
type DataType has no field FactorMargin
in FactorDropMargin(::BN.Factor, ::Array{String,1}) at /Users/Corvus/.julia/v0.5/BN/src/FactorOperations.jl:85
Yichao Yu
Nov 16, 2016, 7:02:48 PM11/16/16
to Julia Users
On Wed, Nov 16, 2016 at 6:50 PM, Kevin Liu <kvt...@gmail.com> wrote:
> From this issue https://github.com/JuliaPy/PyPlot.jl/issues/157 I understand
^^ This is irrelevant unless you are using PyCall
> From this issue https://github.com/JuliaPy/PyPlot.jl/issues/157 I understand
>
> `Factor[:FactorMargin](A, Remove_var, Remain_var, Remove_dims, Remain_dims)`
> (line 85 of FactorOperations.jl) should pass, as it does on Atom, but not on
> the REPL, which throws
>
> Got an exception of type ErrorException outside of a @test
>
> type DataType has no field FactorMargin
>
> in FactorDropMargin(::BN.Factor, ::Array{String,1}) at
> /Users/Corvus/.julia/v0.5/BN/src/FactorOperations.jl:85
>
>
> Help, please.
Kevin Liu
Nov 16, 2016, 7:24:34 PM11/16/16
to julia-users
Hi Yichao!
Here is the function from FactorOperations.jl
function FactorDropMargin(A::Factor, Remove_var::Vector{String})
Remove_dims = indexin(Remove_var, A.var)
if any(Remove_dims==0)
error("Wrong variable!")
end
Remain_var = symdiff(A.var, Remove_var)
Remain_dims = indexin(Remain_var, A.var)
Factor[:FactorMargin](A, Remove_var, Remain_var, Remove_dims, Remain_dims) # line 85
end
runtests.jl:
@testset "Multiply and marginalize factor" begin
A=Factor(["a", "b"],[3, 2],[0.5, 0.1, 0.3, 0.8, 0, 0.9])
B=Factor(["b", "c"],[2, 2],[0.5, 0.1, 0.7, 0.2])
C = FactorProduct(A, B)
FactorDropMargin(C, ["c"]) # line 19
FactorKeepMargin(C, ["b", "a"])
FactorPermute(ans, [2, 1])
FactorKeepMargin(C, ["a", "b"])
end
what I got on the REPL:
julia> Pkg.test("BN")
INFO: Testing BN
Test Summary: |
Define, permute factor, and call (var, card, val) | No tests
Multiply and marginalize factor: Error During Test
Got an exception of type ErrorException outside of a @test
type DataType has no field FactorMargin
in FactorDropMargin(::BN.Factor, ::Array{String,1}) at /Users/Corvus/.julia/v0.5/BN/src/FactorOperations.jl:85
in macro expansion; at /Users/Corvus/.julia/v0.5/BN/test/runtests.jl:19 [inlined]
in macro expansion; at ./test.jl:672 [inlined]
in anonymous at ./<missing>:?
in include_from_node1(::String) at ./loading.jl:488
in include_from_node1(::String) at /Applications/Julia-0.5.app/Contents/Resources/julia/lib/julia/sys.dylib:?
in process_options(::Base.JLOptions) at ./client.jl:262
in _start() at ./client.jl:318
in _start() at /Applications/Julia-0.5.app/Contents/Resources/julia/lib/julia/sys.dylib:?
Test Summary: | Error Total
Multiply and marginalize factor | 1 1
Yichao Yu
Nov 16, 2016, 8:25:24 PM11/16/16
to Julia Users
On Wed, Nov 16, 2016 at 7:24 PM, Kevin Liu <kvt...@gmail.com> wrote:
> Hi Yichao!
In general there's nothing from the code you posted that shows what
you want to do.
>
> Here is the function from FactorOperations.jl
>
> function FactorDropMargin(A::Factor, Remove_var::Vector{String})
> Remove_dims = indexin(Remove_var, A.var)
> if any(Remove_dims==0)
> error("Wrong variable!")
> end
>
> Remain_var = symdiff(A.var, Remove_var)
> Remain_dims = indexin(Remain_var, A.var)
>
> Factor[:FactorMargin](A, Remove_var, Remain_var, Remove_dims,
> Remain_dims) # line 85
Unless you overloaded getindex on this type (which you should include)
you are construction a Vector of `Factor` from a symbol and then
calling it. It's impossible to tell what you actually want to do.
