I am working on a life simulation game based on the game of life in web2py. I have two objects which contain several arrays for the various attributes of the organisms in each space on the grid. I received this error LifeSpace instance has no attribute 'getitem' on the line containing this code nxt[i][j] = processNeighborsEmpty (i, j, cur)
am I referencing the object instance improperly?
ROWS = 60
COLS = 60
GENERATIONS = 100
DELAY = 0.5
#---------------------------------------------------------------------------
class LifeSpace :
def __init__(self):
self.genus = []
self.species = []
self.tribe = []
self.hunger = []
self.wounds = []
self.age = []
self.genus = initGrid (COLS, ROWS, self.genus)
self.species = initGrid (COLS, ROWS, self.species)
self.tribe = initGrid (COLS, ROWS, self.tribe)
self.hunger = initGrid(COLS, ROWS, self.hunger)
self.wounds = initGrid(COLS, ROWS, self.wounds)
self.age = initGrid(COLS, ROWS, self.age)
#---------------------------------------------------------------------------
def initGrid(cols, rows, array):
for i in range(rows):
arrayRow = []
for j in range(cols):
arrayRow+= [0]
array += [arrayRow]
return array
#---------------------------------------------------------------------------
def printGen(cols, rows, array, genNo):
return("Game of Life -- Generation " + str(genNo + 1))
for i in range(rows):
for j in range(cols):
if array.species[i][j] == 0:
return(" ")
elif array.species[i][j] == 'weed':
return("3")
return("\n")
#---------------------------------------------------------------------------
def processNeighborsEmpty(x, y, array):
for j in range(y-1,y+1):
for i in range(x-1,x+1):
if not(i == x and j == y):
if array.genus[i][j] == "plant" :
if array.age[i][j] == "adult" :
array.species[x][y] = array.species[i][j]
array.genus[x][y] = array.genus[i][j]
array.age[x][y] = "seed"
return array[x][y]
#---------------------------------------------------------------------------
def processNeighborsPlant(x, y, array):
if array.age[x][y] != "adult" :
if array.species[x][y] == "tree" :
if array.age[x][y] == "sapling" :
array.age[x][y] = "adult"
elif array.age[x][y] == "seed" :
array.age[x][y] = "sapling"
elif array.age[x][y] == "seed" :
array.age[x][y] = "adult"
#---------------------------------------------------------------------------
def processNextGen(cols, rows, cur, nxt):
for i in range(0,rows):
for j in range(0,cols):
if cur.genus[i][j] == 0 :
nxt[i][j] = processNeighborsEmpty (i, j, cur)
elif cur.genus[i][j] == "plant" :
nxt[i][j] = processNeighborsPlant(i, j, cur)
#---------------------------------------------------------------------------
############################################################################
#---------------------------------------------------------------------------
def index():
thisGen = LifeSpace()
nextGen = LifeSpace()
thisGen.genus[25][25] = "plant"
thisGen.species[25][25] = "weed"
thisGen.age[25][25] = "adult"
for gens in range(GENERATIONS):
printGen(COLS, ROWS, thisGen, gens)
processNextGen(COLS, ROWS, thisGen, nextGen)
time.sleep(DELAY)
thisGen, nextGen = nextGen, thisGen
input("Finished. Press <return> to quit.")
nxt is a LifeSpace object so you cannot reference to it using [] notation.
You need to use nxt.attribute[x][y] with the correct attribute. Maybe nxt.genux[x][y] is what you want to use.
Related
I tried to ask a question regarding nathandrake's #nathandrake post: How do I calculate a word-word co-occurrence matrix with sklearn?
import pandas as pd
def co_occurance_matrix(input_text,top_words,window_size):
co_occur = pd.DataFrame(index=top_words, columns=top_words)
for row,nrow in zip(top_words,range(len(top_words))):
for colm,ncolm in zip(top_words,range(len(top_words))):
count = 0
if row == colm:
co_occur.iloc[nrow,ncolm] = count
else:
for single_essay in input_text:
essay_split = single_essay.split(" ")
max_len = len(essay_split)
top_word_index = [index for index, split in enumerate(essay_split) if row in split]
for index in top_word_index:
if index == 0:
count = count + essay_split[:window_size + 1].count(colm)
elif index == (max_len -1):
count = count + essay_split[-(window_size + 1):].count(colm)
else:
count = count + essay_split[index + 1 : (index + window_size + 1)].count(colm)
if index < window_size:
count = count + essay_split[: index].count(colm)
else:
count = count + essay_split[(index - window_size): index].count(colm)
co_occur.iloc[nrow,ncolm] = count
return co_occur
My question is: what if my words are not one word but bigrams. For example:
corpus = ['ABC DEF IJK PQR','PQR KLM OPQ','LMN PQR XYZ DEF ABC']
words = ['ABC PQR','PQR DEF']
window_size =100
result = co_occurance_matrix(corpus,words,window_size)
result
I changed the word list into a bigram list, then the co_occurance_matrix function is not working. All are showing 0.
