Hoe om beelde te komprimeer met SVD en TensorFlow Core API's

Inhoud oorsig Inleiding Stel SVD basiese beginsels Lae rang benadering met SVD Afbeelding loading Die beeldkompressie algoritme Beoordeling van naderinge Kompressie faktor versus rang Kumulatiewe som van enkele waardes Fout en unieke waardes Konklusie Inleiding Hierdie tutorial ondersoek die tegniek van (SVD) en sy toepassings vir lae-rang benaderingsprobleme. Die SVD word gebruik om werklike of komplekse matrices te factoriseer en het 'n verskeidenheid van gebruik gevalle in data wetenskap, soos beeldkompressie. Die projek. Unieke waarde ontbinding Die beeld Stel import matplotlib from matplotlib.image import imread from matplotlib import pyplot as plt import requests # Preset Matplotlib figure sizes. matplotlib.rcParams['figure.figsize'] = [16, 9] import tensorflow as tf print(tf.__version__) 2024-08-15 02:40:31.007111: E external/local_xla/xla/stream_executor/cuda/cuda_fft.cc:485] Unable to register cuFFT factory: Attempting to register factory for plugin cuFFT when one has already been registered 2024-08-15 02:40:31.028202: E external/local_xla/xla/stream_executor/cuda/cuda_dnn.cc:8454] Unable to register cuDNN factory: Attempting to register factory for plugin cuDNN when one has already been registered 2024-08-15 02:40:31.034794: E external/local_xla/xla/stream_executor/cuda/cuda_blas.cc:1452] Unable to register cuBLAS factory: Attempting to register factory for plugin cuBLAS when one has already been registered 2.17.0 SVD basiese beginsels Die singular waarde ontbinding van 'n matrix, A, word bepaal deur die volgende faktorisering: A = Oorweeg Waar Am×n: die input matrix waar m≥n Um×n: orthogonal matrix, UTU=I, met elke kolom, ui, wat 'n linker enkele vektor van A aanduid Σn×n: diagonaal matrix met elke diagonaal invoer, σi, wat 'n enkele waarde van A verteenwoordig VTn×n: orthogonal matrix, VTV=I, met elke ry, vi, wat 'n regter enkele vektor van A aanduid Wanneer m<n, het U en Σ beide dimensie (m×m), en VT het dimensie (m×n). Die lineêre algebra pakket van TensorFlow het 'n funksie, , wat gebruik kan word om die enkele waarde afbraak van een of meer matrices te bereken. begin deur 'n eenvoudige matriks te definieer en sy SVD-faktorisering te bereken. tf.linalg.svd A = tf.random.uniform(shape=[40,30]) # Compute the SVD factorization s, U, V = tf.linalg.svd(A) # Define Sigma and V Transpose S = tf.linalg.diag(s) V_T = tf.transpose(V) # Reconstruct the original matrix A_svd = U@S@V_T # Visualize plt.bar(range(len(s)), s); plt.xlabel("Singular value rank") plt.ylabel("Singular value") plt.title("Bar graph of singular values"); WARNING: All log messages before absl::InitializeLog() is called are written to STDERR I0000 00:00:1723689633.424919 126589 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1723689633.428762 126589 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1723689633.432420 126589 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1723689633.437707 126589 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1723689633.449819 126589 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1723689633.453529 126589 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1723689633.457079 126589 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1723689633.460600 126589 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1723689633.464019 126589 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1723689633.467451 126589 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1723689633.470857 126589 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1723689633.474376 126589 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1723689634.696864 126589 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1723689634.698995 126589 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1723689634.701043 126589 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1723689634.703182 126589 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1723689634.705235 126589 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1723689634.707223 126589 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1723689634.709163 126589 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1723689634.711162 126589 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1723689634.713114 126589 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1723689634.715090 126589 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1723689634.717036 