Worksheet C

variationalform https://variationalform.github.io/

Just Enough: progress at pace

https://variationalform.github.io/

https://github.com/variationalform

Simon Shaw https://www.brunel.ac.uk/people/simon-shaw.

This work is licensed under CC BY-SA 4.0 (Attribution-ShareAlike 4.0 International) Visit http://creativecommons.org/licenses/by-sa/4.0/ to see the terms.

This document uses python

and also makes use of LaTeX

in Markdown

What this is about:

This worksheet is based on the material in the notebooks

• matrices: matrix concepts and algebra
• systems: systems of linear equations, under- and over-determined cases
• decomp: eigensystem decomposition and SVD.

We look at a specific example of using SVD to compress a photograph.

Note that while the 'lecture' notebooks are prefixed with 1_, 2_ and so on, to indicate the order in which they should be studied, the worksheets are prefixed with A_, B_, ...

# Two new imports here ... PIL and IPython 
import matplotlib.pyplot as plt
import numpy as np
from PIL import Image
import IPython.display

# Use a jpeg photo - ffc.jpg is about 6.2MB (use your own path/filename here)
IPython.display.Image(filename='./gfx/ffc.jpg', width = 150) 

# that is just a display, so ...
# load in the FFC bear - Roy - and visually check him.
img = Image.open('./gfx/ffc.jpg')

# convert him to a numpy array for processing as a matrix
a = np.asarray(img)
im_orig = Image.fromarray(a)
plt.imshow(im_orig)

<matplotlib.image.AxesImage at 0x7fb210317518>

This image is made up of pixels where each pixel has a value for RED, GREEN and BLUE. We can get these colour bands and show them as follows...

# convert band 'bnd' to a numpy array and show them...
for bnd in range(3):
  plt.subplot(1, 3, 1+bnd)
  img_mat = np.array(list(img.getdata(bnd)), float) 
  img_mat = np.matrix(img_mat)
  img_mat.shape = (img.size[1], img.size[0])
  plt.imshow(img_mat)

# get the red, green and blue bands as separate objects...
rband =img.getdata(band=0)
gband =img.getdata(band=1)
bband =img.getdata(band=2)

# and convert each to a numpy arrays for maths processing 
imgr_mat = np.array(list(rband), float) 
imgg_mat = np.array(list(gband), float) 
imgb_mat = np.array(list(bband), float)

# each of these is about 64k elements
print('sizes = ', imgr_mat.size, imgg_mat.size, imgb_mat.size)
print('shapes = ', imgr_mat.shape, imgg_mat.shape, imgb_mat.shape)

sizes =  64144128 64144128 64144128
shapes =  (64144128,) (64144128,) (64144128,)

# get image shape - we can assume they are all the same
imgr_mat.shape = imgg_mat.shape = imgb_mat.shape = (img.size[1], img.size[0])
print('imgr_mat.shape = ', imgr_mat.shape)
print('imgg_mat.shape = ', imgg_mat.shape)
print('imgb_mat.shape = ', imgb_mat.shape)

# convert these 1D-arrays to matrices
imgr_mat1D = np.matrix(imgr_mat)
imgg_mat1D = np.matrix(imgg_mat)
imgb_mat1D = np.matrix(imgb_mat)
print(type(imgb_mat))

imgr_mat.shape =  (9248, 6936)
imgg_mat.shape =  (9248, 6936)
imgb_mat.shape =  (9248, 6936)
<class 'numpy.ndarray'>

• The chained assignment above is OK but you should be aware that there are pitfalls... Take a look here for example.

• https://stackoverflow.com/questions/7601823/how-do-chained-assignments-work

• It depends on what is being chained... Here they are just values and not objects...

print(imgb_mat.shape is imgg_mat.shape)

False

# each of these is about 64 million elements: 9248 by 6936
print('sizes = ', imgr_mat1D.size, imgg_mat1D.size, imgb_mat1D.size)
print('shapes = ', imgr_mat1D.shape, imgg_mat1D.shape, imgb_mat1D.shape)
print('check: 9248 x 6936 = ', 9248 * 6936)

sizes =  64144128 64144128 64144128
shapes =  (9248, 6936) (9248, 6936) (9248, 6936)
check: 9248 x 6936 =  64144128

