Scipy optimize.minimize seems to accept only single-dimension x0. I have a problem where my x0 are shape(n, m). Constraints exist such that each row of x0 should match a certain value.
I could simply iterate through each row and perform the optimization on that; however, I'm hoping to add constraints to the columns at some point.
Is there a known way of handling this? I can't find much discussion of it. I've tried various versions of broadcasting, flattening, etc., but haven't had much luck in creating a reasonable structure.
EDIT: I've added a minimal code example. The constraint condition returns proper zeros when tested with test_x.
import numpy as np
import scipy.optimize
def cost(x, p):
x.reshape(3, 4)
p.reshape(3, 4)
return (x * p).sum()
def demand_constraint(x, d):
x = x.reshape(3, 4)
b = x.sum(axis=0) - d
return np.broadcast_to(b, (3, 4)).flatten()
demand = np.array([10, 14, 8, 26])
prices = np.array([[4, 4, 5, 5], [2, 8, 6, 2], [3, 2, 9, 8]])
x0 = np.zeros_like(prices).flatten()
p0 = prices.flatten()
test_x = np.array([[4, 14, 8, 26], [5, 0, 0, 0], [0, 0, 0, 0]])
cost(x0, p0)
cons = ({'type': 'eq', 'fun': demand_constraint, 'args': (demand,)})
output = scipy.optimize.minimize(cost, x0, args=p0, constraints=cons)
For anyone who may encounter this in a search, the way to handle it is to add a constraint for every individual row. So the demand_constraint above will take a row_index field and a single value will be returned.
This single value can then be incorporated into a constraint that is added. You continue to add constraints (each it's own dictionary) for the shape. My mistake was to assume that the constraint could apply to all of the x. It's better suited to apply to a single x.
Related
I have an optimization problem where I'm trying to find an array that needs to optimize two functions simultaneously.
In the minimal example below I have two known arrays w and x and an unknown array y. I initialize array y to contains only 1s.
I then specify function np.sqrt(np.sum((x-np.array)**2) and want to find the array y where
np.sqrt(np.sum((x-y)**2) approaches 5
np.sqrt(np.sum((w-y)**2) approaches 8
The code below can be used to successfully optimize y with respect to a single array, but I would like to find that the solution that optimizes y with respect to both x and y simultaneously, but am unsure how to specify the two constraints.
y should only consist of values greater than 0.
Any ideas on how to go about this ?
w = np.array([6, 3, 1, 0, 2])
x = np.array([3, 4, 5, 6, 7])
y = np.array([1, 1, 1, 1, 1])
def func(x, y):
z = np.sqrt(np.sum((x-y)**2)) - 5
return np.zeros(x.shape[0],) + z
r = opt.root(func, x0=y, method='hybr')
print(r.x)
# array([1.97522498 3.47287981 5.1943792 2.10120135 4.09593969])
print(np.sqrt(np.sum((x-r.x)**2)))
# 5.0
One option is to use scipy.optimize.minimize instead of root, Here you have multiple solver options and some of them (ie SLSQP) allow you to specify multiple constraints. Note that I changed the variable names so that x is the array you want to optimise and y and z define the constraints.
from scipy.optimize import minimize
import numpy as np
x0 = np.array([1, 1, 1, 1, 1])
y = np.array([6, 3, 1, 0, 2])
z = np.array([3, 4, 5, 6, 7])
constraint_x = dict(type='ineq',
fun=lambda x: x) # fulfilled if > 0
constraint_y = dict(type='eq',
fun=lambda x: np.linalg.norm(x-y) - 5) # fulfilled if == 0
constraint_z = dict(type='eq',
fun=lambda x: np.linalg.norm(x-z) - 8) # fulfilled if == 0
res = minimize(fun=lambda x: np.linalg.norm(x), constraints=[constraint_y, constraint_z], x0=x0,
method='SLSQP', options=dict(ftol=1e-8)) # default 1e-6
print(res.x) # [1.55517124 1.44981672 1.46921122 1.61335466 2.13174483]
print(np.linalg.norm(res.x-y)) # 5.00000000137866
print(np.linalg.norm(res.x-z)) # 8.000000000930026
This is a minimizer so besides the constraints it also wants a function to minimize, I chose just the norm of y, but setting the function to a constant (ie lambda x: 1) would have also worked.