I have two ordered sequences, one (large) is range of positions, one (small) is a sequence of attributes, defined on position_from/position_two which I'd like to join.
So for each element of positions, I need to traverse the other sequences, which is not optimal
def interpolateCurveOnPos(position:Seq[Double], curveAttributes:Seq[CurveSegment]) = {
position.map { pos =>
// range join
val cs = curveAttributes.find(c => pos >= c.position_von && pos < c.position_bis).get
// interpolate curve attribute
val curve = cs.curve_von + (pos - cs.position_von) * (cs.curve_bis - cs.curve_von) / (cs.position_bis - cs.position_von)
return curve
}
What I've tried:
As the index at which the matching curveSegement is found will allways increase, I've introduced a some state variables to reduce the search of the correct entry
def interpolateCurveOnPos(position:Seq[Double], curveAttributes:Seq[CurveSegment]) = {
var idxSave = 0
var csSave : CurveSegment = curveAttributes.head
position.map { pos =>
// range join
val cs = curveAttributes.drop(idxSave).find(c => pos >= c.position_von && pos < c.position_bis).get
if(cs != csSave) {
csSave = cs
idxSave=idxSave+1
}
// interpolate
val curve = cs.curve_von + (pos - cs.position_von) * (cs.curve_bis - cs.curve_von) / (cs.position_bis - cs.position_von)
return curve
}
I wonder if there is a more elegent way to do it?
Below is my implementation of a2c using PyTorch. Upon learning about backpropagation in PyTorch, I have known to zero_grad() the optimizer after each update iteration. However, there is still a RunTime error on second-time backpropagation.
def torchworker(number, model):
worker_env = gym.make("Taxi-v3").env
max_steps_per_episode = 2000
worker_opt = optim.Adam(lr=5e-4, params=model.parameters())
p_history = []
val_history = []
r_history = []
running_reward = 0
episode_count = 0
under = 0
start = time.time()
for i in range(2):
state = worker_env.reset()
episode_reward = 0
penalties = 0
drop = 0
print("Episode {} begins ({})".format(episode_count, number))
worker_env.render()
criterion = nn.SmoothL1Loss()
time_solve = 0
for _ in range(1, max_steps_per_episode):
#worker_env.render()
state = torch.tensor(state, dtype=torch.long)
action_probs = model.forward(state)[0]
critic_value = model.forward(state)[1]
val_history.append((state, critic_value[0]))
# Choose action
action = np.random.choice(6, p=action_probs.detach().numpy())
p_history.append(torch.log(action_probs[action]))
# Apply chosen action
state, reward, done, _ = worker_env.step(action)
r_history.append(reward)
episode_reward += reward
time_solve += 1
if reward == -10:
penalties += 1
elif reward == 20:
drop += 1
if done:
break
# Update running reward to check condition for solving
running_reward = (running_reward * (episode_count) + episode_reward) / (episode_count + 1)
# Calculate discounted returns
returns = deque(maxlen=3500)
discounted_sum = 0
for r in r_history[::-1]:
discounted_sum = r + gamma * discounted_sum
returns.appendleft(discounted_sum)
# Calculate actor losses and critic losses
loss_actor_value = 0
loss_critic_value = 0
history = zip(p_history, val_history, returns)
for log_prob, value, ret in history:
diff = ret - value[1]
loss_actor_value += -log_prob * diff
ret_tensor = torch.tensor(ret, dtype=torch.float32)
loss_critic_value += criterion(value[1], ret_tensor)
loss = loss_actor_value + 0.1 * loss_critic_value
print(loss)
# Update params
loss.backward()
worker_opt.step()
worker_opt.zero_grad()
# Log details
end = time.time()
episode_count += 1
if episode_count % 1 == 0:
worker_env.render()
if running_reward > -50: # Condition to consider the task solved
under += 1
if under > 5:
print("Solved at episode {} !".format(episode_count))
break
I believe there may be something to do with the architecture of my AC model, so I also include it here for reference.
