I cannot solve these two equations using sympy
eq1 = 20*x*y-10*x-4*x**3
eq2 = 10*x**2-8*y-8*y**3
solve([eq1, eq2], [x, y])
My answer is (0,0), (0,-i), (0,i), but the answer of book is (0,0), (+-2.64, 1.90), (+-0.86, 0.65).
book is calculus, 6th edition, James Stewart (section 15-7)
I'm not sure why solve isn't working here. These equations are cubic in each of x and y and since they are non-degenerate that should mean up to 9 roots are possible I think. I can find them by solving eq1 for x, eliminating x from eq2 and then solving eq2 for y. In one line that is
In [101]: sols = [{x: xi.subs(y, yi), y:yi} for xi in solve(eq1, x) for yi in solve(eq2.subs(x, xi), y)]
The full expression for some of these roots is complicated since they come from the cubic formula. I'll show approximate numerical values instead:
In [102]: for s in sols: print('(%s, %s)' % (s[x].n(3, chop=True), s[y].n(3, chop=True)))
(0, 0)
(0, -1.0*I)
(0, 1.0*I)
(-0.857, 0.647)
(-2.64, 1.90)
(-3.9*I, -2.54)
(0.857, 0.647)
(2.64, 1.90)
(3.9*I, -2.54)
Checking these roots with simplify or checksol fails for the (+-3.9*I,-2.54) roots so I'll demonstrate that they are probably solutions numerically instead:
In [103]: [eq1.evalf(subs=s, chop=True) for s in sols]
Out[103]: [0, 0, 0, 0, 0, 0, 0, 0, 0]
In [104]: [eq2.evalf(subs=s, chop=True) for s in sols]
Out[104]: [0, 0, 0, 0, 0, 0, 0, 0, 0]
The procedure recommended by Oscar can be done automatically by using the "manual=True, check=False" flags:
>>> sol = solve((eq1,eq2), check=False, manual=True)
>>> [eq1.subs(s).n(2,chop=True) for s in sol]
[0, 0, 0, 0, 0, 0, 0, 0, 0]
>>> [eq2.subs(s).n(2,chop=True) for s in sol]
[0, 0, 0, 0, 0, 0, 0, 0, 0]
Related
Let's say I have the following (always binary) options:
import numpy as np
a=np.array([1, 1, 0, 0, 1, 1, 1])
b=np.array([1, 1, 0, 0, 1, 0, 1])
c=np.array([1, 0, 0, 1, 0, 0, 0])
d=np.array([1, 0, 1, 1, 0, 0, 0])
And I want to find the optimal combination of the above that get's me to at least, with minimal above:
req = np.array([50,50,20,20,100,40,10])
For example:
final = X1*a + X2*b + X3*c + X4*d
Does this map to a known operational research problem? Or does it fall under mathematical programming?
Is this NP-hard, or exactly solveable in a reasonable amount of time (I've assumed it's combinatorally impossible to solve exactly)
Are there know solutions to this?
Note: The actual length of arrays are longer - think ~50, and the number of options are ~20
My current research has led me to some combination of the assignment problem and knapsack, but not too sure.
It's a covering problem, easily solvable using an integer program solver (I used OR-Tools below). If the X variables can be fractional, substitute NumVar for IntVar. If the X variables are 0--1, substitute BoolVar.
import numpy as np
a = np.array([1, 1, 0, 0, 1, 1, 1])
b = np.array([1, 1, 0, 0, 1, 0, 1])
c = np.array([1, 0, 0, 1, 0, 0, 0])
d = np.array([1, 0, 1, 1, 0, 0, 0])
opt = [a, b, c, d]
req = np.array([50, 50, 20, 20, 100, 40, 10])
from ortools.linear_solver import pywraplp
solver = pywraplp.Solver.CreateSolver("SCIP")
x = [solver.IntVar(0, solver.infinity(), "x{}".format(i)) for i in range(len(opt))]
extra = [solver.NumVar(0, solver.infinity(), "y{}".format(j)) for j in range(len(req))]
for j, (req_j, extra_j) in enumerate(zip(req, extra)):
solver.Add(extra_j == sum(opt_i[j] * x_i for (opt_i, x_i) in zip(opt, x)) - req_j)
solver.Minimize(sum(extra))
status = solver.Solve()
if status == pywraplp.Solver.OPTIMAL:
print("Solution:")
print("Objective value =", solver.Objective().Value())
for i, x_i in enumerate(x):
print("x{} = {}".format(i, x[i].solution_value()))
else:
print("The problem does not have an optimal solution.")
