CODEWITHSUNDEEP
pythonjuliaconvolution
kw3rti
kw3rti

Reputation: 224

Differences between Python and Julia convolutions?

I'm trying to convert a section of code from Python to Julia and I'm having a hard time understanding the output differences between the following convolutions:

Python:

conc = [10,100,1000,5,2000,200,20]
conc_filter = [1,2,3]
sp.ndimage.convolve1d(conc, conc_filter,axis=0,mode='constant')

Output:[ 120, 1230, 2305, 5010, 4215, 6420, 640]

Julia:

conc = [10,100,1000,5,2000,200,20]
conc_filter = [1,2,3]
conv(conc,conc_filter)

Output: [10, 120, 1230, 2305, 5010, 4215, 6420, 640, 60]

Can anyone explain why the output lengths are different? In an attempt to recreate the Python output, I've tried to come up with a logical formula that relates the input/filter sizes with the index range required to slice the conv output and obtain the same result. However, I haven't had much luck. Is there a Julia function that can produce the same output?

Upvotes: 1

Views: 786

Answers (3)

kw3rti
kw3rti

Reputation: 224

For those interested, I worked out the padding scheme used in the 'constant' mode, passed as a parameter to the Python convolution function. The difficulty in finding a relationship between the input/output sizes was due to the way left and right padding is done for symmetric vs unsymmetric convolution filters.

The Julia code below seems to match the Python equivalent for all tested input/outputs. 

conc = [10,100,1000,5,2000,200,20]
conc_filter = [1,2,3]

n=length(conc)
m=length(conc_filter)

padleft=ceil(Int32,m/2)-1
padright=floor(Int32,m/2)

conc =append!(zeros(padleft),conc)
conc = append!(conc,zeros(padright))

out = zeros(n)
for i in 1:n
    for j in 1:m
        out[i] += conc[i+j-1]*conc_filter[m-j+1]
    end
end
out

Result: [ 120, 1230, 2305, 5010, 4215, 6420, 640]

Upvotes: 1

Bill
Bill

Reputation: 6086

What Python conv routine are you using? Here is Python's numpy convolve:

>>> conc = [10,100,1000,5,2000,200,20]
>>> conc_filter = [1,2,3]

>>> numpy.convolve(conc, conc_filter,axis=0,mode='constant')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: convolve() got an unexpected keyword argument 'axis'

>>> numpy.convolve(conc, conc_filter)
array([  10,  120, 1230, 2305, 5010, 4215, 6420,  640,   60])

You might need to peek at the documentation to your Python package and see how to make fft() type routines do what you need.

Upvotes: 2

Julia Learner
Julia Learner

Reputation: 2902

Assuming I understand your question correctly:

According to this answer on stackexchange, the vector length of the Julia example is correct.

N + M - 1 gives an output vector length of 9, as shown by the Julia answer.

Are you sure the Python code and/or copied output is correct?

Upvotes: 0

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