Note also that the constraints are not exactly fulfilled, you can increase the accuracy by setting optional argument ftol to a smaller value ie 1e-10.
For more information see also the documentation and the corresponding sections for each solver.
I have an array of data-points, for example:
[10, 9, 8, 7, 6, 5, 4, 3, 2, 1]
and I need to perform the following sum on the values:
However, the problem is that I need to perform this sum on each value > i. For example, using the last 3 values in the set the sum would be:
and so on up to 10.
If i run something like:
import numpy as np
x = np.array([10, 9, 8, 7, 6, 5, 4, 3, 2, 1])
alpha = 1/np.log(2)
for i in x:
y = sum(x**(alpha)*np.log(x))
print (y)
It returns a single value of y = 247.7827060452275, whereas I need an array of values. I think I need to reverse the order of the data to achieve what I want but I'm having trouble visualising the problem (hope I explained it properly) as a whole so any suggestions would be much appreciated.
The following computes all the partial sums of the grand sum in your formula
import numpy as np
# Generate numpy array [1, 10]
x = np.arange(1, 11)
alpha = 1 / np.log(2)
# Compute parts of the sum
parts = x ** alpha * np.log(x)
# Compute all partial sums
part_sums = np.cumsum(parts)
print(part_sums)
You really do not any explicit loop, or a non-numpy operation (like sum()) here. numpy takes care of all your needs.
Suppose I have an example of numpy array:
import numpy as np
X = np.array([2,5,0,4,3,1])
And I also have a list of arrays, like:
A = [np.array([-2,0,2]), np.array([0,1,2,3,4,5]), np.array([2,5,4,6])]
I want to leave only these items of each list that are also in X. I expect also to do it in a most efficient/common way.
Solution I have tried so far:
Sort X using X.sort().
Find locations of items of each array in X using:
locations = [np.searchsorted(X, n) for n in A]
Leave only proper ones:
masks = [X[locations[i]] == A[i] for i in range(len(A))]
result = [A[i][masks[i]] for i in range(len(A))]
But it doesn't work because locations of third array is out of bounds:
locations = [array([0, 0, 2], dtype=int64), array([0, 1, 2, 3, 4, 5], dtype=int64), array([2, 5, 4, 6], dtype=int64)]
How to solve this issue?
Update
I ended up with idx[idx==len(Xs)] = 0 solution. I've also noticed two different approaches posted between the answers: transforming X into set vs np.sort. Both of them has plusses and minuses: set operations uses iterations which is quite slow in compare with numpy methods; however np.searchsorted speed increases logarithmically unlike acceses of set items which is instant. That why I decided to compare performance using data with huge sizes, especially 1 million items for X, A[0], A[1], A[2].
One idea would be less compute and minimal work when looping. So, here's one with those in mind -
a = np.concatenate(A)
m = np.isin(a,X)
l = np.array(list(map(len,A)))
a_m = a[m]
cut_idx = np.r_[0,l.cumsum()]
l_m = np.add.reduceat(m,cut_idx[:-1])
cl_m = np.r_[0,l_m.cumsum()]
out = [a_m[i:j] for (i,j) in zip(cl_m[:-1],cl_m[1:])]
Alternative #1 :
We can also use np.searchsorted to get the isin mask, like so -
Xs = np.sort(X)
idx = np.searchsorted(Xs,a)
idx[idx==len(Xs)] = 0
m = Xs[idx]==a
Another way with np.intersect1d
If you are looking for the most common/elegant one, think it would be with np.intersect1d -
In [43]: [np.intersect1d(X,A_i) for A_i in A]
Out[43]: [array([0, 2]), array([0, 1, 2, 3, 4, 5]), array([2, 4, 5])]
Solving your issue
You can also solve your out-of-bounds issue, with a simple fix -
for l in locations:
l[l==len(X)]=0
How about this, very simple and efficent:
import numpy as np
X = np.array([2,5,0,4,3,1])
A = [np.array([-2,0,2]), np.array([0,1,2,3,4,5]), np.array([2,5,4,6])]
X_set = set(X)
A = [np.array([a for a in arr if a in X_set]) for arr in A]
#[array([0, 2]), array([0, 1, 2, 3, 4, 5]), array([2, 5, 4])]
According to the docs, set operations all have O(1) complexity, therefore the overall is O(N)
The Problem:
I want to calculate the dot product of a very large set of data. I am able to do this in a nested for-loop, but this is way too slow.