Output:
Solution:
Objective value = 210.0
x0 = 40.0
x1 = 60.0
x2 = -0.0
x3 = 20.0
As an application of Eulers method, I'm trying to implement a code which would compute the recursive matrix product Yn = Yn-1 + A(Yn-1), where Y is a vector and A is a matrix such that the product is defined. This is the current code I have
def f(A, y):
return A.dot(y)
def euler(f, t0, y0, T, dt):
t = np.arange(t0, T + dt, dt)
y = [0,0,0,0]*len(t)
y[0] = y0
for i in range(1, len(t)):
y[i] = y[i - 1] + f(A, y[i - 1])*dt
return t, y
# Define problem specific values
A = np.array([[0, 0, 1, 0],
[0, 0, 0, 1],
[-2, -3, 0, 0],
[-3, -2, 0, 0]])
y1_0 = 1
y2_0 = 2
y3_0 = 0
y4_0 = 0
y0 = [y1_0, y2_0, y3_0, y4_0]
t,y = euler(f,0,y0,2,1)
print(t,y)
For example, the result for points in the range t0 = 0, T = 2 should be the vectors Y1 and Y2. Instead I have
[0 1 2] [[1, 2, 0, 0], array([ 1, 2, -8, -7]), 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
Something is wrong here. While Y1 = [1, 2, -8, -7 ] does show up, there is all of this unnecessary stuff. And Y2 is not printed at all. I suspect this is due to how I define the variable y. For every point in the range of t, I need a vector of 4 zeros - which is then filled up by the function euler, I think. How should correct this?
The computer always does what you tell it to do. In your case y is constructed by repeating 4 zeros len(t) times, giving a list of 12 zeros. The first list entry is replaced by the list y0. The second list entry is replaced by the result of the numpy operations which is a numpy.array. Then the return statement at the level of the loop instructions breaks the loop and returns the t and y arrays. y still contains 10 zeros from its construction that were not replaced.
So construct
y = np.zeros([len(t), len(y0)])
and repair the indentation level.
I need to solve a problem in which I have spent hours, with the data from my excel sheet I have created a 6x36 '' zeros '' matrix of zeros and a 6x6 '' matrix_tran '' coordinate transformation matrix [image 1].
My problem is that I can't find a way to replace the zeros of the '' zeros '' matrix with the values that the matrix '' matrix_tran '' dictates, and whose location must be in the columns (4,5,6, 7,8,9) that are given by the connection vector (4,5,6,7,8,9) of element 15 of the Excel sheet, that is, the last row of the for loop iteration [image 2].
In summary: Below I show how it fits and how it should look [image 3 and 4 respectively].
I would very much appreciate your help, and excuse my English, but it is not my native language, a big greeting.
