I'm sorry if this is basic, I need to make this matrix to define the weight for a Gaussian triple quadrature program I'm trying to make.
def coefraiz(q):
x = sym.symbols('x')
for i in range(1,q+1):
f =1
for j in range(1,q+1):
if i != j:
raizj = np.array(sym.solve(Ps(j),x))
raizi = np.array(sym.solve(Ps(i),x))
f = f*((x-raizj[j])/(raizi[i]-raizj[j]))
coef = sym.integrate(f,-1,1)
M=[coef(raiz)]
return M
Here's the Ps function that the matrix uses, which is functional:
def Ps(n):
x = sym.symbols('x')
a = 1/(2**n)
b = 0
for k in range(n+1):
intro = abs(n-k)
b0 = ((math.factorial(n)) /(math.factorial(k)*math.factorial(intro)))**2
b1 = (x + 1)**(n-k)
b2 = (x - 1)**k
b += b + b0*b1*b2
Pn = a*b
return Pn
Related
I have written this simple convolution function in numpy. But the final array values are all still zero.
Please help me correct this function.
def convolve(a_prev, w, b):
pad = 0
stride = 1
s1 = a_prev.shape
s2 = w.shape
f = s2[1]
m = s1[0]
n_c = s2[0]
n_h = int((s1[1] - f + 2 * pad) / stride) + 1
n_w = int((s1[2] - f + 2 * pad) / stride) + 1
a = np.zeros((m,n_h,n_w,n_c), dtype=np.float32)
for n in range(m):
for z in range(n_c):
y = 0
x = 0
while ((y+f) <= n_h):
# Edit: forget to inialize the x = 0
while ((x+f) <= n_w):
#a[n,y,x,z] = np.sum(a_prev[n,y:y+f,x:x+f]*w[z]) + b[z,0]
a[n,y,x,z] = np.sum(np.multiply(a_prev[n,y:y+f,x:x+f],w[z])) + b[z,0]
x += stride
y += stride
print(a[0,85,:,3])
return a
shape of a_prev is [num_exmamples,height, width, 3] and w is [num_filters,3,3,3]
I found the reason, why it was not working, i make a programming error and forget to initialize the x = 0 before while loop.
Its working fine now.
Below is the correct function.
def convolve(a_prev, kernel, b, pad = 0, stride = 1):
m = a_prev.shape[0]
prev_h = a_prev.shape[1]
prev_w = a_prev.shape[2]
f = kernel.shape[1]
n_c = kernel.shape[0]
new_h = int((prev_h - f + 2 * pad) / stride) + 1
new_w = int((prev_w - f + 2 * pad) / stride) + 1
az = np.zeros((m,new_h,new_w,n_c), dtype=np.float32)
for n in range(m):
for z in range(n_c):
y = 0
while (y+f) <= prev_h:
x = 0
while (x+f) <= prev_w:
az[n,y,x,z] = np.sum(a_prev[n,y:y+f,x:x+f]*kernel[z]) + b[z,0]
x += stride
y += stride
return az
There is an easier way to do convolution without using the previous input as a function input.
import numpy as np
def convolution(x, h):
# x and h are numpy arrays
M, N = np.size(x), np.size(h)
y = np.zeros(M+N-1)
# Initialise y with the length of the output signal
for m in np.arange(M):
for n in np.arange(N):
y[m+n] += x[m]*h[n]
return y
This function uses the basic definition of convolution for discrete signals
I'm trying to solve a constrained minimization problem using SymPy. For a fixed number of variables, say w1, w2, I'm able to do this in the following way:
from sympy import *
w1, w2 = var('w1, w2', real = True)
n1, n2 = symbols('n1, n2', integer = True)
p1, p2 = symbols('p1, p2', real = True)
f = w1**2 / (n1 * p1) + w2**2 / (n2 * p2)
g = w1 + w2 - 1
lam = symbols('lambda', real = True)
L = f - lam * g
gradL = [diff(L, c) for c in [w1, w2]]
KKT_eqs = gradL + [g]
stationary_points = solve(KKT_eqs, [w1, w2, lam], dict = True)
Are we able to solve this problem for a variable number, say k, of variables? I've tried the following:
from sympy import *
i = symbols('i', cls = Idx)
k = symbols('k', integer = True)
w = IndexedBase('w', real = True)
n = IndexedBase('n', integer = True)
p = IndexedBase('p', real = True)
f = summation(w[i]**2 / (n[i] * p[i]), (i, 1, k))
g = summation(w[i], (i, 1, k)) - 1
lam = symbols('lambda', real = True)
L = f - lam * g
However, I wasn't able to figure out how I need to adopt the remainder of the code.
(I'm new to python so please bear with me.)
