I try to add the following constraints to my model. my problem: the function g() expects x as a binary numpy array. So the result arr_a depends on the current value of x in every step of the optimization!
Afterwards, I want the max of this array times x to be smaller than 50.
How can I add this constraint dynamically so that arr_a is always rightfully calculated with the value of x at each iteration while telling the model to keep the constraint arr_a * x <= 50 ? Currently I am getting an error when adding the constraint to the model because g() expects x as numpy array to calculate arr_a, arr_b, arr_c ( g uses np.where(x == 1) within its calculation).
#Init model
from ortools.sat.python import cp_model
model = cp_model.CpModel()
# Declare the variables
x = []
for i in range(self.ds.n_banks):
x.append(model.NewIntVar(0, 1, "x[%i]" % (i)))
#add bool vars
a = model.NewBoolVar('a')
arr_a, arr_b, arr_c = g(df1,df2,df3,x)
model.Add((arr_a.astype('int32') * x).max() <= 50).OnlyEnforceIf(a)
model.Add((arr_a.astype('int32') * x).max() > 50).OnlyEnforceIf(a.Not())
Afterwards i add the target function that naturally also depends on x.
model.Minimize(target(x))
def target(x):
arr_a, arr_b, arr_c = g(df1,df2,df3,x)
return (3 * arr_b * x + 2 * arr_c * x).sum()
EDIT:
My problem changed a bit and i managed to get it work without issues. Nevertheless, I experienced that the constraint is never actually met! self-defined-function is a highly non-linear function that expects the indices where x==1 and where x == 0 and returns a numpy array. Also it is not possible to re-build it with pre-defined functions of the sat.solver.
#Init model
model = cp_model.CpModel()
# Declare the variables
x = [model.NewIntVar(0, 1, "x[%i]" % (i)) for i in range(66)]
# add hints
[model.AddHint(x[i],np.random.choice(2, 1, p=[0.4, 0.6])[0]) for i in range(66)]
open_elements = [model.NewBoolVar("open_elements[%i]" % (i)) for i in range(66)]
closed_elements = [model.NewBoolVar("closed_elements[%i]" % (i)) for i in range(6)]
# open indices as bool vars
for i in range(66):
model.Add(x[i] == 1).OnlyEnforceIf(open_elements[i])
model.Add(x[i] != 1).OnlyEnforceIf(open_elements[i].Not())
model.Add(x[i] != 1).OnlyEnforceIf(closed_elements[i])
model.Add(x[i] == 1).OnlyEnforceIf(closed_elements[i].Not())
model.Add((self-defined-function(np.where(open_elements), np.where(closed_elements), some_array).astype('int32') * x - some_vector).all() <= 0)
Even when I apply a simpler function, it will not work properly.
model.Add((self-defined-function(x, some_array).astype('int32') * x - some_vector).all() <= 0)
I also tried the following:
arr_indices_open = []
arr_indices_closed = []
for i in range(66):
if open_elements[i] == True:
arr_indices_open.append(i)
else:
arr_indices_closed.append(i)
# final Constraint
arr_ = self-defined-function(arr_indices_open, arr_indices_closed, some_array)[0].astype('int32')
for i in range(66):
model.Add(arr_[i] * x[i] <= some_other_vector[i])
Some minimal example for the self-defined-function, with which I simply try to say that n_closed shall be smaller than 10. Even that condition is not met by the solver:
def self_defined_function(arr_indices_closed)
return len(arr_indices_closed)
arr_ = self-defined-function(arr_indices_closed)
for i in range(66):
model.Add(arr_ < 10)
I'm not sure I fully understand the question, but generally, if you want to optimize a function g(x), you'll have to implement it in using the solver's primitives (docs).
It's easier to do when your calculation coincides with an existing solver function, e.g.: if you're trying to calculate a linear expression; but could get harder to do when trying to calculate something more complex. However, I believe that's the only way.
Related
Edit: I have also added a constraint 1.5 to illustrate maybe a different way of approaching the constraint.
