Difference between revisions of "Short Notes on Wavelets"

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== Integer Haar Wavelets, Python implementation ==
 
== Integer Haar Wavelets, Python implementation ==
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The code provide is in no way optimized for speed - it's creating too many temporaries and duplicates.
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This code is for illustration only, and optimization for speed is left to the reader as an exercise.
  
 
=== 1D Case ===
 
=== 1D Case ===
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==== numpy Version ====
 
==== numpy Version ====
  
TBP
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This is highly '''un'''-optimized - it creates copy for each step of transformation, and both directions of transformation use ''np.apply_along_axis'', which I suspect builds the new array row by row.
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Both versions could be parallelised and made into in-place transforms. Enjoy!
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<pre>def haar_int_fwd_2d_np(d):
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    tmp = np.apply_along_axis(haar_int_fwd_1d_np, 0, d)
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    return np.apply_along_axis(haar_int_fwd_1d_np, 1, tmp)
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def haar_int_inv_2d_np(d):
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    tmp = np.apply_along_axis(haar_int_inv_1d_np, 1, d)
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    return np.apply_along_axis(haar_int_inv_1d_np, 0, tmp)</pre>

Latest revision as of 12:11, 27 February 2017

Integer Haar Wavelets, Python implementation

The code provide is in no way optimized for speed - it's creating too many temporaries and duplicates.

This code is for illustration only, and optimization for speed is left to the reader as an exercise.

1D Case

This is a trivial implementation of Haar integer-to-integer wavelets.

The d array (list or numpy.array) has to be power of 2.

Note that resulting values typically use 1 more bit than original ones - if source values are in [0..N) interval, then resulting values are in (-N, N) interval.

Using Lists

def haar_int_fwd_1d(d):
    if len(d) == 1:
        return d
    even = d[::2]
    odd = d[1::2]
    hp = [j - i for i, j in zip(even, odd)]
    lp = [i + (w >> 1) + (w % 2) for i, w in zip(even, hp)]
    return haar_int_fwd_1d(lp) + hp
    
def haar_int_inv_1d(d):
    if len(d) == 1:
        return d
    even = haar_int_inv_1d(d[:len(d) >> 1])
    odd = d[len(d) >> 1:]
    lp = [i - (j >> 1) - (j % 2) for i, j in zip(even, odd)]
    hp = [i + j for i, j in zip(lp, odd)]
    return [x for t in zip(lp, hp) for x in t]

Using numpy Array's

import numpy as np

def haar_int_fwd_1d_np(d):
    if len(d) == 1:
        return d
    hp = d[1::2] - d[::2]
    lp = d[::2] + (hp >> 1) + (hp % 2)
    return np.concatenate((haar_int_fwd_1d_np(lp), hp))

def haar_int_inv_1d_np(d):
    if len(d) == 1:
        return d
    lp = haar_int_inv_1d_np(d[:len(d) >> 1]) 
    hp = d[len(d) >> 1:] 
    even = lp - (hp >> 1) - (hp % 2)
    return np.ravel(np.column_stack((even, even + hp)))

2D Extension

The list-based version is not provided.

numpy Version

This is highly un-optimized - it creates copy for each step of transformation, and both directions of transformation use np.apply_along_axis, which I suspect builds the new array row by row.

Both versions could be parallelised and made into in-place transforms. Enjoy!

def haar_int_fwd_2d_np(d):
    tmp = np.apply_along_axis(haar_int_fwd_1d_np, 0, d)
    return np.apply_along_axis(haar_int_fwd_1d_np, 1, tmp)

def haar_int_inv_2d_np(d):
    tmp = np.apply_along_axis(haar_int_inv_1d_np, 1, d)
    return np.apply_along_axis(haar_int_inv_1d_np, 0, tmp)