Where is my mistake in converting this python code into Julia

New to Julia
#1

I’m trying to convert the python code(left) into Julia (right), python code is working fine, but Julia code gave an error at the highlighted line l1_delta = ...:

DimensionMismatch("A has dimensions (4,1) but B has dimensions (4,1)")

10%20PM

Julia full code is here:

# sigmoid function
sigmoid(x::Real, derive::Bool=false) = 
(derive==true) ? x*(one(x)-x) : one(x)/(one(x) + exp(-x))

x = [0 0 1; 0 1 1; 1 0 1; 1 1 1]

y = [0 0 1 1]'

syn0 = 2 * rand(3,1)

for iter in (:)(1,10000)
    l0 = x
    l1 = sigmoid.(l0 * syn0)
    l1_error = y - l1
    l1_delta = l1_error * sigmoid.(l1,true) # here I get the error
    syn0 += l0' * l1_delta
end

println("Output After Training:")
l1
#2

syn0 = 2 * rand(3,1) - 1?

#3

Thanks for your feedback, the error is in

l1_delta = l1_error * sigmoid.(l1,true)

Regarding your comment, see the below

29%20PM
Screen Shot 2019-10-07 at 10.17.29 PM.png821×366 34.4 KB

#4

You are missing a point in the subtraction:

syn0 = 2 * rand(3,1) .- 1

#5

Thanks, but I still getting error at:

l1_delta = l1_error * sigmoid.(l1,true)
#6

Now that I noticed the comment of the error.
The same broadcast (dot) solution:
l1_delta = l1_error .* sigmoid.(l1,true) # here I get the error

1 Like
#7

The problem, in my opinion, is that you are doing a matrix multiplication, and in that case, both matrixes cannot have the same dimension.

Or you use the transpose of one of them or you use

l1_delta = l1_error .* sigmoid.(l1,true) # Multiply element by element
1 Like
#8

Thanks,
So the python code:

l1_delta = l1_error * sigmoid(l1,True)

Is not a matrix multiplication, and different from the python code using np which is generating the same error got by Julia:

l1_delta = np.dot(l1_error,sigmoid(l1,True))

Now the code is correct, thanks

#9

Transpose can not work here as it will generate array in different dimensions, but the .* worked fine, thanks