And as I previously mentioned, unless you are using PyCall, the issue
you linked is totally unrelated to this.
> Hi Yichao!
In general there's nothing from the code you posted that shows what
you want to do.
>
> Here is the function from FactorOperations.jl
>
> function FactorDropMargin(A::Factor, Remove_var::Vector{String})
> Remove_dims = indexin(Remove_var, A.var)
> if any(Remove_dims==0)
> error("Wrong variable!")
> end
>
> Remain_var = symdiff(A.var, Remove_var)
> Remain_dims = indexin(Remain_var, A.var)
>
> Factor[:FactorMargin](A, Remove_var, Remain_var, Remove_dims,
> Remain_dims) # line 85
you are construction a Vector of `Factor` from a symbol and then
calling it. It's impossible to tell what you actually want to do.
And as I previously mentioned, unless you are using PyCall, the issue
you linked is totally unrelated to this.
Kevin Liu
Nov 17, 2016, 2:27:38 PM11/17/16
to julia-users
I replaced Factor[:FactorMargin]() with Factor.FactorMargin() back again.
Still, for FactorOperations.jl on Atom, I get {UndefVarError: Factor not defined} at the end of each block.
Factor is defined on Factor.jl, and that file evaluates fine.
The main file, BN.jl, includes Factor.jl and FactorOperations.jl and exports Factor, and also evaluates fine.
Kevin Liu
Nov 17, 2016, 2:31:12 PM11/17/16
to julia-users
{Factor.FactorMargin()} simply marginalizes non-stated variables in Factor.
Yichao Yu
Nov 17, 2016, 2:33:05 PM11/17/16
to Julia Users
On Thu, Nov 17, 2016 at 2:27 PM, Kevin Liu <kvt...@gmail.com> wrote:
> I replaced Factor[:FactorMargin]() with Factor.FactorMargin() back again.
>
> Still, for FactorOperations.jl on Atom, I get {UndefVarError: Factor not
> defined} at the end of each block.
>
> Factor is defined on Factor.jl, and that file evaluates fine.
AFAICT the `Factor`is a type so `Factor.FactorMargin` is definitely wrong.
> I replaced Factor[:FactorMargin]() with Factor.FactorMargin() back again.
>
> Still, for FactorOperations.jl on Atom, I get {UndefVarError: Factor not
> defined} at the end of each block.
>
> Factor is defined on Factor.jl, and that file evaluates fine.
Kevin Liu
Nov 17, 2016, 2:39:31 PM11/17/16
to julia-users
Right, I need the instance of Factor
David P. Sanders
Nov 17, 2016, 11:05:20 PM11/17/16
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If you can produce a minimal working example (the *shortest* piece of code that produces the problem and that people can run), you will be more likely to get help. This probably involves removing a lot of stuff from your code that I'd not relevant to this particular problem.
Yichao Yu
Nov 18, 2016, 8:48:58 AM11/18/16
to Julia Users
On Thu, Nov 17, 2016 at 2:39 PM, Kevin Liu <kvt...@gmail.com> wrote:
> Right, I need the instance of Factor
Then use the instance of Factor.
> Right, I need the instance of Factor
Kevin Liu
Nov 18, 2016, 1:07:44 PM11/18/16
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Have a look please https://github.com/hpoit/MLN.jl/tree/master/BN
Kevin Liu
Nov 22, 2016, 2:53:14 PM11/22/16
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A=Factor(["a", "b"],[3, 2],[0.5, 0.1, 0.3, 0.8, 0, 0.9])
B=Factor(["b", "c"],[2, 2],[0.5, 0.1, 0.7, 0.2])
C = FactorProduct(A, B)
FactorDropMargin(C, ["c"])
Kevin Liu
Nov 22, 2016, 3:18:29 PM11/22/16
to julia-users
julia> A=Factor(["a", "b"],[3, 2],[0.5, 0.1, 0.3, 0.8, 0, 0.9]);
julia> B=Factor(["b", "c"],[2, 2],[0.5, 0.1, 0.7, 0.2]);
julia> C = FactorProduct(A, B)
Factor(["a", "b", "c"],[3, 2, 2],[0.25, 0.05, 0.15, 0.08, 0.0, 0.09, 0.35, 0.07, 0.21, 0.16, 0.0, 0.18])
julia> FactorDropMargin(C, ["c"])
Factor(["a", "b"],[3, 2],[0.6, 0.12, 0.36, 0.24, 0.0, 0.27])
julia> FactorKeepMargin(C, ["b", "a"])
Factor(["b", "a"],[2, 3],[0.6, 0.24, 0.12, 0.0, 0.36, 0.27])
julia> FactorPermute(ans, [2, 1])
Factor(["a", "b"],[3, 2],[0.6, 0.12, 0.36, 0.24, 0.0, 0.27])
julia> FactorKeepMargin(C, ["a", "b"])
Factor(["a", "b"],[3, 2],[0.6, 0.12, 0.36, 0.24, 0.0, 0.27])
Yichao Yu
Nov 22, 2016, 3:27:40 PM11/22/16
to Julia Users
> Yichao and DPSanders, I have already used instances of Factor on runtests.jl, instances A, B, and C
AFAICT you are still accessing a non existing field of a type[1] and
it's unclear what you actually want to do.
