I am trying to model assignment problems with the additional constraint that the solution must not contain pairs of anti-parallel arcs, i.e. if in the solution x[i,j]+x[j,i]<=1 must hold for all Binary variables; I have copied an existing solution for assignment and try to add the constraints:
"""
Toy example for testing assignment without anti-parallel arcs
"""
from pyomo.environ import *
from pyomo.opt import *
opt = solvers.SolverFactory("glpk")
# Change to "ipopt" for interior point solver
""" a larger instance that could be activated:
M = ['0', '1', '2','3','4','5','6']
W = ['0', '1', '2','3','4','5','6']
c = {('0','0'):0, ('0','1'):1, ('0','2'):0, ('0','3'):0, ('0','4'):0, ('0','5'):0, ('0','6'):0,
('1','0'):1, ('1','1'):0, ('1','2'):0, ('1','3'):0, ('1','4'):0, ('1','5'):0, ('1','6'):0,
('2','0'):0, ('2','1'):0, ('2','2'):0, ('2','3'):0, ('2','4'):0, ('2','5'):0, ('2','6'):0,
('3','0'):0, ('3','1'):0, ('3','2'):0, ('3','3'):0, ('3','4'):0, ('3','5'):0, ('3','6'):0,
('4','0'):0, ('4','1'):0, ('4','2'):0, ('4','3'):0, ('4','4'):0, ('4','5'):0, ('4','6'):0,
('5','0'):0, ('5','1'):0, ('5','2'):0, ('5','3'):0, ('5','4'):0, ('5','5'):0, ('5','6'):0,
('6','0'):0, ('6','1'):0, ('6','2'):0, ('6','3'):0, ('6','4'):0, ('6','5'):0, ('6','6'):0}
"""
M = ['A', 'B', 'C']
W = ['D', 'E', 'F']
c = {('A','D'):1, ('A','E'):3, ('A','F'):3,
('B','D'):4, ('B','E'):3, ('B','F'):2,
('C','D'):5, ('C','E'):4, ('C','F'):2}
model = ConcreteModel()
model.x = Var(M, W, within=Binary)
model.z = Objective(expr = sum(c[i,j]*model.x[i,j]
for i in M for j in W),
sense=maximize)
def all_m_assigned_rule (model, i):
return sum(model.x[i,j] for j in W) == 1
model.m = Constraint(M, rule=all_m_assigned_rule)
def all_w_assigned_rule (model, j):
return sum(model.x[i,j] for i in M) == 1
model.w = Constraint(W, rule=all_w_assigned_rule)
# additional model constraints that fail:
model.constraints = ConstraintList()
for i in I:
for j in range(i):
model.constraints.add((model.x[i,j]+model.x[j,i]) <= 1)
The essential reported error is
KeyError Traceback (most recent call last)
Input In [12], in <cell line: 48>()
48 for i in I:
49 for j in range(i):
---> 50 model.constraints.add((model.x[i,j]+model.x[j,i]) <= 1)
54 results = opt.solve(model)
56 model.x.get_values()
Can anyone provide an explanation of the failure?
A suggestion that fixes the problem would be great!
model.x is indexed by W and M in your example and not by I (which is not defined) or integers (for j in range(i)). model.x[i,j] and model.x[j,i] don't make sense with how model.x is indexed since W and M have different values.
Related
I searched about this issue, I got more questions "the same error" but different code and different reason. So, I was hesitant more to put my issue here. After reading the majority of answers, I didn't find a solution for my issue.