I am trying to write the following constraints in Pyomo for each (i,j) pair on an MxN grid:
The code that I have thus far is as follows, and I am just hoping I can get some feedback on whether or not the constraint definition is written properly to meet the intention.The idea is that each (i,j) cell on the 6x6 grid will have the following two constraints.
model = AbstractModel()
#Define the index sets for the grid, time horizions, and age classes:
model.Iset = RangeSet(6)
model.Jset = RangeSet(6)
model.Tset = RangeSet(7)
model.Kset = RangeSet(50)
#Define model parameters:
model.s = Param(within=NonNegativeIntegers)
#Define model variables:
model.juvenille = Var(model.Iset, model.Jset, model.Tset, model.Kset,
within=NonNegativeReals, initialize = "some expression"
#Constraints:
# Constraint #1
def juv_advance(model, i, j, t, k):
return model.juvenille[i,j,t+1,k+1] == model.juvenille[i,j,t,k]*model.juvsurv
# Constraint #1.5
def juv_advance(model, t, k):
return model.juvenille[t+1,k+1] == model.juvenille[t,k]*model.s \\
for i in model.Iset for j in model.Jset
# Constraint #2
def juv_total(model, i, j, t, k):
return sum(model.juvenille[k] for k in range(1,50))
Additionally, if anybody feels like answering this... how could you save the calculated j_t+1 value to use as the initial value in the next time period.
I would try something like this:
model = AbstractModel()
#Define the index sets for the grid, time horizions, and age classes:
model.Iset = RangeSet(6)
model.Jset = RangeSet(6)
model.Tset = RangeSet(7)
model.Kset = RangeSet(50)
#Define model parameters:
model.s = Param(within=NonNegativeIntegers)
#Define model variables:
model.juvenille = Var(model.Iset, model.Jset, model.Tset, model.Kset,
within=NonNegativeReals, initialize="some expression")
# As far as I see your problem in you second constraint the big J is a new variable ?
If that is the case than you have to create it:
model.J_big =Var(model.Iset, model.Jset, model.Tset, within=NonNegativeReals)
#Constraints:
# Constraint #1
def juv_advance(model, i, j, t, k):
k_len = len(model.Kset)
t_len = len(model.Tset)
if k == 1 and t == 1:
return "some expression"
elif t < t_len and k < k_len:
return model.juvenille[i,j,t+1,k+1] == model.juvenille[i,j,t,k]*model.s
else:
return "Here has to come a statement what should happen with the last index (because if you are iterating to k=50 there is no k=51) "
model.ConstraintNumber1 = Constraint(model.Iset, model.Jset, model.Tset, model.Kset, rule=juv_advance)
# Constraint #2
def juv_total(model, i, j, t, k):
return model.J_big[i,k,j] == sum(model.juvenille[i,j,t,k] for k in model.Kset)
model.ConstraintNumber2 = Constraint(model.Iset, model.Jset, model.Tset, rule=juv_total)
It is important that you not only define the rule of the constraint but also the constraint itself. Also you have to have in mind that you K and T sets are ending somewhere and that an expression of k+1 does not work if there is no k+1. Another point that could be mentioned is that if you start with k+1 == something the first k-value that is taken into account is k = 2.
I hope this helps, maybe someone also knows something smarter, I am also quite new to pyomo.
I have to optimize the coefficients for three numpy arrays which maximizes my evaluation function.
I have a target array called train['target'] and three predictions arrays named array1, array2 and array3.
I want to put the best linear coefficients i.e., x,y,z for these three arrays which will maximize the function
roc_aoc_curve(train['target'], xarray1 + yarray2 +z*array3)
the above function would be maximum when prediction is closer to the target.
i.e, xarray1 + yarray2 + z*array3 should be closer to train['target'].
The range of x,y,z >=0 and x,y,z <= 1
Basically I am trying to put the weights x,y,z for each of the three arrays which would make the function
xarray1 + yarray2 +z*array3 closer to the train['target']
Any help in getting this would be appreciated.
I used pulp.LpProblem('Giapetto', pulp.LpMaximize) to do the maximization. It works for normal numbers, integers etc, however failing while trying to do with arrays.
import numpy as np
import pulp
# create the LP object, set up as a maximization problem
prob = pulp.LpProblem('Giapetto', pulp.LpMaximize)
# set up decision variables
x = pulp.LpVariable('x', lowBound=0)
y = pulp.LpVariable('y', lowBound=0)
z = pulp.LpVariable('z', lowBound=0)
score = roc_auc_score(train['target'],x*array1+ y*array2 + z*array3)
prob += score
coef = x+y+z
prob += (coef==1)
# solve the LP using the default solver
optimization_result = prob.solve()
# make sure we got an optimal solution
assert optimization_result == pulp.LpStatusOptimal
# display the results
for var in (x, y,z):
print('Optimal weekly number of {} to produce: {:1.0f}'.format(var.name, var.value()))
Getting error at the line
score = roc_auc_score(train['target'],x*array1+ y*array2 + z*array3)
TypeError: unsupported operand type(s) for /: 'int' and 'LpVariable'
Can't progress beyond this line when using arrays. Not sure if my approach is correct. Any help in optimizing the function would be appreciated.