[1] https://github.com/hpoit/MLN.jl/blob/1c13725666f34587e57c4a1757e6222cacaeab73/BN/src/FactorOperations.jl#L87
Kevin Liu
Nov 22, 2016, 3:36:17 PM11/22/16
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julia> A=Factor(["a", "b"],[3, 2],[0.5, 0.1, 0.3, 0.8, 0, 0.9]);
julia> B=Factor(["b", "c"],[2, 2],[0.5, 0.1, 0.7, 0.2]);
julia> B=Factor(["b", "c"],[2, 2],[0.5, 0.1, 0.7, 0.2]);
julia> C = FactorProduct(A, B)
Factor(["a", "b", "c"],[3, 2, 2],[0.25, 0.05, 0.15, 0.08, 0.0, 0.09, 0.35, 0.07, 0.21, 0.16, 0.0, 0.18])
julia> FactorDropMargin(C, ["c"]) # Yichao, this is what I want to do
Factor(["a", "b"],[3, 2],[0.6, 0.12, 0.36, 0.24, 0.0, 0.27]) # Yichao, this is what I want to do
julia> FactorKeepMargin(C, ["b", "a"])
Factor(["a", "b"],[3, 2],[0.6, 0.12, 0.36, 0.24, 0.0, 0.27]) # Yichao, this is what I want to do
julia> FactorKeepMargin(C, ["b", "a"])
Factor(["b", "a"],[2, 3],[0.6, 0.24, 0.12, 0.0, 0.36, 0.27])
julia> FactorPermute(ans, [2, 1])
Factor(["a", "b"],[3, 2],[0.6, 0.12, 0.36, 0.24, 0.0, 0.27])
julia> FactorKeepMargin(C, ["a", "b"])
Factor(["a", "b"],[3, 2],[0.6, 0.12, 0.36, 0.24, 0.0, 0.27])
Kevin Liu
Nov 22, 2016, 3:45:42 PM11/22/16
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Yichao, I used a hashtag in the last message to show you what I want to do. Is it clear?
Yichao Yu
Nov 22, 2016, 3:54:34 PM11/22/16
to Julia Users
On Tue, Nov 22, 2016 at 3:45 PM, Kevin Liu <kvt...@gmail.com> wrote:
> Yichao, I used a hashtag in the last message to show you what I want to do. Is it clear?
No
> Yichao, I used a hashtag in the last message to show you what I want to do. Is it clear?
I'm just talking about the `Factor.` in the line I linked. I don't
know what you want to access. Do you just want `FactorMargin`? What's
the extra `Factor.` for?
Kevin Liu
Nov 22, 2016, 4:23:59 PM11/22/16
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I would like to remove variable "c" from factor C. I tried removing `Factor.` but it didn't work.
Yichao Yu
Nov 22, 2016, 4:39:46 PM11/22/16
to Julia Users
On Tue, Nov 22, 2016 at 4:23 PM, Kevin Liu <kvt...@gmail.com> wrote:
> I would like to remove variable "c" from factor C. I tried removing
> `Factor.` but it didn't work.
There might be (almost certainly) multiple mistakes in the code so
> I would like to remove variable "c" from factor C. I tried removing
> `Factor.` but it didn't work.
fixing one won't fix all of them.
Kevin Liu
Nov 22, 2016, 6:34:11 PM11/22/16
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Thanks. Is there anything that sticks out to you?