The original and full code here
chapter6.py:
#!/usr/bin/env python
# -*- coding: utf-8 -*-
import numpy as np
import matplotlib.pyplot as plt
from datasets import gtsrb
from classifiers import MultiClassSVM
def main():
strategies = ['one-vs-one', 'one-vs-all']
features = [None, 'gray', 'rgb', 'hsv', 'surf', 'hog']
accuracy = np.zeros((2, len(features)))
precision = np.zeros((2, len(features)))
recall = np.zeros((2, len(features)))
for f in xrange(len(features)):
print "feature", features[f]
(X_train, y_train), (X_test, y_test) = gtsrb.load_data(
"datasets/gtsrb_training",
feature=features[f],
test_split=0.2,
seed=42)
# convert to numpy
X_train = np.squeeze(np.array(X_train)).astype(np.float32)
y_train = np.array(y_train)
X_test = np.squeeze(np.array(X_test)).astype(np.float32)
y_test = np.array(y_test)
# find all class labels
labels = np.unique(np.hstack((y_train, y_test)))
for s in xrange(len(strategies)):
print " - strategy", strategies[s]
# set up SVMs
MCS = MultiClassSVM(len(labels), strategies[s])
# training phase
print " - train"
MCS.fit(X_train, y_train)
# test phase
print " - test"
acc, prec, rec = MCS.evaluate(X_test, y_test)
accuracy[s, f] = acc
precision[s, f] = np.mean(prec)
recall[s, f] = np.mean(rec)
print " - accuracy: ", acc
print " - mean precision: ", np.mean(prec)
print " - mean recall: ", np.mean(rec)
# plot results as stacked bar plot
f, ax = plt.subplots(2)
for s in xrange(len(strategies)):
x = np.arange(len(features))
ax[s].bar(x - 0.2, accuracy[s, :], width=0.2, color='b',
hatch='/', align='center')
ax[s].bar(x, precision[s, :], width=0.2, color='r', hatch='\\',
align='center')
ax[s].bar(x + 0.2, recall[s, :], width=0.2, color='g', hatch='x',
align='center')
ax[s].axis([-0.5, len(features) + 0.5, 0, 1.5])
ax[s].legend(('Accuracy', 'Precision', 'Recall'), loc=2, ncol=3,
mode='expand')
ax[s].set_xticks(np.arange(len(features)))
ax[s].set_xticklabels(features)
ax[s].set_title(strategies[s])
plt.show()
if __name__ == '__main__':
main()
classifiers.py
import cv2
import numpy as np
from abc import ABCMeta, abstractmethod
from matplotlib import pyplot as plt
__author__ = "Michael Beyeler"
__license__ = "GNU GPL 3.0 or later"
class Classifier:
"""
Abstract base class for all classifiers
A classifier needs to implement at least two methods:
- fit: A method to train the classifier by fitting the model to
the data.
- evaluate: A method to test the classifier by predicting labels of
some test data based on the trained model.
A classifier also needs to specify a classification strategy via
setting self.mode to either "one-vs-all" or "one-vs-one".
The one-vs-all strategy involves training a single classifier per
class, with the samples of that class as positive samples and all
other samples as negatives.
The one-vs-one strategy involves training a single classifier per
class pair, with the samples of the first class as positive samples
and the samples of the second class as negative samples.
This class also provides method to calculate accuracy, precision,
recall, and the confusion matrix.
"""
__metaclass__ = ABCMeta
#abstractmethod
def fit(self, X_train, y_train):
pass
#abstractmethod
def evaluate(self, X_test, y_test, visualize=False):
pass
def _accuracy(self, y_test, Y_vote):
"""Calculates accuracy
This method calculates the accuracy based on a vector of
ground-truth labels (y_test) and a 2D voting matrix (Y_vote) of
size (len(y_test), num_classes).
:param y_test: vector of ground-truth labels
:param Y_vote: 2D voting matrix (rows=samples, cols=class votes)
:returns: accuracy e[0,1]
"""
# predicted classes
y_hat = np.argmax(Y_vote, axis=1)
# all cases where predicted class was correct
mask = y_hat == y_test
return np.float32(np.count_nonzero(mask)) / len(y_test)
def _precision(self, y_test, Y_vote):
"""Calculates precision
This method calculates precision extended to multi-class
classification by help of a confusion matrix.
:param y_test: vector of ground-truth labels
:param Y_vote: 2D voting matrix (rows=samples, cols=class votes)
:returns: precision e[0,1]
"""
# predicted classes
y_hat = np.argmax(Y_vote, axis=1)
if self.mode == "one-vs-one":
# need confusion matrix
conf = self._confusion(y_test, Y_vote)
# consider each class separately
prec = np.zeros(self.num_classes)
for c in xrange(self.num_classes):
# true positives: label is c, classifier predicted c
tp = conf[c, c]
# false positives: label is c, classifier predicted not c
fp = np.sum(conf[:, c]) - conf[c, c]
if tp + fp != 0:
prec[c] = tp * 1. / (tp + fp)
elif self.mode == "one-vs-all":
# consider each class separately
prec = np.zeros(self.num_classes)
for c in xrange(self.num_classes):
# true positives: label is c, classifier predicted c
tp = np.count_nonzero((y_test == c) * (y_hat == c))
# false positives: label is c, classifier predicted not c
fp = np.count_nonzero((y_test == c) * (y_hat != c))
if tp + fp != 0:
prec[c] = tp * 1. / (tp + fp)
return prec
def _recall(self, y_test, Y_vote):
"""Calculates recall
This method calculates recall extended to multi-class
classification by help of a confusion matrix.