126589 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1723689634.719041 126589 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1723689634.757916 126589 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1723689634.759983 126589 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1723689634.761969 126589 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1723689634.764105 126589 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1723689634.766234 126589 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1723689634.768229 126589 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1723689634.770180 126589 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1723689634.772176 126589 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1723689634.774158 126589 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1723689634.776614 126589 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1723689634.779022 126589 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1723689634.781451 126589 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 die funksie kan gebruik word om die matrix rekonstruksie direk te bereken van die uitkomste van . tf.einsum tf.linalg.svd A_svd = tf.einsum('s,us,vs -> uv',s,U,V) print('\nReconstructed Matrix, A_svd', A_svd) Reconstructed Matrix, A_svd tf.Tensor( [[0.819255 0.95082176 0.30085772 ... 0.34306473 0.9112067 0.45277762] [0.71988434 0.20786765 0.3098724 ... 0.82478774 0.82917416 0.19994356] [0.23278469 0.17072585 0.10488278 ... 0.60508984 0.8121238 0.10594818] ... [0.10953172 0.72077066 0.30288345 ... 0.5790409 0.15389094 0.11265922] [0.44894537 0.04313891 0.6947845 ... 0.41589245 0.6668094 0.53494996] [0.79140556 0.47814542 0.24980234 ... 0.32851136 0.86007696 0.32692394]], shape=(40, 30), dtype=float32) Lae rang benadering met die SVD Die rang van 'n matrix, A, word bepaal deur die dimensie van die vektorruimte wat deur sy kolommen versprei word.Die SVD kan gebruik word om 'n matrix met 'n laer rang te benader, wat uiteindelik die dimensionaliteit van data verminder wat nodig is om die inligting wat deur die matrix verteenwoordig word, te stoor. Die rang-r-aanpak van A in terme van die SVD word gedefinieer deur die formule: Ar=UrΣrVrT Waar Urm×r: matrix wat bestaan uit die eerste r-kolomme van U Σrr×r: diagonaal matrix wat bestaan uit die eerste r-singuliere waardes in Σ Vrr×nT: matrix wat bestaan uit die eerste r reëls van VT Begin deur 'n funksie te skryf om die rang-r-benadering van 'n gegewe matrix te bereken. Hierdie lae-rang-benaderingsprocedure word gebruik vir beeldkompressie; daarom is dit ook nuttig om die fisiese data groottes vir elke benadering te bereken. Vir eenvoud, veronderstel dat die data groottes vir 'n rang-r-benadering matrix gelyk is aan die totale aantal elemente wat nodig is om die benadering te bereken. Volgende, skryf 'n funksie om die oorspronklike matrix te visualiseer, A se rang-r-benadering, Ar en die foutmatrix. def rank_r_approx(s, U, V, r, verbose=False): # Compute the matrices necessary for a rank-r approximation s_r, U_r, V_r = s[..., :r], U[..., :, :r], V[..., :, :r] # ... implies any number of extra batch axes # Compute the low-rank approximation and its size A_r = tf.einsum('...s,...us,...vs->...uv',s_r,U_r,V_r) A_r_size = tf.size(U_r) + tf.size(s_r) + tf.size(V_r) if verbose: print(f"Approximation Size: {A_r_size}") return A_r, A_r_size def viz_approx(A, A_r): # Plot A, A_r, and A - A_r vmin, vmax = 0, tf.reduce_max(A) fig, ax = plt.subplots(1,3) mats = [A, A_r, abs(A - A_r)] titles = ['Original A', 'Approximated A_r', 'Error |A - A_r|'] for i, (mat, title) in enumerate(zip(mats, titles)): ax[i].pcolormesh(mat, vmin=vmin, vmax=vmax) ax[i].set_title(title) ax[i].axis('off') print(f"Original Size of A: {tf.size(A)}") s, U, V = tf.linalg.svd(A) Original Size of A: 1200 # Rank-15 approximation A_15, A_15_size = rank_r_approx(s, U, V, 15, verbose = True) viz_approx(A, A_15) Approximation Size: 1065 # Rank-3 approximation A_3, A_3_size = rank_r_approx(s, U, V, 3, verbose = True) viz_approx(A, A_3) Approximation Size: 213 Soos verwag word, het die gebruik van laer range minder akkurate benaderinge. Maar die gehalte van hierdie lae-rang benaderinge is dikwels goed genoeg in werklike wêreld scenario's. Let ook daarop dat die hoofdoel van lae-rang benadering met SVD is om die dimensionaliteit van die data te verminder, maar nie om die disk ruimte van die data self te verminder nie. Afbeelding loading Die volgende foto is beskikbaar op die Imagen is 'n teks-tot-beeld-diffusie model wat deur Google Research se Brain-span ontwikkel is. 