Here is a small example:
import numpy as np
points = np.array([[0.5, 2, 3, 5.5, 8, 11], [1, 2, -1.5, 0.5, 4, 5]])
lines = np.array([[0, 2, 4, 6, 10, 10, 0, 0], [0, 0, 0, 0, 0, 4, 4, 0]])
x1 = lines[0][0:-1]
y1 = lines[1][0:-1]
L1 = np.asarray([x1, y1])
# calculate the relative length of the projection
# of each point onto each line
a = np.diff(lines)
b = points[:,:,None] - L1[:,None,:]
print(a.shape)
print(b.shape)
[rows, cols, pages] = np.shape(b)
Z = np.zeros((cols, pages))
for k in range(cols):
for l in range(pages):
Z[k][l] = a[0][l]*b[0][k][l] + a[1][l]*b[1][k][l]
N = np.linalg.norm(a, axis=0)**2
relativeProjectionLength = np.squeeze(np.asarray(Z/N))
In this example, the first two dimensions of both a and b represent the x- and y-coordinates that I need for the dot product.
The shape of a is (2,7) and b has (2,6,7). Since the dot product reduces the first dimension I would expect the result to be of the shape (6,7). How can I calculate this without the slow loops?
What I have tried:
I think that numpy.dot with correct broadcasting could do the job, however I have trouble setting up the dimensions correctly.
a = a[:, None, :]
Z = np.dot(a,b)
This on gives me the following error:
shapes (2,1,7) and (2,6,7) not aligned: 7 (dim 2) != 6 (dim 1)
You can use np.einsum -
np.einsum('ij,ikj->kj',a,b)
Explanation :
Keep the last axes aligned for the two inputs.
Sum-reduce the first from those.
Let the rest stay, which is the second axis of b.
Usual rules on whether to use einsum or stick to a loopy-dot based method apply here.
numpy.dot does not reduce the first dimension. From the docs:
For N dimensions it is a sum product over the last axis of a and the second-to-last of b:
dot(a, b)[i,j,k,m] = sum(a[i,j,:] * b[k,:,m])
That is exactly what the error is telling you: it is attempting to match axis 2 in the first vector to axis 1 in the second.
You can fix this using numpy.rollaxis or better yet numpy.moveaxis. Instead of a = a[:, None, :], do
a = np.movesxis(a, 0, -1)
b = np.moveaxis(b, 0, -2)
Z = np.dot(a, b)
Better yet, you can construct your arrays to have the correct shape up front. For example, transpose lines and do a = np.diff(lines, axis=0).
I'm trying to solve a simple equation: dM/dr = r*p(r) in python.
I have the values of p at certain values of r:
p(0)=1, p(1)=3, p(2)=5, p(3)=7, p(4)=9, p(5)=11.
I tried using the following code but I get the error
The size of the array returned by func (6) does not match the size of
y0 (1).
I think the problem is that I'm not matching the p values with the r values correctly. There should only be one initial condition since I am only trying to solve one equation. Any help would be greatly appreciated.
This is my code:
from scipy import integrate
import numpy as np
r = np.array([0, 1, 2, 3, 4, 5])
p = np.array([1, 3, 5, 7, 9, 11])
def deriv (z, r, data):
M = r*p
return M
init = np.array([0])
soln = integrate.odeint(deriv, init, p, (r,), full_output=True)
print soln
You are seeing this error because the size of init does not match the size of the array returned by deriv().
To solve the problem, change the following line
init = np.array([0])
to
init = np.array([0, 0, 0, 0, 0, 0])
For more examples on using 'odeint', see:
http://scipy-cookbook.readthedocs.org/items/numpy_scipy_ordinary_differential_equations.html
http://docs.scipy.org/doc/scipy/reference/generated/scipy.integrate.odeint.html