:param y_test: vector of ground-truth labels
:param Y_vote: 2D voting matrix (rows=samples, cols=class votes)
:returns: recall e[0,1]
"""
# predicted classes
y_hat = np.argmax(Y_vote, axis=1)
if self.mode == "one-vs-one":
# need confusion matrix
conf = self._confusion(y_test, Y_vote)
# consider each class separately
recall = np.zeros(self.num_classes)
for c in xrange(self.num_classes):
# true positives: label is c, classifier predicted c
tp = conf[c, c]
# false negatives: label is not c, classifier predicted c
fn = np.sum(conf[c, :]) - conf[c, c]
if tp + fn != 0:
recall[c] = tp * 1. / (tp + fn)
elif self.mode == "one-vs-all":
# consider each class separately
recall = np.zeros(self.num_classes)
for c in xrange(self.num_classes):
# true positives: label is c, classifier predicted c
tp = np.count_nonzero((y_test == c) * (y_hat == c))
# false negatives: label is not c, classifier predicted c
fn = np.count_nonzero((y_test != c) * (y_hat == c))
if tp + fn != 0:
recall[c] = tp * 1. / (tp + fn)
return recall
def _confusion(self, y_test, Y_vote):
"""Calculates confusion matrix
This method calculates the confusion matrix based on a vector of
ground-truth labels (y-test) and a 2D voting matrix (Y_vote) of
size (len(y_test), num_classes).
Matrix element conf[r,c] will contain the number of samples that
were predicted to have label r but have ground-truth label c.
:param y_test: vector of ground-truth labels
:param Y_vote: 2D voting matrix (rows=samples, cols=class votes)
:returns: confusion matrix
"""
y_hat = np.argmax(Y_vote, axis=1)
conf = np.zeros((self.num_classes, self.num_classes)).astype(np.int32)
for c_true in xrange(self.num_classes):
# looking at all samples of a given class, c_true
# how many were classified as c_true? how many as others?
for c_pred in xrange(self.num_classes):
y_this = np.where((y_test == c_true) * (y_hat == c_pred))
conf[c_pred, c_true] = np.count_nonzero(y_this)
return conf
class MultiClassSVM(Classifier):
"""
Multi-class classification using Support Vector Machines (SVMs)
This class implements an SVM for multi-class classification. Whereas
some classifiers naturally permit the use of more than two classes
(such as neural networks), SVMs are binary in nature.
However, we can turn SVMs into multinomial classifiers using at least
two different strategies:
* one-vs-all: A single classifier is trained per class, with the
samples of that class as positives (label 1) and all
others as negatives (label 0).
* one-vs-one: For k classes, k*(k-1)/2 classifiers are trained for each
pair of classes, with the samples of the one class as
positives (label 1) and samples of the other class as
negatives (label 0).
Each classifier then votes for a particular class label, and the final
decision (classification) is based on a majority vote.
"""
def __init__(self, num_classes, mode="one-vs-all", params=None):
"""
The constructor makes sure the correct number of classifiers is
initialized, depending on the mode ("one-vs-all" or "one-vs-one").
:param num_classes: The number of classes in the data.
:param mode: Which classification mode to use.
"one-vs-all": single classifier per class
"one-vs-one": single classifier per class pair
Default: "one-vs-all"
:param params: SVM training parameters.
For now, default values are used for all SVMs.
Hyperparameter exploration can be achieved by
embedding the MultiClassSVM process flow in a
for-loop that classifies the data with
different parameter values, then pick the
values that yield the best accuracy.
Default: None
"""
self.num_classes = num_classes
self.mode = mode
self.params = params or dict()
# initialize correct number of classifiers
self.classifiers = []
if mode == "one-vs-one":
# k classes: need k*(k-1)/2 classifiers
for _ in xrange(num_classes*(num_classes - 1) / 2):
self.classifiers.append(cv2.ml.SVM_create())
elif mode == "one-vs-all":
# k classes: need k classifiers
for _ in xrange(num_classes):
self.classifiers.append(cv2.ml.SVM_create())
else:
print "Unknown mode ", mode
def fit(self, X_train, y_train, params=None):
"""Fits the model to training data
This method trains the classifier on data (X_train) using either
the "one-vs-one" or "one-vs-all" strategy.