When you add sums of array elements to a PuLP model, you have to use built-in PuLP constructs like lpSum to do it -- you can't just add arrays together (as you discovered).
So your score definition should look something like this:
score = pulp.lpSum([train['target'][i] - (x * array1[i] + y * array2[i] + z * array3[i]) for i in arr_ind])
A few notes about this:
[+] You didn't provide the definition of roc_auc_score so I just pretended that it equals the sum of the element-wise difference between the target array and the weighted sum of the other 3 arrays.
[+] I suspect your actual calculation for roc_auc_score is nonlinear; more on this below.
[+] arr_ind is a list of the indices of the arrays, which I created like this:
# build array index
arr_ind = range(len(array1))
[+] You also didn't include the arrays, so I created them like this:
array1 = np.random.rand(10, 1)
array2 = np.random.rand(10, 1)
array3 = np.random.rand(10, 1)
train = {}
train['target'] = np.ones((10, 1))
Here is my complete code, which compiles and executes, though I'm sure it doesn't give you the result you are hoping for, since I just guessed about target and roc_auc_score:
import numpy as np
import pulp
# create the LP object, set up as a maximization problem
prob = pulp.LpProblem('Giapetto', pulp.LpMaximize)
# dummy arrays since arrays weren't in OP code
array1 = np.random.rand(10, 1)
array2 = np.random.rand(10, 1)
array3 = np.random.rand(10, 1)
# build array index
arr_ind = range(len(array1))
# set up decision variables
x = pulp.LpVariable('x', lowBound=0)
y = pulp.LpVariable('y', lowBound=0)
z = pulp.LpVariable('z', lowBound=0)
# dummy roc_auc_score since roc_auc_score wasn't in OP code
train = {}
train['target'] = np.ones((10, 1))
score = pulp.lpSum([train['target'][i] - (x * array1[i] + y * array2[i] + z * array3[i]) for i in arr_ind])
prob += score
coef = x + y + z
prob += coef == 1
# solve the LP using the default solver
optimization_result = prob.solve()
# make sure we got an optimal solution
assert optimization_result == pulp.LpStatusOptimal
# display the results
for var in (x, y,z):
print('Optimal weekly number of {} to produce: {:1.0f}'.format(var.name, var.value()))
Output:
Optimal weekly number of x to produce: 0
Optimal weekly number of y to produce: 0
Optimal weekly number of z to produce: 1
Process finished with exit code 0
Now, if your roc_auc_score function is nonlinear, you will have additional troubles. I would encourage you to try to formulate the score in a way that is linear, possibly using additional variables (for example, if you want the score to be an absolute value).
I am reading a book, and found an error as below:
def relu(x):
return (x>0)*x
def relu2dev(x):
return (x>0)
street_lights = np.array([[1,0,1],[0,1,1],[0,0,1],[1,1,1]])
walk_stop = np.array([[1,1,0,0]]).T
alpha = 0.2
hidden_size = 4
weights_0_1 = 2*np.random.random((3,hidden_size))-1
weights_1_2 = 2*np.random.random((hidden_size,1))-1
for it in range(60):
layer_2_error = 0;
for i in range(len(street_lights)):
layer_0 = street_lights[i:i+1]
layer_1 = relu(np.dot(layer_0,weights_0_1))
layer_2 = np.dot(layer_1,weights_1_2)
layer_2_delta = (layer_2-walk_stop[i:i+1])
# -> layer_2_delta's shape is (1,1), so why np.sum?
layer_2_error += np.sum((layer_2_delta)**2)
layer_1_delta = layer_2_delta.dot(weights_1_2.T) * relu2dev(layer_1)
weights_1_2 -= alpha * layer_1.T.dot(layer_2_delta)
weights_0_1 -= alpha * layer_0.T.dot(layer_1_delta)
if(it % 10 == 9):
print("Error: " + str(layer_2_error))
The error place is commented with # ->:
layer_2_delta's shape is (1,1), so why would one use np.sum? I think np.sum can be removed, but not quite sure, since it comes from a book.
As you say, layer_2_delta has a shape of (1,1). This means it is a 2 dimensional array with one element: layer_2_delta = np.array([[X]]). However, layer_2_error is a scalar. So you can get the scalar from the array by either selecting the value at the first index (layer_2_delta[0,0]) or by summing all the elements (which in this case is just the one). As the book seems to use "sum of square errors", it seems natural to keep the notation which is square each element in array and then add all of these up (for instruction purposes): this would be more general (e.g., to cases where the layer has more than one element) than the index approach. But you're right, there could be other ways to do this :).