import pandas as pd
import numpy as np
ex = pd.ExcelFile('matrix_tr.xlsx')
hoja = ex.parse('Hoja1')
cols = 36
for n in range(0,len(hoja)):
A = hoja['ELEMENT #'][n]
B = hoja['1(i)'][n]
C = hoja['2(i)'][n]
D = hoja['3(i)'][n]
E = hoja['1(j)'][n]
F = hoja['2(j)'][n]
G = hoja['3(j)'][n]
H = hoja['X(i)'][n]
I = hoja['Y(i)'][n]
J = hoja['X(j)'][n]
K = hoja['Y(j)'][n]
L = np.sqrt((J-H)**2+(K-I)**2)
lx = (J-H)/L
ly = (K-I)/L
zeros = np.zeros((6, cols))
counters = hoja.loc[:, ["1(i)", "2(i)", "3(i)", "1(j)", "2(j)", "3(j)"]]
for _, i1, i2, i3, j1, j2, j3 in counters.itertuples():
matrix_tran = np.array([[lx, ly, 0, 0, 0, 0],
[-ly, lx, 0, 0, 0, 0],
[0, 0, 1, 0, 0, 0],
[0, 0, 0, lx, ly, 0],
[0, 0, 0, -ly, lx, 0],
[0, 0, 0, 0, 0, 1]])
zeros[:, [i1 - 1, i2 - 1, i3 - 1, j1 - 1, j2 - 1 , j3 - 1]] = matrix_tran
Try with a transposed zeros matrix
import pandas as pd
import numpy as np
ex = pd.ExcelFile('c:/tmp/SO/matrix_tr.xlsx')
hoja = ex.parse('Hoja1')
counters = hoja.loc[:, ["1(i)", "2(i)", "3(i)", "1(j)", "2(j)", "3(j)"]]
# zeros matrix transposed
cols = 36
zeros_trans = np.zeros((cols,6))
# last row only
for n in range(14,len(hoja)):
Xi = hoja['X(i)'][n]
Yi = hoja['Y(i)'][n]
Xj = hoja['X(j)'][n]
Yj = hoja['Y(j)'][n]
X = Xj-Xi
Y = Yj-Yi
L = np.sqrt(X**2+Y**2)
lx = X/L
ly = Y/L
matrix_tran = np.array([[lx, ly, 0, 0, 0, 0],
[-ly, lx, 0, 0, 0, 0],
[0, 0, 1, 0, 0, 0],
[0, 0, 0, lx, ly, 0],
[0, 0, 0, -ly, lx, 0],
[0, 0, 0, 0, 0, 1]])
i = 0
for r in counters.iloc[n]:
zeros_trans[r-1] = matrix_tran[i]
i += 1
print(np.transpose(zeros_trans))
Given some SymPy matrix M
M = Matrix([
[0.000111334436666596, 0.00114870370895408, -0.000328330524152990, 5.61388353859808e-6, -0.000464532588930332, -0.000969955779635878, 1.70579589853818e-5, -5.77891177019884e-6, -0.000186812539472235, -2.37115911398055e-5],
[-0.00105346453420510, 0.000165063406707273, -0.00184449574409890, 0.000658080565333929, 0.00197652092300241, 0.000516180213512589, 9.53823860082390e-5, 0.000189858427211978, -3.80494288487685e-5, 0.000188984043643408],
[-0.00102465075104153, -0.000402915220398109, 0.00123785300884241, -0.00125808154543978, 0.000126618511490838, 0.00185985865307693, 0.000123626008509804, 0.000211557638637554, 0.000407232404255796, 1.89851719447102e-5],
[0.230813497584639, -0.209574389008468, 0.742275067362657, -0.202368828927654, -0.236683258718819, 0.183258819107153, 0.180335891933511, -0.530606389541138, -0.379368598768419, 0.334800403899511],
[-0.00102465075104153, -0.000402915220398109, 0.00123785300884241, -0.00125808154543978, 0.000126618511490838, 0.00185985865307693, 0.000123626008509804, 0.000211557638637554, 0.000407232404255796, 1.89851719447102e-5],
[0.00105346453420510, -0.000165063406707273, 0.00184449574409890, -0.000658080565333929, -0.00197652092300241, -0.000516180213512589, -9.53823860082390e-5, -0.000189858427211978, 3.80494288487685e-5, -0.000188984043643408],
[0.945967255845168, -0.0468645728473480, 0.165423896937049, -0.893045423193559, -0.519428986944650, -0.0463256408085840, -0.0257001217930424, 0.0757328764368606, 0.0541336731317414, -0.0477734271777646],
[-0.0273371493900004, -0.954100482348723, -0.0879282784854250, 0.100704543595514, -0.243312734473589, -0.0217088779350294, 0.900584332231093, 0.616061129532614, 0.0651163853434486, -0.0396603397583054],
[0.0967584768347089, -0.0877680087304911, -0.667679934757176, -0.0848411039101494, -0.0224646387789634, -0.194501966574153, 0.0755161040544943, 0.699388977592066, 0.394125039254254, -0.342798611994521],