:param X_train: input data (rows=samples, cols=features)
:param y_train: vector of class labels
:param params: dict to specify training options for cv2.SVM.train
leave blank to use the parameters passed to the
constructor
"""
if params is None:
params = self.params
if self.mode == "one-vs-one":
svm_id = 0
for c1 in xrange(self.num_classes):
for c2 in xrange(c1 + 1, self.num_classes):
# indices where class labels are either `c1` or `c2`
data_id = np.where((y_train == c1) + (y_train == c2))[0]
# set class label to 1 where class is `c1`, else 0
y_train_bin = np.where(y_train[data_id] == c1, 1,
0).flatten()
self.classifiers[svm_id].train(X_train[data_id, :],
y_train_bin,
params=self.params)
svm_id += 1
elif self.mode == "one-vs-all":
for c in xrange(self.num_classes):
# train c-th SVM on class c vs. all other classes
# set class label to 1 where class==c, else 0
y_train_bin = np.where(y_train == c, 1, 0).flatten()
# train SVM
self.classifiers[c].train(X_train, y_train_bin,
params=self.params)
def evaluate(self, X_test, y_test, visualize=False):
"""Evaluates the model on test data
This method evaluates the classifier's performance on test data
(X_test) using either the "one-vs-one" or "one-vs-all" strategy.
:param X_test: input data (rows=samples, cols=features)
:param y_test: vector of class labels
:param visualize: flag whether to plot the results (True) or not
(False)
:returns: accuracy, precision, recall
"""
# prepare Y_vote: for each sample, count how many times we voted
# for each class
Y_vote = np.zeros((len(y_test), self.num_classes))
if self.mode == "one-vs-one":
svm_id = 0
for c1 in xrange(self.num_classes):
for c2 in xrange(c1 + 1, self.num_classes):
data_id = np.where((y_test == c1) + (y_test == c2))[0]
X_test_id = X_test[data_id, :]
y_test_id = y_test[data_id]
# set class label to 1 where class==c1, else 0
# y_test_bin = np.where(y_test_id==c1,1,0).reshape(-1,1)
# predict labels
y_hat = self.classifiers[svm_id].predict_all(X_test_id)
for i in xrange(len(y_hat)):
if y_hat[i] == 1:
Y_vote[data_id[i], c1] += 1
elif y_hat[i] == 0:
Y_vote[data_id[i], c2] += 1
else:
print "y_hat[", i, "] = ", y_hat[i]
# we vote for c1 where y_hat is 1, and for c2 where y_hat
# is 0 np.where serves as the inner index into the data_id
# array, which in turn serves as index into the results
# array
# Y_vote[data_id[np.where(y_hat == 1)[0]], c1] += 1
# Y_vote[data_id[np.where(y_hat == 0)[0]], c2] += 1
svm_id += 1
elif self.mode == "one-vs-all":
for c in xrange(self.num_classes):
# set class label to 1 where class==c, else 0
# predict class labels
# y_test_bin = np.where(y_test==c,1,0).reshape(-1,1)
# predict labels
y_hat = self.classifiers[c].predict_all(X_test)
# we vote for c where y_hat is 1
if np.any(y_hat):
Y_vote[np.where(y_hat == 1)[0], c] += 1
# with this voting scheme it's possible to end up with samples
# that have no label at all...in this case, pick a class at
# random...
no_label = np.where(np.sum(Y_vote, axis=1) == 0)[0]
Y_vote[no_label, np.random.randint(self.num_classes,
size=len(no_label))] = 1
accuracy = self._accuracy(y_test, Y_vote)
precision = self._precision(y_test, Y_vote)
recall = self._recall(y_test, Y_vote)
return accuracy, precision, recall
when running chapter6.py
The output is:
feature None
- strategy one-vs-one
- train
Traceback (most recent call last):
File "/home/redhwan/Downloads/opencv-python-blueprints-master/chapter6/chapter6.py", line 77, in <module>
main()
File "/home/redhwan/Downloads/opencv-python-blueprints-master/chapter6/chapter6.py", line 44, in main
MCS.fit(X_train, y_train)
File "/home/redhwan/Downloads/opencv-python-blueprints-master/chapter6/classifiers.py", line 258, in fit
params=self.params)
TypeError: only length-1 arrays can be converted to Python scalars
please help me or your suggestion
Thank you in advance!
