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im trying to write a program that gives the integral approximation of e(x^2) between 0 and 1 based on this integral formula:
Formula
i've done this code so far but it keeps giving the wrong answer (Other methods gives 1.46 as an answer, this one gives 1.006).
I think that maybe there is a problem with the two for cycles that does the Riemman sum, or that there is a problem in the way i've wrote the formula. I also tried to re-write the formula in other ways but i had no success
Any kind of help is appreciated.
import math
import numpy as np
def f(x):
y = np.exp(x**2)
return y
a = float(input("¿Cual es el limite inferior? \n"))
b = float(input("¿Cual es el limite superior? \n"))
n = int(input("¿Cual es el numero de intervalos? "))
x = np.zeros([n+1])
y = np.zeros([n])
z = np.zeros([n])
h = (b-a)/n
print (h)
x[0] = a
x[n] = b
suma1 = 0
suma2 = 0
for i in np.arange(1,n):
x[i] = x[i-1] + h
suma1 = suma1 + f(x[i])
alfa = (x[i]-x[i-1])/3
for i in np.arange(0,n):
y[i] = (x[i-1]+ alfa)
suma2 = suma2 + f(y[i])
z[i] = y[i] + alfa
int3 = ((b-a)/(8*n)) * (f(x[0])+f(x[n]) + (3*(suma2+f(z[i]))) + (2*(suma1)))
print (int3)
I'm not a math major but I remember helping a friend with this rule for something about waterplane area for ships.
Here's an implementation based on Wikipedia's description of the Simpson's 3/8 rule:
# The input parameters
a, b, n = 0, 1, 10
# Divide the interval into 3*n sub-intervals
# and hence 3*n+1 endpoints
x = np.linspace(a,b,3*n+1)
y = f(x)
# The weight for each points
w = [1,3,3,1]
result = 0
for i in range(0, 3*n, 3):
# Calculate the area, 4 points at a time
result += (x[i+3] - x[i]) / 8 * (y[i:i+4] * w).sum()
# result = 1.4626525814387632
You can do it using numpy.vectorize (Based on this wikipedia post):
a, b, n = 0, 1, 10**6
h = (b-a) / n
x = np.linspace(0,n,n+1)*h + a
fv = np.vectorize(f)
(
3*h/8 * (
f(x[0]) +
3 * fv(x[np.mod(np.arange(len(x)), 3) != 0]).sum() + #skip every 3rd index
2 * fv(x[::3]).sum() + #get every 3rd index
f(x[-1])
)
)
#Output: 1.462654874404461
If you use numpy's built-in functions (which I think is always possible), performance will improve considerably:
a, b, n = 0, 1, 10**6
x = np.exp(np.square(np.linspace(0,n,n+1)*h + a))
(
3*h/8 * (
x[0] +
3 * x[np.mod(np.arange(len(x)), 3) != 0].sum()+
2 * x[::3].sum() +
x[-1]
)
)
#Output: 1.462654874404461
I have a code that works perfectly well but I wish to speed up the time it takes to converge. A snippet of the code is shown below:
def myfunction(x, i):
y = x + (min(0, target[i] - data[i, :]x))*data[i]/(norm(data[i])**2))
return y
rows, columns = data.shape
start = time.time()
iterate = 0
iterate_count = []
norm_count = []
res = 5
x_not = np.ones(columns)
norm_count.append(norm(x_not))
iterate_count.append(0)
while res > 1e-8:
for row in range(rows):
y = myfunction(x_not, row)
x_not = y
iterate += 1
iterate_count.append(iterate)
norm_count.append(norm(x_not))
res = abs(norm_count[-1] - norm_count[-2])
print('Converge at {} iterations'.format(iterate))
print('Duration: {:.4f} seconds'.format(time.time() - start))
I am relatively new in Python. I will appreciate any hint/assistance.
Ax=b is the problem we wish to solve. Here, 'A' is the 'data' and 'b' is the 'target'
Ugh! After spending a while on this I don't think it can be done the way you've set up your problem. In each iteration over the row, you modify x_not and then pass the updated result to get the solution for the next row. This kind of setup can't be vectorized easily. You can learn the thought process of vectorization from the failed attempt, so I'm including it in the answer. I'm also including a different iterative method to solve linear systems of equations. I've included a vectorized version -- where the solution is updated using matrix multiplication and vector addition, and a loopy version -- where the solution is updated using a for loop to demonstrate what you can expect to gain.
