train a neural net with these data.
On Tue, Jul 5, 2016 at 12:43 PM, Andre Bieler
wrote:
> Can you tell me what you will be doing with this tree after everything is
> set up correctly?
Can you tell me what you will be doing with this tree after everything is set
up correctly?
the objective is to get 100 (leafs) vectors (which are not equal in length)
of data from child and root.
initial data is in the root
then the data is divided between childs (the size of each vector may varies
from 1 child to another)
each child share it's data with its own leaf (the size of each
Ahmed,
is it necessary that the data is stored in all 3 levels? (root has data,
child has data, leaf has data)
Is it not good enough if the data is stored in the leafs only?
Maybe you can explain what your "end goal" is?
But l have encountered a problem when distributing data from root to child
then to leafs. l'm looking for something like this:
*level 0root : *data= rand(1) # generate 1 random
values
*level 1 10 child * for each child attribute a certain number of
generated values ,
hello,
l fixed the problem by adding a function called data_node() as follows :
function data_node(node)
println("data : ", node.data)
println("index ", node.index)
end
for i in 1:nChildren
for k in 1:root.child[i].nchild
data_node(root.child[i].child[k])
end
end
On Monday, July
The idea that comes to my mind is the following :
add a new variable to MyNode type wich is index,
type MyNode{T}
data::Vector{T}
level::Int
child :: Vector{MyNode}
nchild :: Int
index :: Int
end
function node_info(node)
println("Level: ", node.level)
Well for 1) this is pretty much exactly what is being done in my example
code.
The difference is that it prints out the content of "data". (But I put the
index of
each child into that "data" varialbe)
My suggestion is that you insert a new field in the MyNode type, this
variable
can then hold
Hello Andre thantk you a lot , when l run the code l think it's gives what
l'm looking for. but l have trouble to understand the code in order to add
the following instruction :
1) l need to print a list of leaf of each child by retrieving the indexes
: for example
child 1 : [leaf 1, leaf 5 , ]
and maybe someday I will be smart enough for syntax highlighting... maybe..
Hi Ahmed, sorry for the late answer, but
it is a busy weekend... I am not sure I understand
your question, but the following for loop connects the leafs
to the children.
for i in 1:nChildren
root.child[i].child = all_leafs[i]
end
Here a full example code where some information is printed in the
l don't understand your code!! how do you connect each child to its leafs ?
On Fri, Jul 1, 2016 at 4:42 PM, Ahmed Mazari
wrote:
> ha you should just reply from the google group rather than from your
> inbox mail. then click on { } to highlight the syntax
>
> On Fri,
ha you should just reply from the google group rather than from your inbox
mail. then click on { } to highlight the syntax
On Fri, Jul 1, 2016 at 4:38 PM, Andre Bieler
wrote:
> ah for &^%$@@ sake how do you enable syntax highlighting?? I thought
> ```julia would do
1) yes they are of the same type as root and child . the leaf is the last
level of the tree so it doesn't have any child
2) yes no children with zero leaf at least one leaf
On Fri, Jul 1, 2016 at 4:37 PM, Andre Bieler
wrote:
> I think I start seeing what you want. A
ah for &^%$@@ sake how do you enable syntax highlighting?? I thought
```julia would do the trick??
If you tell me the secret I can re-post my previous message more nicely :)
I think I start seeing what you want. A few questions:
1) Can the leafs be of the same type as the root and child?
2) Does every child need to have one child at least? (= no child with zero
leafs)
you can do the following:
```julia
# this generates a list of empty vectors that can hold MyNode
Let's define the function that allows to associate randomly a certain
number of leafs for each child under a constraint : the total number of
leafs is 100 and the function that allows to add_link between each child
with its leafs. Please don't hesitate to modify the code, there is some
Well the child can also have children, just like in your graph you attached.
This is then one level down in your graph.
Note that the root is of the same type as the children. (MyNode can be parent,
child, grandchild etc.) The children are just put inside the root. Then you can
continue and
hi,
l don't understand what you're doing here. add_children!() need to have as
a parameter a root (node parent) not child, isn't it ?
*for child in root.child*
* add_children!(child, 10)*
*end*
On Wednesday, June 29, 2016 at 4:53:55 PM UTC+2, Ahmed Mazari wrote:
>
> Hello ,
>
>
> I need to
On Thursday, June 30, 2016 at 2:12:30 PM UTC+2, Ahmed Mazari wrote:
>
> Hi Andre.
> You will fnd attached my files and a picture that illustrates the tree
> structure l'm looking for.
> file 1 : somme comments on your codes
> file 2 : my code to be improved.
>
> For the part of your code. We
Hi Andre.
You will fnd attached my files and a picture that illustrates the tree
structure l'm looking for.
file 1 : somme comments on your codes
file 2 : my code to be improved.
For the part of your code. We should add child, number of child and data.
the structure seems to be appropriate from
Not very elegant, but maybe something like this?
type MyNode
level::Int
nLeafs::Int
leafs::Vector{MyNode}
end
MyNode() = MyNode(0,0,MyNode[])
N = 100
root = MyNode(0,N, [MyNode() for i in 1:N])
then you can go on and define the 1st level with a for loop and so on.
Because it is
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