I wanna get all integer solutions in a limited time, is it possible?
This is a linear, integer constraint satisfaction problem, which can be solved efficiently by OR Tools' CP-SAT. I've modified their example to solve your problem in Python:
from ortools.sat.python import cp_model
class VarArraySolutionPrinter(cp_model.CpSolverSolutionCallback):
"""Print intermediate solutions."""
def __init__(self, variables):
cp_model.CpSolverSolutionCallback.__init__(self)
self.__variables = variables
self.__solution_count = 0
def on_solution_callback(self):
self.__solution_count += 1
for v in self.__variables:
print('%s=%i' % (v, self.Value(v)), end=' ')
print()
def solution_count(self):
return self.__solution_count
def SearchForAllSolutionsSampleSat():
"""Showcases calling the solver to search for all solutions."""
# Creates the model.
model = cp_model.CpModel()
p = [1, 2, 3, 4]
ceq = 30
cgeq = 2
N = len(p)
# Creates the variables
x = [model.NewIntVar(0, 100, f'x{i}') for i in range(N)]
# Create the constraints.
model.Add(sum([xi*pi for xi, pi in zip(x, p)]) == ceq)
model.Add(sum(x) >= cgeq)
# Create a solver and solve.
solver = cp_model.CpSolver()
solution_printer = VarArraySolutionPrinter(x)
status = solver.SearchForAllSolutions(model, solution_printer)
print('Status = %s' % solver.StatusName(status))
print('Number of solutions found: %i' % solution_printer.solution_count())
SearchForAllSolutionsSampleSat()
I'm experimenting with sympy to reproduce an example where a box has three marbles:
Red
White
Blue
Two marbles will be drawn at random without replacement.
Q: What is the chance of drawing the Red marble and then the White marble?
I have been able to calculate this using the multiplication rule by hard-coding P() instances wrapping the initial distribution before the first marble is selected and then the distribution before the second marble is selected:
from sympy.stats import DiscreteUniform, density, P
from sympy import symbols, Eq
# Coloured marbles
R, W, B = symbols('R W B')
# Select first marble without replacement
PFirstSelection = P(Eq(DiscreteUniform('FirstSeletion', (R, W, B)), R))
# Select second marble - Red is not longer available because it was selected without replacement
PSecondSelection = P(Eq(DiscreteUniform('SecondSelection', (W, B) ), W))
print(PFirstSelection)
# 1/3
print(PSecondSelection)
# 1/2
# Multiplication rule
print(PFirstSelection * PSecondSelection)
# 1/6
Is there a better way that I can achieve this with sympy?
In this case you'd better to use combination functions.
DiscreteUniform seems not for changing elements after creation.
from sympy.functions.combinatorial.numbers import nC, nP
print(1 / nP(3, 2)) # 1/6
If you don't care about order,
print(nP(2, 2) / nP(3, 2)) # 1/3
Edited. (and also modified for python3)
For N of M things, you can simply do like below
from sympy.functions.combinatorial.numbers import nC, nP
def pickProb(candidates, picks, ordered=False):
picks_num = len(picks)
numerator = nP(picks_num, picks_num) if ordered else 1
denominator = nP(len(candidates), picks_num)
return numerator / denominator
print(pickProb('RWB', 'RW')) # 1/6
print(pickProb('RWBrwba', 'Ra')) # 1/42
print(pickProb('RWBrwba', 'RWa')) # 1/210
print(pickProb('RWBrwba', 'RWa', ordered=True)) # 1/35
And combination functions can also handle duplicates, like 'R', 'R', 'W', 'B'.
from operator import mul
from sympy.functions.combinatorial.numbers import nC, nP
def pickProb(candidates, picks):
picks_num = len(picks)
c_counts = {}
for c in candidates:
c_counts[c] = c_counts[c] + 1 if c in c_counts else 1
p_counts = {}
for p in picks:
p_counts[p] = p_counts[p] + 1 if p in p_counts else 1
combinations = reduce(mul, [nP(c_counts[x], p_counts[x]) for x in p_counts.keys()], 1)
denominator = nP(len(candidates), picks_num) / combinations
return 1 / denominator
print(pickProb('RWBra', 'RWa')) # 1/60
print(pickProb('RRRWa', 'RWa')) # 1/20
print(pickProb('RRRWa', 'RRa')) # 1/10
But DiscreteUniform cannot, because this case is not "uniform".
from sympy.stats import DiscreteUniform, density, P, Hypergeometric
from sympy import Symbol, Eq
deck = DiscreteUniform('M', 'RRWB')
print(density(deck).dict) # {W: 1/4, R: 1/4, B: 1/4}
print(P(Eq(deck, Symbol('R')))) # 1/4
I think you're using correctly sympy, but you can improve your way to use python (eg., more generic, more functional, more generic, no hardcoding).