One thing you can do is look for a pattern within the concrete cases:
>>> from sympy import *
... from sympy.abc import i
... w = IndexedBase('w')
... np = IndexedBase('np')
... lam = symbols('lambda', real = True)
... def go(n):
... ww = [w[i] for i in range(n)]
... f = Add(*[wi**2/np[i] for i,wi in enumerate(ww)])
... g = Add(*ww) - 1
... L = f - lam * g
... gradL = [diff(L, c) for c in ww]
... KKT_eqs = gradL + [g]
... return solve(KKT_eqs, ww + [lam], dict = True)
>>> go(2)
[{lambda: 2/(np[0] + np[1]), w[0]: np[0]/(np[0] + np[1]), w[1]: np[1]/(np[0] + np[1])}]
>>> go(3)
[{lambda: 2/(np[0] + np[1] + np[2]), w[0]: np[0]/(np[0] + np[1] + np[2]), w[1]: np[1]/(np[0] + np[1] + np[2]), w[2]: np[2]/(np[0] + np[1] + np[2])}]
Note: since n[i]*p[i] always appear together, those two variables have been combined into one. Do you see the pattern for the solutions? Try go(4) if you don't.
To generalize without concrete insight you might be able to do something with MatrixExpr.
I want to implement subgradient and Stochastic descent using a cost function, calculate the number of iterations that it takes to find a perfect classifier for the data and also the weights (w) and bias (b).
the dataset is in four dimension
this is my cost function
i have take the derivative of the cost function and here it is:
When i run my code i get a lot of errors, can someone please help.
Here is my Code in python
import numpy as np
learn_rate = 1
w = np.zeros((4,1))
b = 0
M = 1000
data = '/Users/labuew/Desktop/dataset.data'
#calculating the gradient
def cal_grad_w(data, w, b):
for i in range (M):
sample = data[i,:]
Ym = sample[-1]
Xm = sample[0:4]
if -Ym[i]*(w*Xm+b) >= 0:
tmp = 1.0
else:
tmp = 0
value = Ym[i]*Xm*tmp
sum = sum +value
return sum
def cal_grad_b(data, w, b):
for i in range (M):
sample = data[i,:]
Ym = sample[-1]
Xm = sample[0:4]
if -Ym*(w*Xm+b) >= 0:
tmp = 1.0
else:
tmp = 0
value = Ym[i]*x*tmp
sum = sum +value
return sum
if __name__ == '__main__':
counter = 0
while 1:
counter +=1
dw = cal_grad_w(data, w, b)
db = cal_grad_b(data, w, b)
if dw == 0 and db == 0:
break
w = w - learn_rate*dw
b = b - learn_rate *dw
print(counter,w,b)
are you missing the numpy load function?
data = np.load('/Users/labuew/Desktop/dataset.data')
It looks like you're doing the numerics on the string.
also
Ym = sample[-1]
Xm = sample[0:4]
Also 4 dimensions implies that Ym = Xm[3]? Is your data rank 2 with the second rank being dimension 5? [0:4] includes the forth dimension i.e.
z = [1,2,3,4]
z[0:4] = [1,2,3,4]
This would be my best guess. I'm taking a few educated guesses about your data format.
import numpy as np
learn_rate = 1
w = np.zeros((1,4))
b = 0
M = 1000
#Possible format
#data = np.load('/Users/labuew/Desktop/dataset.data')
#Assumed format
data = np.ones((1000,5))
#calculating the gradient
def cal_grad_w(data, w, b):
sum = 0
for i in range (M):
sample = data[i,:]
Ym = sample[-1]
Xm = sample[0:4]
if -1*Ym*(np.matmul(w,Xm.reshape(4,1))+b) >= 0:
tmp = 1.0
else:
tmp = 0
value = Ym*Xm*tmp
sum = sum +value
return sum.reshape(1,4)
def cal_grad_b(data, w, b):
sum = 0
for i in range (M):
sample = data[i,:]
Ym = sample[-1]
Xm = sample[0:4]
if -1*Ym*(np.matmul(w,Xm.reshape(4,1))+b) >= 0:
tmp = 1.0
else:
tmp = 0
value = Ym*tmp
sum = sum +value
return sum
if __name__ == '__main__':
counter = 0
while 1:
counter +=1
dw = cal_grad_w(data, w, b)
db = cal_grad_b(data, w, b)
if dw.all() == 0 and db == 0:
break
w = w - learn_rate*dw
b = b - learn_rate*db
print([counter,w,b])
Put in dummy data because I don't know the format.