1. The failed attempt
Let's take a look at what you're doing here.
def myfunction(x, i):
y = x + (min(0, target[i] - data[i, :] # x)) * (data[i] / (norm(data[i])**2))
return y
You subtract
the dot product of (the ith row of data and x_not)
from the ith row of target,
limited at zero.
You multiply this result with the ith row of data divided my the norm of that row squared. Let's call this part2
Then you add this to the ith element of x_not
Now let's look at the shapes of the matrices.
data is (M, N).
target is (M, ).
x_not is (N, )
Instead of doing these operations rowwise, you can operate on the entire matrix!
1.1. Simplifying the dot product.
Instead of doing data[i, :] # x, you can do data # x_not and this gives an array with the ith element giving the dot product of the ith row with x_not. So now we have data # x_not with shape (M, )
Then, you can subtract this from the entire target array, so target - (data # x_not) has shape (M, ).
So far, we have
part1 = target - (data # x_not)
Next, if anything is greater than zero, set it to zero.
part1[part1 > 0] = 0
1.2. Finding rowwise norms.
Finally, you want to multiply this by the row of data, and divide by the square of the L2-norm of that row. To get the norm of each row of a matrix, you do
rownorms = np.linalg.norm(data, axis=1)
This is a (M, ) array, so we need to convert it to a (M, 1) array so we can divide each row. rownorms[:, None] does this. Then divide data by this.
part2 = data / (rownorms[:, None]**2)
1.3. Add to x_not
Finally, we're adding each row of part1 * part2 to the original x_not and returning the result
result = x_not + (part1 * part2).sum(axis=0)
Here's where we get stuck. In your approach, each call to myfunction() gives a value of part1 that depends on target[i], which was changed in the last call to myfunction().
2. Why vectorize?
Using numpy's inbuilt methods instead of looping allows it to offload the calculation to its C backend, so it runs faster. If your numpy is linked to a BLAS backend, you can extract even more speed by using your processor's SIMD registers
The conjugate gradient method is a simple iterative method to solve certain systems of equations. There are other more complex algorithms that can solve general systems well, but this should do for the purposes of our demo. Again, the purpose is not to have an iterative algorithm that will perfectly solve any linear system of equations, but to show what kind of speedup you can expect if you vectorize your code.
Given your system
data # x_not = target
Let's define some variables:
A = data.T # data
b = data.T # target
And we'll solve the system A # x = b
x = np.zeros((columns,)) # Initial guess. Can be anything
resid = b - A # x
p = resid
while (np.abs(resid) > tolerance).any():
Ap = A # p
alpha = (resid.T # resid) / (p.T # Ap)
x = x + alpha * p
resid_new = resid - alpha * Ap
beta = (resid_new.T # resid_new) / (resid.T # resid)
p = resid_new + beta * p
resid = resid_new + 0
To contrast the fully vectorized approach with one that uses iterations to update the rows of x and resid_new, let's define another implementation of the CG solver that does this.
def solve_loopy(data, target, itermax = 100, tolerance = 1e-8):
A = data.T # data
b = data.T # target
rows, columns = data.shape
x = np.zeros((columns,)) # Initial guess. Can be anything
resid = b - A # x
resid_new = b - A # x
p = resid
niter = 0
while (np.abs(resid) > tolerance).any() and niter < itermax:
Ap = A # p
alpha = (resid.T # resid) / (p.T # Ap)
for i in range(len(x)):
x[i] = x[i] + alpha * p[i]
resid_new[i] = resid[i] - alpha * Ap[i]
# resid_new = resid - alpha * A # p
beta = (resid_new.T # resid_new) / (resid.T # resid)
p = resid_new + beta * p
resid = resid_new + 0
niter += 1
return x
And our original vector method:
def solve_vect(data, target, itermax = 100, tolerance = 1e-8):
A = data.T # data
b = data.T # target
rows, columns = data.shape
x = np.zeros((columns,)) # Initial guess. Can be anything
resid = b - A # x
resid_new = b - A # x
p = resid
niter = 0
while (np.abs(resid) > tolerance).any() and niter < itermax:
Ap = A # p
alpha = (resid.T # resid) / (p.T # Ap)
x = x + alpha * p
resid_new = resid - alpha * Ap
beta = (resid_new.T # resid_new) / (resid.T # resid)
p = resid_new + beta * p
resid = resid_new + 0
niter += 1
return x
Let's solve a simple system to see if this works first:
2x1 + x2 = -5
−x1 + x2 = -2
should give a solution of [-1, -3]
data = np.array([[ 2, 1],
[-1, 1]])
target = np.array([-5, -2])
print(solve_loopy(data, target))
print(solve_vect(data, target))
Both give the correct solution [-1, -3], yay! Now on to bigger things:
data = np.random.random((100, 100))
target = np.random.random((100, ))
Let's ensure the solution is still correct:
sol1 = solve_loopy(data, target)
np.allclose(data # sol1, target)
# Output: False
sol2 = solve_vect(data, target)
np.allclose(data # sol2, target)
# Output: False
Hmm, looks like the CG method doesn't work for badly conditioned random matrices we created. Well, at least both give the same result.