For instance:
from sympy.stats import DiscreteUniform, density, P
from sympy import symbols, Eq
from itertools import accumulate
def ToSet(value):
return set(value.split(' '))
def ProbaOfPick(pickSet, fromSet, operationTag):
return P(Eq(DiscreteUniform(operationTag, symbols(fromSet)), symbols(pickSet)))
def PickWithoutReplacement(allset, picklist, probaFunc):
currentSet = allset
probaSeq = []
operationSeq = []
for pick in picklist:
operationTag = "picking: " + pick
newP = probaFunc(pick, currentSet, operationTag)
operationSeq.append(operationTag + " from " + str(currentSet))
probaSeq.append(newP)
currentSet -= set(pick)
return (operationSeq, probaSeq)
allset = ToSet('R W B Y Ma G1 G2')
picks = 'R', 'W', 'G2'
operationSeq, probaSeq = PickWithoutReplacement(allset, picks, ProbaOfPick)
probas = list(accumulate(probaSeq, lambda a, b: a*b))
for op in operationSeq:
print(op)
print(probas)
Also your can change uniform distribution to anything non-uniform.
EDIT: dependency injection (ProbaOfPick -> probaFunc) added.
This code is only a starter.
Result:
picking: R from {'G2', 'Ma', 'Y', 'B', 'R', 'G1', 'W'}
picking: W from {'G2', 'Ma', 'Y', 'B', 'G1', 'W'}
picking: G2 from {'G2', 'Ma', 'Y', 'B', 'G1'}
[1/7, 1/42, 1/210]
Next steps: allow to pick more than 1 each step, allow non uniform probability distribution, etc
I have a script as follows:
import numpy as np
import pandas as pd
import pdb
# conventions: W = fitness, A = affinity ; sex: 1=M, 0=F; alien: 1=alien,
# 0=native
# pop array order: W, A, sex, alien
def mkpop(n):
W = np.repeat(a=1, repeats=n)
A = np.random.normal(1, 0.1, size=n)
A[A < 0] = 0
alien = np.repeat(a=False, repeats=n)
sex = np.random.randint(0, 2, n)
pop = np.array([W, A, sex, alien])
pop = np.transpose(pop)
return pop
def migrate(pop, n=10, gParams=[1, 0.1]):
W = np.random.gamma(shape=gParams[0], scale=gParams[1], size=n)
A = np.repeat(1, n)
# 0 is native; 1 is alien
alien = np.repeat(True, n)
# 0 is female
sex = np.random.randint(0, 2, n)
popAlien = np.array([W, A, sex, alien])
popAlien = np.transpose(popAlien)
pop = np.vstack((pop, popAlien))
return pop
def mate(pop):
# split into male and female
f = pop[pop[:, 2] == 0]
m = pop[pop[:, 2] == 1]
# create transition matricies for native and alien mates
# m with native = m.!alien.transpose * f.alien
# negate alien
naLog = list(np.asarray(m[:, 3]) == False)
naPdMat = np.outer(naLog, f[:, 1])
# mate with alien = m.alien.transpose * affinity
alPdMat = np.outer(m[:, 3], f[:, 1])
# add transition matrices for probability density matrix
pdMat = alPdMat + naPdMat
# transition matrix is equal to the pd matrix / column sumso
colSums = np.sum(pdMat, axis=0)
pMat = pdMat / colSums
# select mates
def choice(x):
ch = np.random.choice(a=range(0, len(x)), p=x)
return ch
mCh = np.apply_along_axis(choice, 0, pMat)
mCh = m[mCh, :]
WMid = (f[:, 0] + mCh[:, 0]) / 2
AMid = (f[:, 1] + mCh[:, 1]) / 2
# assign fitness based on group affiliation; only native/alien matings have
# modified fitness
# reassign fitness and affinity based on group id and midparent vals
W1 = np.where(
(f[:, 3] == mCh[:, 3]) |
((f[:, 3] == 1) & (mCh[:, 3] == 0))
)
WMid[W1] = 1
# number of offspring is a poisson-distributed variable with lambda=2W
nOff = map(lambda x: np.random.poisson(lam=x), 2 * WMid)
# generate offspring
# expand list of nOff to numbers of offspring per pair
# realized offspring is index posisions of W and A vals to be replicated
# for offspring
# this can be rewritten to return a matrix of the appropriate length. This
# should work
midVals = np.array([WMid, AMid]).T