I am using cvxpy to work on some simple portfolio optimisation problem. The only constraint I can't get my head around is the cardinality constraint for the number non-zero portfolio holdings. I tried two approaches, a MIP approach and a traditional convex one.
here is some dummy code for a working traditional example.
import numpy as np
import cvxpy as cvx
np.random.seed(12345)
n = 10
k = 6
mu = np.abs(np.random.randn(n, 1))
Sigma = np.random.randn(n, n)
Sigma = Sigma.T.dot(Sigma)
w = cvx.Variable(n)
ret = mu.T*w
risk = cvx.quad_form(w, Sigma)
objective = cvx.Maximize(ret - risk)
constraints = [cvx.sum_entries(w) == 1, w>= 0, cvx.sum_smallest(w, n-k) >= 0, cvx.sum_largest(w, k) <=1 ]
prob = cvx.Problem(objective, constraints)
prob.solve()
print prob.status
output = []
for i in range(len(w.value)):
output.append(round(w[i].value,2))
print 'Number of non-zero elements : ',sum(1 for i in output if i > 0)
I had the idea to use, sum_smallest and sum_largest (cvxpy manual) my thought was to constraint the smallest n-k entries to 0 and let my target range k sum up to one, I know I can't change the direction of the inequality in order to stay convex, but maybe anyone knows about a clever way of constraining the problem while still keeping it simple.
My second idea was to make this a mixed integer problem, s.th along the lines of
import numpy as np
import cvxpy as cvx
np.random.seed(12345)
n = 10
k = 6
mu = np.abs(np.random.randn(n, 1))
Sigma = np.random.randn(n, n)
Sigma = Sigma.T.dot(Sigma)
w = cvx.Variable(n)
binary = cvx.Bool(n)
integer = cvx.Int(n)
ret = mu.T*w
risk = cvx.quad_form(w, Sigma)
objective = cvx.Maximize(ret - risk)
constraints = [cvx.sum_entries(w) == 1, w>= 0, cvx.sum_entries(binary) == k ]
prob = cvx.Problem(objective, constraints)
prob.solve()
print prob.status
output = []
for i in range(len(w.value)):
output.append(round(w[i].value,2))
print sum(1 for i in output if i > 0)
for i in range(len(w.value)):
print round(binary[i].value,2)
print output
looking at my binary vector it seems to be doing the right thing but the sum_entries constraint doesn't work, looking into the binary vector values I noticed that 0 isn't 0 it's very small e.g xxe^-20 I assume this will mess things up. Anyone can give me any guidance if this is the right way to go? I can use the standard solvers, as well as Mosek if that helps. I would prefer to have a non MIP implementation as I understand this is a combinatorial problem and will get very slow for larger problems. Ultimately I would like to either constraint on exact number of target holdings or a range e.g. 20-30.
Also the documentation in cvxpy around MIP is very short. thanks
A bit chaotic, this question.
So first: this kind of cardinality-constraint is NP-hard. This means, you can't express it using cvxpy without using Integer-programming (or else it would implicate P=NP)!
That beeing said, it would have been nicer, if there would be a pure version of the code without trying to formulate this constraint. I just assume it's the first code without the sum_smallest and sum_largest constraints.
So let's tackle the MIP-approach:
Your code trying to do this makes no sense at all
You introduce some binary-vars, but they have no connection to any other variable at all (so a constraint on it's sum is useless)!
You introduce some integer-vars, but they don't have any use at all!
So here is a MIP-approach:
import numpy as np
import cvxpy as cvx
np.random.seed(12345)
n = 10
k = 6
mu = np.abs(np.random.randn(n, 1))
Sigma = np.random.randn(n, n)
Sigma = Sigma.T.dot(Sigma)
w = cvx.Variable(n)
ret = mu.T*w
risk = cvx.quad_form(w, Sigma)
objective = cvx.Maximize(ret - risk)
binary = cvx.Bool(n) # !!!
constraints = [cvx.sum_entries(w) == 1, w>= 0, w - binary <= 0., cvx.sum_entries(binary) == k] # !!!
prob = cvx.Problem(objective, constraints)
prob.solve(verbose=True)
print(prob.status)
output = []
for i in range(len(w.value)):
output.append(round(w[i].value,2))
print('Number of non-zero elements : ',sum(1 for i in output if i > 0))
So we just added some binary-variables and connected them to w to indicate if w is nonzero or not.