np.allclose(sol1, sol2)
# Output: True
But let's not get discouraged! We don't really care if it works perfectly, the point of this is to demonstrate how amazing vectorization is. So let's time this:
import timeit
timeit.timeit('solve_loopy(data, target)', number=10, setup='from __main__ import solve_loopy, data, target')
# Output: 0.25586539999994784
timeit.timeit('solve_vect(data, target)', number=10, setup='from __main__ import solve_vect, data, target')
# Output: 0.12008900000000722
Nice! A ~2x speedup simply by avoiding a loop while updating our solution!
For larger systems, this will be even better.
for N in [10, 50, 100, 500, 1000]:
data = np.random.random((N, N))
target = np.random.random((N, ))
t_loopy = timeit.timeit('solve_loopy(data, target)', number=10, setup='from __main__ import solve_loopy, data, target')
t_vect = timeit.timeit('solve_vect(data, target)', number=10, setup='from __main__ import solve_vect, data, target')
print(N, t_loopy, t_vect, t_loopy/t_vect)
This gives us:
N t_loopy t_vect speedup
00010 0.002823 0.002099 1.345390
00050 0.051209 0.014486 3.535048
00100 0.260348 0.114601 2.271773
00500 0.980453 0.240151 4.082644
01000 1.769959 0.508197 3.482822
I'm trying to write a program that will convert a 480x480 image into 240x240 quality, by taking the average color of 4 pixels and replacing all 4 with that average color. Then I repeat this for all 2x2 squares in the image. My program so far keeps replacing the entire image with the average color from the top left corner 4 squares, which is bgr [150, 138, 126]. Can someone help me see what I'm doing wrong?
def get_square(x1, x2, y1, y2):
square = image[x1:x2, y1:y2]
return square
def bgr_vals(img):
b_vals, g_vals, r_vals = [], [], []
x, y = 0, 0
for i in range(4):
b, g, r = image[x][y]
b_vals.append(b)
g_vals.append(g)
r_vals.append(r)
if y != 1:
y += 1
elif x == 0 and y == 1:
x += 1
elif x == 1 and y == 1:
y -= 1
return b_vals, g_vals, r_vals
def avg_color(bgr_vals):
b_avg = np.average(bgr_vals[0])
g_avg = np.average(bgr_vals[1])
r_avg = np.average(bgr_vals[2])
return [b_avg, g_avg, r_avg]
image = cv2.imread('src.jpg')
y1, y2 = 0, 2
for i in range(240):
x1, x2 = 0, 2
for i in range(240):
patch = get_square(x1, x2, y1, y2)
bgr_colors = bgr_vals(patch)
color = avg_color(bgr_colors)
image[x1:x2, y1:y2] = color
x1 += 2
x2 += 2
y1 += 2
y2 += 2
Thanks!
In your function bgr_vals you take argument img (which is the current patch), but within the method you inadvertently access image (the whole image) in this line:
b, g, r = image[x][y]
Your script works fine when you fix this typo.
Here are some tips for the future:
Images in OpenCV (and most libraries) are stored in row-first order, this means that you should write image[y][x] or img[y][x]. In your case it doesn't matter but in future work it might.
In general try to test your programs with non-square images, it is a common pitfall to only test with square-sized images
When you have cascaded loops, don't use the same variable name in the loop (you used i both times). Again, in your case it doesn't matter but as soon as you would use the value of i the confusion would start
Instead of increasing x, y within the loop you can use the loop variable directly, by getting a range with a step:
for y in range(0, 480, 2):
for x in range(0, 480, 2):
…
Use array shape instead of hard-coding dimensions:
for y in range(0, image.shape[0], 2):
for x in range(0, image.shape[1], 2):
…
The sum of two numbers is 20. If each number is added to its square root, the product of the two sums is 155.55. Use Secant Method to approximate, to within 10^(-4), the value of the two numbers.