realOff = np.array([0, 0])
for i in range(0, len(nOff)):
sibs = np.repeat([np.array(midVals[i])], [nOff[i]], axis=0)
realOff = np.vstack((realOff, sibs))
offspring = np.delete(realOff, 0, 0)
sex = np.random.randint(0, 2, len(offspring))
alien = np.repeat(0, len(offspring))
otherStats = np.array([sex, alien]).T
offspring = np.hstack([offspring, otherStats])
return offspring # should return offspring
def sim(nInit, nGen=100, nAlien=10, gParams=[1, 0.1]):
gen = 0
pop = mkpop
stats = pd.DataFrame(columns=('gen', 'W', 'WMean', 'AMean', 'WVar', 'AVar'))
while gen < nGen:
pop = migrate(pop, nAlien, gParams)
offspring = mate(pop)
var = np.var(offspring, axis=0)
mean = np.mean(offspring, axis=0)
N = len(offspring)
W = N / nInit
genStats = N.append(W, gen, mean, var)
stats = stats.append(genStats)
print(N, gen)
gen = gen + 1
return stats
print mkpop(100)
print mate(mkpop(100))
#
sim(100, 100, 10, [1, 0.1])
Running this script, outputs NameError: name 'sim' is not defined. It is apparent from the commands before the final one that all the other functions defined within this script work without a hitch. I'm not sure what is going on here, and there is probably some very easy fix that I'm overlooking. Ctags recognizes this function just fine. It's entirely possibe that sim() doesn't actually work yet, as I haven't been able to debug it.
Your sim function defined in mate function scope so it's invisible to global scope. You need to fix your indentation for sim function
I'm new to Pymc3 and I'm trying to create the Categorical Mixture Model shown in https://en.wikipedia.org/wiki/Mixture_model#Categorical_mixture_model . I'm having difficulty hooking up the 'x' variable. I think it's because I have to make the z variable Deterministic, but I'm getting an error message at the line where 'x' is assigned : "ValueError: We expected 3 inputs but got 2.". It looks like the p function only accepts 2 inputs so I'm stuck. I've tried a bunch of different things, but haven't been able to get this to work yet.
import numpy as np
from pymc3 import *
import theano.tensor as t
K = 3 #NUMBER OF TOPICS
V = 20 #NUMBER OF WORDS
N = 15 #NUMBER OF DOCUMENTS
#GENERAETE RANDOM CATEGORICAL MIXTURES
data = np.ones([N,V])
#theano.compile.ops.as_op(itypes=[t.lscalar, t.dscalar, t.dscalar],otypes=[t.dvector])
def p(z=z, phi=phi):
return [phi[z[i,j]] for i in range(D) for j in range(W)]
model = Model()
with model:
alpha = np.ones(V)
beta = np.ones(K)
theta = [Dirichlet('theta_%i' % i, alpha, shape=V) for i in range(K)]
phi = Dirichlet('phi', beta, shape=K)
z = [Categorical('z_%i' % i, p = phi, shape=V) for i in range(N)]
x = [Categorical('x_%i_%i' % (i,j), p=p(z[i][j],phi), observed=data[i,j]) for i in range(N) for j in range(V)]
#x = [Categorical('x_%i_%i' % (i,j), p=theta[z[i][j]], observed=data[i,j]) for i in range(N) for j in range(V)]
print "Created model. Now begin sampling"
step = Slice()
trace = sample(n, step)
trace.get_values('phi')
For starters, in your example above, z and phi have no value which would allow them to be used as default values. We also don't have values for D and W.
As for the number of arguments, the function you define has 2 but your theano decorator above it has 3. I'd suggest
#theano.compile.ops.as_op(itypes=[t.lscalar, t.dvector],otypes=[t.dvector])
def p(z, phi):
return [phi[z[i,j]] for i,j in zip(range(D),range(W))]