class ActorCriticNetwork(nn.Module):
def __init__(self, num_inputs, num_hidden, num_actions):
super(ActorCriticNetwork, self).__init__()
self.embed = nn.Embedding(500, 10)
self.fc1 = nn.Linear(10, num_hidden * 2)
self.fc2 = nn.Linear(num_hidden * 2, num_hidden)
self.c = nn.Linear(num_hidden, 1)
self.fc3 = nn.Linear(num_hidden, num_hidden)
self.a = nn.Linear(num_hidden, num_actions)
def forward(self, x):
out = F.relu(self.embed(x))
out = F.relu(self.fc1(out))
out = F.relu(self.fc2(out))
critic = self.c(out)
out = F.relu(self.fc3(out.detach()))
actor = F.softmax(self.a(out), dim=-1)
return actor, critic
Would you please tell me what the mistake here is? Thank you in advance.
SOLVED: I forgot to clear the history of probabilities, action-values and rewards after iterations. It is clear why that would cause the issue, as the older elements would cause propagating through old dcgs.
Here is my code.
import UIKit
var str = "Hello, playground"
//There are two sorted arrays nums1 and nums2 of size m and n respectively.
//Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).
//Example 1:
//nums1 = [1, 3]
//nums2 = [2]
//
//The median is 2.0
//Example 2:
//nums1 = [1, 2]
//nums2 = [3, 4]
//
//The median is (2 + 3)/2 = 2.5
var num1 = [1,2,2,5]
var num2 = [2,3,9,9]
class Solution {
func findMedianSortedArrays(_ nums1: [Int], _ nums2: [Int]) -> Double {
var A = nums1
var B = nums2
var m = nums1.count
var n = nums2.count
var max_of_left : Int = 0
var min_of_right = 0
if n < m {
var temp : [Int]
var tempt : Int
temp = nums1
tempt = m
A = nums2
B = temp
m = n
n = tempt
}
if n == 0{
fatalError("Arrays must be fulfilled")
}
var imin = 0
var imax = m
let half_len = Int((m+n+1)/2)
while imin <= imax {
let i = Int((imin + imax) / 2)
let j = half_len - i
if i > 0 && A[i-1] > B[j]{
imax = i - 1
}
else if i < m && A[i] < B[j-1]{
imin = i + 1
}
else
{
if i == 0{
max_of_left = B[j-1]
}
else if j == 0{
max_of_left = A[i-1]
}
else
{
max_of_left = max(A[i-1], B[j-1])
}
if m+n % 2 == 1{
return Double(max_of_left)
}
if i==m{
min_of_right = B[j]
}
else if j == n{
min_of_right = A[i]
}
else{
min_of_right = min(A[i], B[j])
//editor indicates error here
}
return Double((Double(max_of_left+min_of_right) / 2.0))
}
}
}
}
var a = Solution()
print(a.findMedianSortedArrays(num1, num2))
error: day4_Median_of_Two_Sorted_Arrays.playground:86:13: error: missing return in a function expected to return 'Double'
Since I put my return out of if statement, I think it will be okay because it will stop while looping when it meets return.
But editor says it's not.
I want to know why. Please explain me why.
Every code path through your findMedianSortedArrays() must return a Double.
So you need a return of a Double placed outside of your while loop. Even if you had every code path within the while loop have a return double, if imin > imax you wouldn't even enter the while loop, and so would need a return of a double outside it.
I fixed it by putting another return out of while loop.
//: Playground - noun: a place where people can play
import UIKit
var str = "Hello, playground"
//There are two sorted arrays nums1 and nums2 of size m and n respectively.
//Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).
//Example 1:
//nums1 = [1, 3]
//nums2 = [2]
//
//The median is 2.0
//Example 2:
//nums1 = [1, 2]
//nums2 = [3, 4]
//
//The median is (2 + 3)/2 = 2.5
var num1 = [1,2,2,5]
var num2 = [2,3,9,9]
class Solution {
func findMedianSortedArrays(_ nums1: [Int], _ nums2: [Int]) -> Double {
var A = nums1
var B = nums2
var m = nums1.count
var n = nums2.count
var max_of_left : Int = 0
var min_of_right = 0
if n < m {
var temp : [Int]
var tempt : Int
temp = nums1
tempt = m
A = nums2
B = temp
m = n
n = tempt
}
if n == 0{
fatalError("Arrays must be fulfilled")
}
var imin = 0
var imax = m
let half_len = Int((m+n+1)/2)
while imin <= imax {
let i = Int((imin + imax) / 2)
let j = half_len - i
if i > 0 && A[i-1] > B[j]{
imax = i - 1
}
else if i < m && A[i] < B[j-1]{
imin = i + 1
}
else
{
if i == 0{
max_of_left = B[j-1]
}
else if j == 0{
max_of_left = A[i-1]
}
else
{
max_of_left = max(A[i-1], B[j-1])
}
if m+n % 2 == 1{
return Double(max_of_left)
}
if i==m{
min_of_right = B[j]
}
else if j == n{
min_of_right = A[i]
}
else{
min_of_right = min(A[i], B[j])
}
return Double((Double(max_of_left+min_of_right) / 2.0))
}
}
return Double((Double(max_of_left+min_of_right) / 2.0))
}
}
var a = Solution()
print(a.findMedianSortedArrays(num1, num2))
I'm trying to solve a car problem.