I have the following code:
import numpy as np
import cvxpy as cp
import math
import sys
def solve05( p, a ):
m,n,ids,inv,k = 0,len(p),{},{},0
for i in range(n):
for j in range(n):
ids[(i,j)] = k
inv[k] = (i,j)
k = k+1
# Problem data
A = np.zeros((2*n,n*n+n))
D = np.zeros((2*n,n*n+n))
b = np.zeros(2*n)
B = np.zeros(2*n)
c = np.zeros(2*n)
for j in range(n):
for i in range(n):
idx = ids[(i,j)]
A[j,idx] = 1
b[j] = 1
for i in range(n):
for j in range(n):
idx = ids[(i,j)]
A[i+n,idx] = p[j]
A[i+n,n*n+i] = -1
b[i+n] = p[i]
# Construct the problem
x = cp.Variable(n*n+n)
print("M = ",A)
print("b = ",b)
CF = 1e3
print("Now scaling M by ",CF)
A = A*CF
print(A)
b = b*CF
constraints = [0 <= x, A*x == b]
pex = x[n*n]+x[n*n+1]+x[n*n+2]+1
constraints.append(x[n*n] <= a[0]*CF)
constraints.append(x[n*n+1] <= a[1]*CF)
constraints.append(x[n*n+2] <= a[2]*CF)
constraints.append(x[n*n] >= 0.01)
constraints.append(x[n*n+1] >= 0.01)
constraints.append(x[n*n+2] >= 0.01)
ex = pex.__pow__(-1)
print("Dummy variables: ",x[n*n],x[n*n+1],x[n*n+2])
print("Objective function: ",ex)
print("[should be convex] Curvature: ",ex.curvature)
objective = cp.Minimize(ex)
prob = cp.Problem(objective,constraints)
result = prob.solve(verbose=True)
print('problem state: ', prob.status)
alpha = np.zeros((n,n))
for i in range(n):
for j in range(n):
alpha[i,j] = x.value[ids[(i,j)]]
dummy = [x.value[j] for j in range(n*n,n*n+n)]
return (x,alpha)
if __name__ == '__main__':
p = [0.0005,0.0001,0.0007]
a = [900,500,700]
n = len(a)
(sl,alpha) = solve05(p,a)
for row in alpha:
for x in row:
print("%.4f " % (x), end=" "),
print("")
It fails with "Problem UNFEASIBLE" verdict, and I am eager to know why.
Is there any way to know more? I am not a convex programming expert, so any comments on why this is a bad model is appreciated. I have also tried scaling the problem, because I thought some numerical instability may be what is causing problems, but alas.
The answer ecos+cvxpy was giving is correct. The problem is unfeasible, which can be shown by summing up all the equations and observing that the LHS is some quantity F, whereas the RHS is F+e, for some e > 0.
I'm currently trying to implement the GDT described by Felzenszwalb and Huttenlocher (http://www.cs.cornell.edu/~dph/papers/dt.pdf) inside of Python for an image processing algorithm. However I used the algorithm described in the paper they published a few years back but got faulty results. I found a C# implementation here: https://dsp.stackexchange.com/questions/227/fastest-available-algorithm-for-distance-transform/29727?noredirect=1#comment55866_29727
And converted it to Python (which is pretty much the same I had before).
This is my code:
def of_column(dataInput):
output = zeros(dataInput.shape)
n = len(dataInput)
k = 0
v = zeros((n,))
z = zeros((n + 1,))
v[0] = 0
z[0] = -inf
z[1] = +inf
s = 0
for q in range(1, n):
while True:
s = (((dataInput[q] + q * q) - (dataInput[v[k]] + v[k] * v[k])) / (2.0 * q - 2.0 * v[k]))
if s <= z[k]:
k -= 1
else:
break
k += 1
v[k] = q
z[k] = s
z[k + 1] = +inf
k = 0
for q in range(n):
while z[k + 1] < q:
k += 1
output[q] = ((q - v[k]) * (q - v[k]) + dataInput[v[k]])
return output
I still can't find my error. When giving the algorithm a binary (boolean) numpy array it just returns the array itself not the Distance Transform. Why is this not working in Python?
I got it working after hours and hours. The answer given in the link above implementing the code in C# suggests putting up the "white" areas to a very large number. My dataInput array was a boolean array (0, 1). I replaced all 1s with 2^32 and it works just fine. The higher the number the more blurry it gets. The lower the more similar to the source it gets.
I would like to add the function for 2D that works with the 1D function described previously:
###############################################################################
# distance transform of 1d function using squared distance
###############################################################################
def dt_1d(dataInput, n):
output = np.zeros(dataInput.shape)
k = 0
v = np.zeros((n,))
z = np.zeros((n + 1,))
v[0] = 0
z[0] = -np.inf
z[1] = +np.inf
for q in range(1, n):
s = (((dataInput[q] + q * q) - (dataInput[v[k]] + v[k] * v[k])) / (2.0 * q - 2.0 * v[k]))
while s <= z[k]:
k -= 1
s = (((dataInput[q] + q * q) - (dataInput[v[k]] + v[k] * v[k])) / (2.0 * q - 2.0 * v[k]))
k += 1
v[k] = q
z[k] = s
z[k + 1] = +np.inf
k = 0
for q in range(n):
while z[k + 1] < q:
k += 1
value = ((q - v[k]) * (q - v[k]) + dataInput[v[k]])
if value > 255: value = 255
if value < 0: value = 0
output[q] = value
print output
return output
###############################################################################
# distance transform of 2d function using squared distance
###############################################################################
def dt_2d(dataInput):
height, width = dataInput.shape
f = np.zeros(max(height, width))
# transform along columns
for x in range(width):
f = dataInput[:,x]
dataInput[:,x] = dt_1d(f, height)
# transform along rows
for y in range(height):
f = dataInput[y,:]
dataInput[y,:] = dt_1d(f, width)
return dataInput
I hope it helps.