Let's look at these matrices - the following code introduces subplot

fontsize=20
fig=plt.figure(figsize=(8, 3))
fig.suptitle('Horizontally stacked subplots', fontsize=fontsize)

axs=fig.add_subplot(1,3,1); axs.imshow(imgr_mat)
axs.set_xlabel('x', fontsize=fontsize); axs.set_ylabel('y', fontsize=fontsize)
axs.set_title('RED', fontsize=fontsize)

axs=fig.add_subplot(1,3,2); axs.imshow(imgg_mat)
axs.set_xlabel('x', fontsize=fontsize); axs.set_ylabel('y', fontsize=fontsize)
axs.set_title('GREEN', fontsize=fontsize)

axs=fig.add_subplot(1,3,3); axs.imshow(imgb_mat)
axs.set_xlabel('x', fontsize=fontsize); axs.set_ylabel('y', fontsize=fontsize)
axs.set_title('BLUE', fontsize=fontsize)
# use fractions of fontsize
plt.tight_layout(pad=0.3, w_pad=2.5, h_pad=0.3); plt.show()

Let's look at how much memory these photo layers occupy...

print('Each matrix contains ', imgr_mat.size, ' elements')
print('Each element occupies ', imgr_mat.itemsize, ' bytes')
print('So each matrix occupies ', imgr_mat.size * imgg_mat.itemsize, ' bytes in memory')
print('This is ', 3*imgr_mat.size * imgg_mat.itemsize, ' bytes for all three')

print('This is {:e} bytes for all three'.format(3*imgr_mat.size * imgg_mat.itemsize))

Each matrix contains  64144128  elements
Each element occupies  8  bytes
So each matrix occupies  513153024  bytes in memory
This is  1539459072  bytes for all three
This is 1.539459e+09 bytes for all three

Compression...

Now that we know about the Singular Value Decomposition, we can hope to compress these objects using mathematics.

First get the SVD's of the R, G and B layers... (takes a while)

Ur, Sr, VTr = np.linalg.svd(imgr_mat) 
Ug, Sg, VTg = np.linalg.svd(imgg_mat) 
Ub, Sb, VTb = np.linalg.svd(imgb_mat) 

print(f'RED:   shapes of Ur, Sr, VTr = {Ur.shape}, {Sr.shape}, {VTr.shape}')
print(f'GREEN: shapes of Ug, Sg, VTg = {Ug.shape}, {Sg.shape}, {VTg.shape}')
print(f'BLUE:  shapes of Ub, Sb, VTb = {Ub.shape}, {Sb.shape}, {VTb.shape}')

RED:   shapes of Ur, Sr, VTr = (9248, 9248), (6936,), (6936, 6936)
GREEN: shapes of Ug, Sg, VTg = (9248, 9248), (6936,), (6936, 6936)
BLUE:  shapes of Ub, Sb, VTb = (9248, 9248), (6936,), (6936, 6936)

# choose the number of components to use in the reconstruction
nc = 5 # 1387    
rec_imgr = np.matrix(Ur[:, :nc]) * np.diag(Sr[:nc]) * np.matrix(VTr[:nc, :])
rec_imgg = np.matrix(Ug[:, :nc]) * np.diag(Sg[:nc]) * np.matrix(VTg[:nc, :])
rec_imgb = np.matrix(Ub[:, :nc]) * np.diag(Sb[:nc]) * np.matrix(VTb[:nc, :])

img_all = np.array([rec_imgr, rec_imgg, rec_imgb]).T
img_all = np.swapaxes(img_all,0,1)

PIL_image = Image.fromarray(np.uint8(img_all)).convert('RGB')
# uncomment this to spawn an external viewer
#PIL_image.show()
# save the reconstruction
PIL_image.save("ffc_recon.jpg")

fig=plt.figure(figsize=(7, 5))
fig.suptitle('Comparison', fontsize=30)

plt.subplot(1,2,1)
ax = plt.gca()
im = Image.fromarray(np.uint8(img_all)).convert('RGB')
ax.imshow(im)
ax.set_title(f'Recon, nc = {nc}', fontsize=20)

plt.subplot(1,2,2)
ax = plt.gca()
ax.imshow(im_orig)
ax.set_title('original', fontsize=20)

Text(0.5, 1.0, 'original')

Review

We have seen a few examples now of how the SVD can play a very important role in Data Science. It is able to take data in matrix-form and distill it to its essence.

Exercises

1. Create logarithmic scree plots for each of the colour bands - what value of nc do these suggest?
2. Try to create non-log scree plots for each of the colour bands - what happens?
3. Calculate the percentage in memory saving resulting from the chosen value of nc
4. Look at each 'photo' corresponding to each singular component. Can you see the original coming through?
5. Overlay the first few modes for each colour. What can you see now?