first, I have an original code of car problem:
# Ampl Car Example
#
# Shows how to convert a minimize final time optimal control problem
# to a format pyomo.dae can handle by removing the time scaling from
# the ContinuousSet.
#
# min tf
# dxdt = 0
# dvdt = a-R*v^2
# x(0)=0; x(tf)=L
# v(0)=0; v(tf)=0
# -3<=a<=1
from pyomo.environ import *
from pyomo.dae import *
m = ConcreteModel()
m.R = Param(initialize=0.001) # Friction factor
m.L = Param(initialize=100.0) # Final position
m.tau = ContinuousSet(initialize=[0.0, 0.80, 1.0]) # Unscaled time
m.time = Var(m.tau) # Scaled time
m.tf = Var()
m.x = Var(m.tau,bounds=(0,None))
m.v = Var(m.tau,bounds=(0,None))
m.a = Var(m.tau, bounds=(-3.0,1.0),initialize=0)
m.dtime = DerivativeVar(m.time)
m.dx = DerivativeVar(m.x)
m.dv = DerivativeVar(m.v)
m.obj = Objective(expr=m.tf)
def _ode1(m,i):
if i == 0 :
return Constraint.Skip
return m.dx[i] == m.tf * m.v[i]
m.ode1 = Constraint(m.tau, rule=_ode1)
def _ode2(m,i):
if i == 0 :
return Constraint.Skip
return m.dv[i] == m.tf*(m.a[i] - m.R*m.v[i]**2)
m.ode2 = Constraint(m.tau, rule=_ode2)
def _ode3(m,i):
if i == 0:
return Constraint.Skip
return m.dtime[i] == m.tf
m.ode3 = Constraint(m.tau, rule=_ode3)
def _init(m):
yield m.x[0] == 0
yield m.x[1] == m.L
yield m.v[0] == 0
yield m.v[1] == 0
yield m.time[0] == 0
m.initcon = ConstraintList(rule=_init)
discretizer = TransformationFactory('dae.collocation')
discretizer.apply_to(m,ncp=1, scheme='LAGRANGE-RADAU')
solver = SolverFactory('ipopt')
solver.solve(m, tee=True)
print("final time = %6.2f" %(value(m.tf)))
Now, I want to use class to express a car,then I could instantiate two cars.
So I write like this:
from pyomo.environ import *
from pyomo.dae import *
m = ConcreteModel()
class Car():
def __init__(self,friction):
self.friction = friction
self.R = Param(initialize = self.friction) # Friction factor
self.tau = ContinuousSet(bounds=(0, 1)) # Unscaled time
self.time = Var(self.tau) # Scaled time
self.tf = Var()
self.x = Var(self.tau, bounds=(0, None), initialize=0)
self.v = Var(self.tau, bounds=(0, None))
self.a = Var(self.tau, bounds=(-3.0, 1.0), initialize=0)
self.dtime = DerivativeVar(self.time)
self.dx = DerivativeVar(self.x)
self.dv = DerivativeVar(self.v)
def _ode1(m, i):
if i == 0:
return Constraint.Skip
return self.dx[i] == m.tf * self.v[i]
self.ode1 = Constraint(self.tau, rule=_ode1)
def _ode2(m, i):
if i == 0:
return Constraint.Skip
return self.dv[i] == m.tf * (self.a[i] - self.R * self.v[i] ** 2)
self.ode2 = Constraint(self.tau, rule=_ode2)
def _ode3(m, i):
if i == 0:
return Constraint.Skip
return self.dtime[i] == m.tf
self.ode3 = Constraint(self.tau, rule=_ode3)
m.car1 = Car(0.001)
m.obj = Objective(expr=m.car1.tf)
def _init(m):
yield m.car1.x[0] == 0
yield m.car1.x[1] == 100
yield m.car1.v[0] == 0
yield m.car1.v[1] == 0
yield m.car1.time[0] == 0
m.car1.initcon = ConstraintList(rule=_init)
discretizer = TransformationFactory('dae.finite_difference')
discretizer.apply_to(m, nfe=10, scheme='BACKWARD')
solver = SolverFactory('ipopt')
solver.solve(m, tee=True)
print("final time = %6.2f" % (value(m.car1.tf)))
However, I get this:
Traceback (most recent call last):
File "D:/pyo/pyomoceshi/ceshi3/car/classcar3.py", line 79, in <module>