Yichao Yu
Nov 23, 2016, 6:44:34 PM11/23/16
to Kevin Liu, Julia Users
On Wed, Nov 23, 2016 at 4:02 PM, Kevin Liu <kvt...@gmail.com> wrote:
> Yichao, would you give me some direction? I am a bit lost.
Post and/or identify the error after you've fixed the `Factor.` problem
>
> On Tue, Nov 22, 2016 at 7:58 PM, Kevin Liu <kvt...@gmail.com> wrote:
>>
>> Do you want a cut in the profits for helping me get it to work? It's a
>> marathon. I still have Markov Random Field and Markov Logic Network in
>> line... and two of the largest private Brazilian banks on standby.
> Yichao, would you give me some direction? I am a bit lost.
Post and/or identify the error after you've fixed the `Factor.` problem
>
> On Tue, Nov 22, 2016 at 7:58 PM, Kevin Liu <kvt...@gmail.com> wrote:
>>
>> Do you want a cut in the profits for helping me get it to work? It's a
>> marathon. I still have Markov Random Field and Markov Logic Network in
>> line... and two of the largest private Brazilian banks on standby.
Yichao Yu
Nov 23, 2016, 6:55:00 PM11/23/16
to Julia Users
On Wed, Nov 23, 2016 at 6:50 PM, Kevin Liu <kvt...@gmail.com> wrote:
> Attached!
```
help?> permutedims
search: permutedims permutedims! ipermutedims
permutedims(A, perm)
Permute the dimensions of array A. perm is a vector specifying a permutation
of length ndims(A). This is a generalization of transpose for
multi-dimensional arrays. Transpose is equivalent to permutedims(A, [2,1]).
julia> A = reshape(collect(1:8), (2,2,2))
2×2×2 Array{Int64,3}:
[:, :, 1] =
1 3
2 4
<BLANKLINE>
[:, :, 2] =
5 7
6 8
julia> permutedims(A, [3, 2, 1])
2×2×2 Array{Int64,3}:
[:, :, 1] =
1 3
5 7
<BLANKLINE>
[:, :, 2] =
2 4
6 8
```
You are not giving `permutedims` the correct second parameters
(https://github.com/hpoit/MLN.jl/blob/1c13725666f34587e57c4a1757e6222cacaeab73/BN/src/FactorOperations.jl#L66).
> Attached!
```
help?> permutedims
search: permutedims permutedims! ipermutedims
permutedims(A, perm)
Permute the dimensions of array A. perm is a vector specifying a permutation
of length ndims(A). This is a generalization of transpose for
multi-dimensional arrays. Transpose is equivalent to permutedims(A, [2,1]).
julia> A = reshape(collect(1:8), (2,2,2))
2×2×2 Array{Int64,3}:
[:, :, 1] =
1 3
2 4
<BLANKLINE>
[:, :, 2] =
5 7
6 8
julia> permutedims(A, [3, 2, 1])
2×2×2 Array{Int64,3}:
[:, :, 1] =
1 3
5 7
<BLANKLINE>
[:, :, 2] =
2 4
6 8
```
You are not giving `permutedims` the correct second parameters
(https://github.com/hpoit/MLN.jl/blob/1c13725666f34587e57c4a1757e6222cacaeab73/BN/src/FactorOperations.jl#L66).
Kevin Liu
Nov 23, 2016, 7:31:25 PM11/23/16
to Julia Users
In `permute_dims = [Remain_dims,Remove_dims]`, both Remain and Remove_dims are vectors. Even if I `permute_dims = [Remain_dims]`, I still get the same error.
Yichao Yu
Nov 23, 2016, 7:59:30 PM11/23/16
to Julia Users
On Wed, Nov 23, 2016 at 7:31 PM, Kevin Liu <kvt...@gmail.com> wrote:
> In `permute_dims = [Remain_dims,Remove_dims]`, both Remain and Remove_dims
Which is why it is wrong.
> In `permute_dims = [Remain_dims,Remove_dims]`, both Remain and Remove_dims
> are vectors. Even if I `permute_dims = [Remain_dims]`, I still get the same
> error.
1-element Array{Array{Int64,1},1}:
[1]
julia> [[1];]
1-element Array{Int64,1}:
1
Kevin Liu
Nov 23, 2016, 8:22:21 PM11/23/16
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Hey Yichao, it's late here, I will continue tomorrow. Thanks for your help!
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