Outline Solutions

1. For the scree plots - here are the red ones.

plt.bar(range(Sr.shape[0]),Sr, log=True)

<BarContainer object of 6936 artists>

# this fails - there are too many and each bar is too thin to show
#plt.figure(figsize=(12, 12)) # you can try a bigger plot, doesn't help
plt.bar(range(Sr.shape[0]),Sr)

<BarContainer object of 6936 artists>

plt.bar(range(50),Sr[:50])

<BarContainer object of 50 artists>

1. The memory ratio is the new size of the first nc components divided by the size of the original

print('Each matrix contains ', imgr_mat.size, ' elements')
print('Each element occupies ', imgr_mat.itemsize, ' bytes')
print('So each matrix occupies ', imgr_mat.size * imgg_mat.itemsize, ' bytes in memory')

num_bytes_orig = 3*imgr_mat.size * imgg_mat.itemsize
print(f'This is {num_bytes_orig} bytes for all three')

Each matrix contains  64144128  elements
Each element occupies  8  bytes
So each matrix occupies  513153024  bytes in memory
This is 1539459072 bytes for all three

For nc components, for each colour, we had this:

rec_imgr = np.matrix(Ur[:, :nc]) * np.diag(Sr[:nc]) * np.matrix(VTr[:nc, :])

So, for this colour we need nc columns in Ur, nc scalars in Sr and nc rows in VTr. And the same for each of the other colours.

print(f'RED:   shapes of Ur, Sr, VTr = {Ur.shape}, {Sr.shape}, {VTr.shape}')
num_bytes_recon  = 3*nc*(Ur.shape[1] + 1 + VTr.shape[0])*Ur.itemsize
print(f'This is {num_bytes_recon} bytes for all three')

RED:   shapes of Ur, Sr, VTr = (9248, 9248), (6936,), (6936, 6936)
This is 1942200 bytes for all three

# The percentage memory savings ratio is then... 
print(f'percentage savings in bytes = {100*num_bytes_recon/num_bytes_orig} %')

percentage savings in bytes = 0.12616119748326768 %

1. This is a complicated bit of code. There is a lot of value in understanding it

The graphics are unlikely to fit on the slide - use the notebook

# cell toggle scrolling from menu to prevent small window and scroll bar
fontsize=20; fig=plt.figure(figsize=(15, 15))
fig.suptitle('First Few Modes', fontsize=fontsize)

c = 0; max_m = 5; step_m = 2
for m in range(0,max_m*step_m, step_m):
  n = m+1
  c += 1
  mode_imgr = np.matrix(Ur[:, m:n]) * np.diag(Sr[m:n]) * np.matrix(VTr[m:n, :])
  axs=fig.add_subplot(max_m,3,c)
  axs.imshow(mode_imgr)
  axs.set_title('red, mode = '+str(m), fontsize=fontsize)
  c += 1
  mode_imgg = np.matrix(Ug[:, m:n]) * np.diag(Sg[m:n]) * np.matrix(VTg[m:n, :])
  axs=fig.add_subplot(max_m,3,c)
  axs.imshow(mode_imgg)
  axs.set_title('green, mode = '+str(m), fontsize=fontsize)
  c += 1
  mode_imgb = np.matrix(Ub[:, m:n]) * np.diag(Sb[m:n]) * np.matrix(VTb[m:n, :])
  axs=fig.add_subplot(max_m,3,c)
  axs.imshow(mode_imgb)

  axs.set_title('blue, mode = '+str(m), fontsize=fontsize)

1. This code is also of value.

The graphics are unlikely to fit on the slide - use the notebook

# cell toggle scrolling from menu to prevent small window and scroll bar
fontsize=20
fig=plt.figure(figsize=(15,7))
fig.suptitle('First Few Modes Combined', fontsize=fontsize)

m = 0; n = max_m*step_m
c = 1
mode_imgr = np.matrix(Ur[:, m:n]) * np.diag(Sr[m:n]) * np.matrix(VTr[m:n, :])
axs=fig.add_subplot(1,3,c)
axs.imshow(mode_imgr)
axs.set_title('red, modes up to = '+str(n), fontsize=fontsize)
c += 1
mode_imgg = np.matrix(Ug[:, m:n]) * np.diag(Sg[m:n]) * np.matrix(VTg[m:n, :])
axs=fig.add_subplot(1,3,c)
axs.imshow(mode_imgg)
axs.set_title('green, modes up to = '+str(n), fontsize=fontsize)
c += 1
mode_imgb = np.matrix(Ub[:, m:n]) * np.diag(Sb[m:n]) * np.matrix(VTb[m:n, :])
axs=fig.add_subplot(1,3,c)
axs.imshow(mode_imgb)
axs.set_title('blue, modes up to = '+str(n), fontsize=fontsize)

Text(0.5, 1.0, 'blue, modes up to = 10')