If w is nonzero:
w will be > 0 because of constraint w>= 0
binary needs to be 1, or else constraint w - binary <= 0. is not fulfilled
So it's just introducing these binaries and this one indicator-constraint.
Now the cvx.sum_entries(binary) == k does what it should do.
Be careful with the implication-direction we used here. It might be relevant when chaging the constraint on k (like <=).
Keep in mind, that the default MIP-solver is awful. I also fear that Mosek's interface (sub-optimal within cvxpy) won't solve this, but i might be wrong.
Edit: Your in-range can easily be formulated using two more indicators for:
(k >= a) <= ind_0
(k <= b) <= ind_1
and adding a constraint which equals a logical_and:
ind_0 + ind_1 >= 2
I've had a similar problem where my weights could be negative and did not need to sum to 1 (but still need to be bounded), so I've modified sascha's example to accommodate relaxing these constraints using the CVXpy absolute value function. This should allow for a more general approach to tackling cardinality constraints with MIP
import numpy as np
import cvxpy as cvx
np.random.seed(12345)
n = 10
k = 6
mu = np.abs(np.random.randn(n, 1))
Sigma = np.random.randn(n, n)
Sigma = Sigma.T.dot(Sigma)
w = cvx.Variable(n)
ret = mu.T*w
risk = cvx.quad_form(w, Sigma)
objective = cvx.Maximize(ret - risk)
binary = cvx.Variable(n,boolean=True) # !!!
maxabsw=2
constraints = [ w>= -maxabsw,w<=maxabsw, cvx.abs(w)/maxabsw - binary <= 0., cvx.sum(binary) == k] # !!!
prob = cvx.Problem(objective, constraints)
prob.solve(verbose=True)
print(prob.status)
output = []
for i in range(len(w.value)):
output.append(round(w[i].value,2))
print('Number of non-zero elements : ',sum(1 for i in output if i > 0))
I need the indices (as numpy array) of the rows matching a given condition in a table (with billions of rows) and this is the line I currently use in my code, which works, but is quite ugly:
indices = np.array([row.nrow for row in the_table.where("foo == 42")])
It also takes half a minute, and I'm sure that the list creation is one of the reasons why.
I could not find an elegant solution yet and I'm still struggling with the pytables docs, so does anybody know any magical way to do this more beautifully and maybe also a bit faster? Maybe there is special query keyword I am missing, since I have the feeling that pytables should be able to return the matched rows indices as numpy array.
tables.Table.get_where_list() gives indices of the rows matching a given condition
I read the source of pytables, where() is implemented in Cython, but it seems not fast enough. Here is a complex method that can speedup:
Create some data first:
from tables import *
import numpy as np
class Particle(IsDescription):
name = StringCol(16) # 16-character String
idnumber = Int64Col() # Signed 64-bit integer
ADCcount = UInt16Col() # Unsigned short integer
TDCcount = UInt8Col() # unsigned byte
grid_i = Int32Col() # 32-bit integer
grid_j = Int32Col() # 32-bit integer
pressure = Float32Col() # float (single-precision)
energy = Float64Col() # double (double-precision)
h5file = open_file("tutorial1.h5", mode = "w", title = "Test file")
group = h5file.create_group("/", 'detector', 'Detector information')
table = h5file.create_table(group, 'readout', Particle, "Readout example")
particle = table.row
for i in range(1001000):
particle['name'] = 'Particle: %6d' % (i)
particle['TDCcount'] = i % 256
particle['ADCcount'] = (i * 256) % (1 << 16)
particle['grid_i'] = i
particle['grid_j'] = 10 - i
particle['pressure'] = float(i*i)
particle['energy'] = float(particle['pressure'] ** 4)
particle['idnumber'] = i * (2 ** 34)
# Insert a new particle record
particle.append()
table.flush()
h5file.close()
Read the column in chunks and append the indices into a list and concatenate the list to array finally. You can change the chunk size according to your memory size:
h5file = open_file("tutorial1.h5")
table = h5file.get_node("/detector/readout")
size = 10000
col = "energy"
buf = np.zeros(batch, dtype=table.coldtypes[col])
res = []
for start in range(0, table.nrows, size):
length = min(size, table.nrows - start)
data = table.read(start, start + batch, field=col, out=buf[:length])
tmp = np.where(data > 10000)[0]
tmp += start
res.append(tmp)
res = np.concatenate(res)