Based on http://campus.murraystate.edu/academic/faculty/wlyle/420/Secant.htm
#inital guess
x1 = 10
x2 = 50
Epsilon = 1e-4
#given function
def func(x):
return abs(x)**0.5 * (abs(x)+20)**0.5 - 155.55
y1 = func(x1)
y2 = func(x2)
#loop max 20 times
for i in range(20):
ans = x2 - y2 * (x2-x1)/(y2-y1)
y3 = func(ans)
print("Try:{}\tx1:{:0.3f}\tx2:{:0.3f}\ty3:{:0.3f}".format(i,x1, x2, y3))
if (abs(y3) < Epsilon):
break
x1, x2 = x2, ans
y1, y2 = y2, y3
print("\n\nThe numbers are: {:0.3f} and {:0.3f}".format(ans, ans+20))
Based on Your Title
This code works well in most of the cases. Taken from Secant Method Using Python (Output Included)
# Defining Function
def f(x):
return x**3 - 5*x - 9
# Implementing Secant Method
def secant(x0,x1,e,N):
print('\n\n*** SECANT METHOD IMPLEMENTATION ***')
step = 1
condition = True
while condition:
if f(x0) == f(x1):
print('Divide by zero error!')
break
x2 = x0 - (x1-x0)*f(x0)/( f(x1) - f(x0) )
print('Iteration-%d, x2 = %0.6f and f(x2) = %0.6f' % (step, x2, f(x2)))
x0 = x1
x1 = x2
step = step + 1
if step > N:
print('Not Convergent!')
break
condition = abs(f(x2)) > e
print('\n Required root is: %0.8f' % x2)
# Input Section
x0 = input('Enter First Guess: ')
x1 = input('Enter Second Guess: ')
e = input('Tolerable Error: ')
N = input('Maximum Step: ')
# Converting x0 and e to float
x0 = float(x0)
x1 = float(x1)
e = float(e)
# Converting N to integer
N = int(N)
#Note: You can combine above three section like this
# x0 = float(input('Enter First Guess: '))
# x1 = float(input('Enter Second Guess: '))
# e = float(input('Tolerable Error: '))
# N = int(input('Maximum Step: '))
# Starting Secant Method
secant(x0,x1,e,N)
firstly, apologize for little cryptic title to my question. Let me try to explain my need:-
I am reading two features namely X1, X2 from a CSV file. I have a training set of data in a csv file containing 1000 records with each line corresponding to the value of X1, X2. To make my training set fit better to my machine learning code, I want to do feature mapping that would take X1, X2 and create polynomial terms to the power of 4. for example if X1 =a, X2=b, I want to add newer features a^2, a*b, b^2, a^3,a^2*b,a*b^2,a^4...and so on.
Now if I read them as a numpy matrix , I want to see the data like this:
[ [ 1 a b a^2 a*b, b^2 a^3 a^2*b......]
[.... ............ ............ ]
[ ..
..] ]
Note that the number of rows are fixed , but the number of columns are determined by the degree selected. Also first three columns need to be
[[1 a b ..]
[1 c d ..]
..
..]
The pseudo code I am thinking of is as follows:-
def poly(X): # where X is a numpy matrix with X1, X2 columns,
degree = 4;
r= X.shape[0]
c=1 # number of columns
val_matrix= np.ones(shape=(r,c)) # creating a (r,1) matrix init with 1s
# *start of psuedo code*
while i<=degree:
while j <=i:
val_matrix[:, c+1] = (X1.^(i-j)).*(X2.^j)
I am not sure how to get this working in python?. would appreciate some suggestion. Note that ^ refers to the power of.
Starting with two vectors X1 and X2 you could create the monomials:
X1p = X1[:, None]**np.arange(max_deg + 1)
X2p = X2[:, None]**np.arange(max_deg + 1)
and then combine them using mgrid
i, j = np.mgrid[:max_deg + 1,:max_deg + 1]
m = i+j <= max_deg
result = X1p[:, i[m]]*X2p[:, j[m]]
Alternatively you could apply the indices directly to X1 and X2:
result = X1[:, None]**i[m] * X2[:, None]**j[m]
This requires fewer lines of code but uses more multiplications.
If the number of multiplications is a concern, X1p and X2p could also be computed cheaper; X1p:
X1p = np.empty((len(X1), max_deg + 1), X1.dtype)
X1p[:, 0] = 1
X1p[:, 1:] = X1[:, None]
np.multiply.accumulate(X1p[:,1:], axis=-1, out=X1p[:, 1:])
and similar for X2p