'N AI het hierdie beeld geskep op grond van die oproep: "'n foto van 'n Corgi-hond wat op 'n fiets in Times Square ry. Dit dra sonbril en 'n strandhoed." Hoe cool is dit! Jy kan ook die URL hieronder verander na enige .jpg skakel om in 'n persoonlike beeld van keuse te laai. Die beeld Na die lees van 'n JPEG-lêer, outputs Matplotlib 'n matrix, I, van vorm (m×n×3) wat 'n 2-dimensionele beeld met 3 kleurkanale vir rooi, groen en blou verteenwoordig. img_link = "https://imagen.research.google/main_gallery_images/a-photo-of-a-corgi-dog-riding-a-bike-in-times-square.jpg" img_path = requests.get(img_link, stream=True).raw I = imread(img_path, 0) print("Input Image Shape:", I.shape) Input Image Shape: (1024, 1024, 3) def show_img(I): # Display the image in matplotlib img = plt.imshow(I) plt.axis('off') return show_img(I) Die beeldkompressie algoritme Nou gebruik die SVD om lae-rang benaderinge van die monsterbeeld te bereken. Onthou dat die beeld vorm (1024×1024×3) is en dat die SVD-teorie slegs vir 2-dimensionele matrices toepas. Dit beteken dat die monsterbeeld in 3 gelykgrootte matrices wat elk van die 3 kleurkanale ooreenstem, moet gepas word. Dit kan gedoen word deur die matriks om vorm te wees (3×1024×1024). Om die benaderingfout duidelik te visualiseer, moet die RGB-waardes van die beeld van [0,255] na [0,1] omskaleer. Onthou om die benaderinge waardes binne hierdie interval te klip voordat jy hulle visualiseer. Funksie is daarvoor nuttig. tf.clip_by_value def compress_image(I, r, verbose=False): # Compress an image with the SVD given a rank I_size = tf.size(I) print(f"Original size of image: {I_size}") # Compute SVD of image I = tf.convert_to_tensor(I)/255 I_batched = tf.transpose(I, [2, 0, 1]) # einops.rearrange(I, 'h w c -> c h w') s, U, V = tf.linalg.svd(I_batched) # Compute low-rank approximation of image across each RGB channel I_r, I_r_size = rank_r_approx(s, U, V, r) I_r = tf.transpose(I_r, [1, 2, 0]) # einops.rearrange(I_r, 'c h w -> h w c') I_r_prop = (I_r_size / I_size) if verbose: # Display compressed image and attributes print(f"Number of singular values used in compression: {r}") print(f"Compressed image size: {I_r_size}") print(f"Proportion of original size: {I_r_prop:.3f}") ax_1 = plt.subplot(1,2,1) show_img(tf.clip_by_value(I_r,0.,1.)) ax_1.set_title("Approximated image") ax_2 = plt.subplot(1,2,2) show_img(tf.clip_by_value(0.5+abs(I-I_r),0.,1.)) ax_2.set_title("Error") return I_r, I_r_prop Nou, bereken rang-r benaderinge vir die volgende rande : 100, 50, 10 I_100, I_100_prop = compress_image(I, 100, verbose=True) Original size of image: 3145728 Number of singular values used in compression: 100 Compressed image size: 614700 Proportion of original size: 0.195 I_50, I_50_prop = compress_image(I, 50, verbose=True) Original size of image: 3145728 Number of singular values used in compression: 50 Compressed image size: 307350 Proportion of original size: 0.098 I_10, I_10_prop = compress_image(I, 10, verbose=True) Original size of image: 3145728 Number of singular values used in compression: 10 Compressed image size: 61470 Proportion of original size: 0.020 Beoordeling van naderinge Daar is 'n verskeidenheid interessante metodes om doeltreffendheid te meet en meer beheer te hê oor matriksbenaderinge. Kompressie faktor versus rang Vir elkeen van die bovenstaande benaderinge, sien hoe die data groottes verander met die rang. plt.figure(figsize=(11,6)) plt.plot([100, 50, 10], [I_100_prop, I_50_prop, I_10_prop]) plt.xlabel("Rank") plt.ylabel("Proportion of original image size") plt.title("Compression factor vs rank"); Gebaseer op hierdie plot, is daar 'n lineêre verhouding tussen die kompressiefaktor van 'n benaderde beeld en sy rang. Om dit verder te verken, onthou dat die data grootte van 'n benaderde matrix, Ar, gedefinieer word as die totale aantal elemente wat nodig is vir sy berekening. x=(m×r)+r+(r×n)=r×(m+n+1) c=xy=r×(m+n+1)m×n Waar x: grootte van die lug A: Die grootte van A c = xy: die kompressiefaktor r: rang van die benadering m en n: reeks- en kolomdimensies van A Om die rang, r, wat nodig is om 'n beeld te komprimeer na 'n gewenste faktor, c, te vind, kan die bovenstaande vergelyking herarrangeer word om vir r op te los: r = c × m × nm + n + 1 Let daarop dat hierdie formule onafhanklik is van die