[-0.000222668873333193, -0.00229740741790816, 0.000656661048305981, -1.12277670771962e-5, 0.000929065177860663, 0.00193991155927176, -3.41159179707635e-5, 1.15578235403977e-5, 0.000373625078944470, 4.74231822796110e-5]
])
I have calculated SymPy rank() and rref() of the matrix. Rank is 7 and rref() result is:
Matrix([
[1, 0, 0, 0, 0, 0, 0, -5.14556976678473, -3.72094268951566, 3.48581267477014],
[0, 1, 0, 0, 0, 0, 0, -5.52930150663022, -4.02230308325653, 3.79193678096199],
[0, 0, 1, 0, 0, 0, 0, 2.44893308665325, 1.83777402439421, -1.87489784909824],
[0, 0, 0, 1, 0, 0, 0, -7.33732284392352, -5.25036238623229, 4.97256759287563],
[0, 0, 0, 0, 1, 0, 0, 5.48049237370489, 3.90091366576548, -3.83642187384021],
[0, 0, 0, 0, 0, 1, 0, -10.6826798792866, -7.56560803870182, 7.45974067056387],
[0, 0, 0, 0, 0, 0, 1, -3.04726210012149, -2.66388837034592, 2.48327234504403],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]])
Weird thing is that if I calculate rank with either NumPy or MATLAB I get value 6 and calculating rref with MATLAB I get the expected result - last 4 rows are all zero (instead of only last 3).
Does any one know where does this difference comes from and why am I unable to get correct results with SymPy? I know that rank 6 is correct because it is system of the equations where some linear dependency exist.
Looking at the eigenvalues of your matrix, the rank is indeed 6:
array([ 1.14550481e+00+0.00000000e+00j, -1.82137718e-01+6.83443168e-01j,
-1.82137718e-01-6.83443168e-01j, 2.76223053e-03+0.00000000e+00j,
-3.51138883e-04+8.61508469e-04j, -3.51138883e-04-8.61508469e-04j,
5.21160131e-17+0.00000000e+00j, -2.65160469e-16+0.00000000e+00j,
-2.67753616e-18+9.70937977e-18j, -2.67753616e-18-9.70937977e-18j])
With the sympy version I have, I get even a rank of 8, compared to the rank 6 that numpy returns.
But actually, Sympy cannot solve the eigenvalues of this matrix due to the size of the matrix (probably related to SymPy could not compute the eigenvalues of this matrix).
So one of them, Sympy, is trying to solve symbolically the equations and find the rank (based on imperfect floating point numbers), whereas the other one, numpy, uses approximations (lapack IIRC) to find the eigenvalues. By having an adequate threshold, numpy finds the proper rank, but it could have said differently with a different threshold. Sympy tried to find the rank based on an approximate system of a perfect 6 rank system and finds that it is of rank 7 or 8. It's not surprising due to the floating point difference (Sympy moves to integers to try to find the eigenvalues, for instance, instead of staying in floating point realm).
I'd like to solve a multivariate linear regression equation for vector X with m elements while I have n observations, Y. If I assume the measurements have Gaussian random errors. How can I solve this problem using python? My problem looks like this:
A simple example of W when m=5, is given as follows:
P.S. I would like to consider the effect of errors and precisely I want to measure the standard deviation of errors.
You can do it like this
def myreg(W, Y):
from numpy.linalg import pinv
m, n = Y.shape
k = W.shape[1]
X = pinv(W.T.dot(W)).dot(W.T).dot(Y)
Y_hat = W.dot(X)
Residuals = Y_hat - Y
MSE = np.square(Residuals).sum(axis=0) / (m - 2)
X_var = (MSE[:, None] * pinv(W.T.dot(W)).flatten()).reshape(n, k, k)
Tstat = X / np.sqrt(X_var.diagonal(axis1=1, axis2=2)).T
return X, Tstat
demo
W = np.array([
[ 1, -1, 0, 0, 1],
[ 0, -1, 1, 0, 0],
[ 1, 0, 0, 1, 0],
[ 0, 1, -1, 0, 1],
[ 1, -1, 0, 0, -1],
])
Y = np.array([2, 4, 1, 5, 3])[:, None]
X, V = myreg(W, Y)