Commit 1097541b authored by Pierre-antoine Comby's avatar Pierre-antoine Comby

devoirs de 455

parent c9283f84
...@@ -7,12 +7,15 @@ ...@@ -7,12 +7,15 @@
\subsection{version objet} \subsection{version objet}
\inputminted{python}{../algo_code/huffman2.py} \inputminted{python}{../algo_code/huffman2.py}
\section{Codage arithétique} \section{Codage arithmétique}
\inputminted{python}{../algio_code/code_arithmetique.py}
\section{Codage LZW} \section{Codage LZW}
\inputminted{python}{../algo_code/LZW.py} \inputminted{python}{../algo_code/LZW.py}
\section{Quantification} \section{Quantification}
\subsection{Quantification uniforme} \subsection{Quantification uniforme}
\inputminted{python}{../algo_code/quantif.py}
\subsection{Algorithme de Llyod-max} \subsection{Algorithme de Llyod-max}
\inputminted{python}{../algo_code/llyod_max.py}
\subsection{Algorithme LBG} \subsection{Algorithme LBG}
en 2D , ne pas essayer de tracer les cellule de voronoi en 2D , ne pas essayer de tracer les cellule de voronoi
\section{Codeur prédictif} \section{Codeur prédictif}
......
...@@ -74,6 +74,5 @@ def arithm_pratique(X,p): ...@@ -74,6 +74,5 @@ def arithm_pratique(X,p):
l[-1] = 2*l[-1]-0.5 l[-1] = 2*l[-1]-0.5
h[-1] = 2*h[-1]-0.5 h[-1] = 2*h[-1]-0.5
return c return c
#print(arithm(X,p)) #print(arithm(X,p))
print(arithm_pratique(X,p)) print(arithm_pratique(X,p))
#!/usr/bin/env python3
import numpy as np
from sipy import integrate
from scipy import norm
M = 8
X = np.random.normal(0,1,1000)
def ddp(x):
mean = 0,
sigma = 1
return norm.pdf(x,mean,sigma)
def init_thres_vec(M,X):
step = (np.max(X)-np.min(X))/M
thres_intervals = np.array([])
mid = np.mean(X)
for i in range(int(M/2)):
thres_intervals = np.append(thres_vec,mid+(i+1)*step)
thres_intervals = np.insert(thtres_vec,0,mid-(1+1)*step)
return thres_intervals
def quant(x,thres,intervals):
thres= np.append(thres, np.inf)
thres= np.insert(thres, 0, -np.inf)
x_hat_q = np.zeros(np.shape(x))
for i in range(len(thres)-1):
if i == 0:
x_hat_q = np.where(np.logical_and(x > thres[i], x <= thres[i+1]),
np.full(np.size(x_hat_q), intervals[i]), x_hat_q)
elif i == range(len(thres))[-1]-1:
x_hat_q = np.where(np.logical_and(x > thres[i], x <= thres[i+1]),
np.full(np.size(x_hat_q), intervals[i]), x_hat_q)
else:
x_hat_q = np.where(np.logical_and(x > thres[i], x < thres[i+1]),
np.full(np.size(x_hat_q), intervals[i]), x_hat_q)
return x_hat_q
def LlyodMax(X,intervals, max_iter=1000,eps=1e-5):
err_min = np.inf
for i in range(max_iter):
for j in range(len(x_hat_q)):
centroids[i] = integrate.quad(lambda x : x*ddp(x),
intervals[j],intervals[j+1])[0]/
integrate.quad(lambda x : ddp(x),
intervals[j],intervals[j+1])[0]
intervals = 0.5*(centroids[1:]+centroids[:-1])
x_hat = quant(X,centroids,intervals)
err = np.linalg.norm(X-x_hat)
if err < err_min:
err_min =err
intervals_min = intervals
centroids_min = centroids
if err_min< 1e-5:
break
best_x_hat = quant(X,centroids_min,intervals_min)
return best_x_hat
#!/usr/bin/env python3
import numpy as np
import matplotlib.pyplot as plt
N = 1000
X = np.random.rand(N)
X_c = (X - 0.5)*10
def quantif_uniforme(M,X,xmin=-1,xmax=1,d=0):
"""
réalise la quantification uniforme d'un vecteur sur M niveau
"""
delta = 2 * xmax/M # pas de quantification
Q = np.zeros(len(X))
for k in range(len(X)):
q = (X[k]/ delta)
if abs(q)<d: #seuil
Q[k] = 0
continue
elif abs(q)<2*delta:
if q <0:
Q[k] =-1
else:
Q[k] = 1
continue
else:
Q[k] = int(q)
return Q,delta
def reverse_quantif(Q,delta):
return Q*delta
Q,delta = quantif_uniforme(4,X_c)
Q_2,delta = quantif_uniforme(4,X_c,d=0.5):
print(len(Q),len(X_c))
plt.figure()
plt.grid()
plt.plot(X_c,Q,'.')
plt.plot(X_c,Q_2,'.')
plt.show()
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