kleurkanaaldimensie aangesien elkeen van die RGB-aanpassings mekaar nie beïnvloed nie. def compress_image_with_factor(I, compression_factor, verbose=False): # Returns a compressed image based on a desired compression factor m,n,o = I.shape r = int((compression_factor * m * n)/(m + n + 1)) I_r, I_r_prop = compress_image(I, r, verbose=verbose) return I_r Komprimeer 'n beeld tot 15% van sy oorspronklike grootte. compression_factor = 0.15 I_r_img = compress_image_with_factor(I, compression_factor, verbose=True) Original size of image: 3145728 Number of singular values used in compression: 76 Compressed image size: 467172 Proportion of original size: 0.149 Kumulatiewe som van enkele waardes Die kumulatiewe som van enkele waardes kan 'n nuttige aanduiding wees van die hoeveelheid energie wat deur 'n rang-r-approximasie gevang word. Funksie kan hiervoor nuttig wees. tf.cumsum def viz_energy(I): # Visualize the energy captured based on rank # Computing SVD I = tf.convert_to_tensor(I)/255 I_batched = tf.transpose(I, [2, 0, 1]) s, U, V = tf.linalg.svd(I_batched) # Plotting average proportion across RGB channels props_rgb = tf.map_fn(lambda x: tf.cumsum(x)/tf.reduce_sum(x), s) props_rgb_mean = tf.reduce_mean(props_rgb, axis=0) plt.figure(figsize=(11,6)) plt.plot(range(len(I)), props_rgb_mean, color='k') plt.xlabel("Rank / singular value number") plt.ylabel("Cumulative proportion of singular values") plt.title("RGB-averaged proportion of energy captured by the first 'r' singular values") viz_energy(I) Dit lyk asof meer as 90% van die energie in hierdie prentjie binne die eerste 100 enkele waardes gevang word. def compress_image_with_energy(I, energy_factor, verbose=False): # Returns a compressed image based on a desired energy factor # Computing SVD I_rescaled = tf.convert_to_tensor(I)/255 I_batched = tf.transpose(I_rescaled, [2, 0, 1]) s, U, V = tf.linalg.svd(I_batched) # Extracting singular values props_rgb = tf.map_fn(lambda x: tf.cumsum(x)/tf.reduce_sum(x), s) props_rgb_mean = tf.reduce_mean(props_rgb, axis=0) # Find closest r that corresponds to the energy factor r = tf.argmin(tf.abs(props_rgb_mean - energy_factor)) + 1 actual_ef = props_rgb_mean[r] I_r, I_r_prop = compress_image(I, r, verbose=verbose) print(f"Proportion of energy captured by the first {r} singular values: {actual_ef:.3f}") return I_r Komprimeer 'n beeld om 75% van sy energie te behou. energy_factor = 0.75 I_r_img = compress_image_with_energy(I, energy_factor, verbose=True) Original size of image: 3145728 Number of singular values used in compression: 35 Compressed image size: 215145 Proportion of original size: 0.068 Proportion of energy captured by the first 35 singular values: 0.753 Fout en unieke waardes Daar is ook 'n interessante verhouding tussen die benadering fout en die enkele waardes. Dit blyk dat die vierkante Frobenius norm van die benadering gelyk is aan die som van die vierkante van sy enkele waardes wat uitgesluit is: Oor die oorspronklike oorspronklike oorspronklike oorspronklike oorspronklike Testeer hierdie verhouding met 'n rang-10 benadering van die voorbeeld matrix aan die begin van hierdie tutorial. s, U, V = tf.linalg.svd(A) A_10, A_10_size = rank_r_approx(s, U, V, 10) squared_norm = tf.norm(A - A_10)**2 s_squared_sum = tf.reduce_sum(s[10:]**2) print(f"Squared Frobenius norm: {squared_norm:.3f}") print(f"Sum of squared singular values left out: {s_squared_sum:.3f}") Squared Frobenius norm: 33.021 Sum of squared singular values left out: 33.021 Konklusie Hierdie nootboek het die proses geïmplementeer om die enkele waarde afbreek met TensorFlow te implementeer en dit toe te pas om 'n beeldkompressie-algoritme te skryf. Die TensorFlow Core API's kan gebruik word vir 'n verskeidenheid hoëprestasie wetenskaplike rekenaargebruik gevalle. Om meer te leer oor TensorFlow se lineêre algebra funksionaliteite, besoek die dokumente vir die linalg module. Die SVD kan ook toegepas word om aanbevelingstelsels te bou. Vir meer voorbeelde van die gebruik van die TensorFlow Core API's, kyk na die As jy meer wil leer oor die oplaai en voorbereiding van data, sien die tutoriale op of . Die gids Beeld data laai CSV data laai Oorspronklik gepubliseer op die TensorFlow webwerf, hierdie artikel verskyn hier onder 'n nuwe kop en is gelisensieer onder CC BY 4.0. Oorspronklik gepubliseer op die TensorFlow webwerf, hierdie artikel verskyn hier onder 'n nuwe kop en is gelisensieer onder CC BY 4.0. Die TensorFlow Die TensorFlow