solver.solve(m, tee=True)
File "D:\python\m\lib\site-packages\pyomo\opt\base\solvers.py", line 582, in solve
self._presolve(*args, **kwds)
File "D:\python\m\lib\site-packages\pyomo\opt\solver\shellcmd.py", line 196, in _presolve
OptSolver._presolve(self, *args, **kwds)
File "D:\python\m\lib\site-packages\pyomo\opt\base\solvers.py", line 661, in _presolve
**kwds)
File "D:\python\m\lib\site-packages\pyomo\opt\base\solvers.py", line 729, in _convert_problem
**kwds)
File "D:\python\m\lib\site-packages\pyomo\opt\base\convert.py", line 110, in convert_problem
problem_files, symbol_map = converter.apply(*tmp, **tmpkw)
File "D:\python\m\lib\site-packages\pyomo\solvers\plugins\converter\model.py", line 164, in apply
io_options=io_options)
File "D:\python\m\lib\site-packages\pyomo\core\base\block.py", line 1646, in write
io_options)
File "D:\python\m\lib\site-packages\pyomo\repn\plugins\ampl\ampl_.py", line 357, in __call__
include_all_variable_bounds=include_all_variable_bounds)
File "D:\python\m\lib\site-packages\pyomo\repn\plugins\ampl\ampl_.py", line 783, in _print_model_NL
list(self_varID_map[id(var)] for var in ampl_repn._linear_vars),
File "D:\python\m\lib\site-packages\pyomo\repn\plugins\ampl\ampl_.py", line 783, in <genexpr>
list(self_varID_map[id(var)] for var in ampl_repn._linear_vars),
KeyError: 68767416L
I want to know how to solve it or use other ways.
Below is a working version of your script. I changed things so that instead of a Car class there is a Car function that returns a Pyomo Block representing the car. By having a Car class you were essentially trying to create a subclass of Block and running into several subtle challenges that go along with that. You can see the blog post here for more information. The second change I made was in your declaration of the initial conditions, I changed the name of the ConstraintList from m.car1.initcon to m.car1_initcon. The difference is whether you want the ConstraintList to live on the car1 Block or the model. In your code, the 'dot' in the name meant you were trying to put it on the car1 Block but the constraints yielded in the rule were relative to the model. I changed the name to resolve this inconsistency.
from pyomo.environ import *
from pyomo.dae import *
m = ConcreteModel()
def Car(model, friction):
def construct_car_block(b):
b.R = Param(initialize = friction) # Friction factor
b.tau = ContinuousSet(bounds=(0, 1)) # Unscaled time
b.time = Var(b.tau) # Scaled time
b.tf = Var()
b.x = Var(b.tau, bounds=(0, None), initialize=0)
b.v = Var(b.tau, bounds=(0, None))
b.a = Var(b.tau, bounds=(-3.0, 1.0), initialize=0)
b.dtime = DerivativeVar(b.time)
b.dx = DerivativeVar(b.x)
b.dv = DerivativeVar(b.v)
def _ode1(b, i):
if i == 0:
return Constraint.Skip
return b.dx[i] == b.tf * b.v[i]
b.ode1 = Constraint(b.tau, rule=_ode1)
def _ode2(b, i):
if i == 0:
return Constraint.Skip
return b.dv[i] == b.tf * (b.a[i] - b.R * b.v[i] ** 2)
b.ode2 = Constraint(b.tau, rule=_ode2)
def _ode3(m, i):
if i == 0:
return Constraint.Skip
return b.dtime[i] == b.tf
b.ode3 = Constraint(b.tau, rule=_ode3)
return Block(rule=construct_car_block)
m.car1 = Car(m, friction=0.001)
m.obj = Objective(expr=m.car1.tf)
def _init(m):
yield m.car1.x[0] == 0
yield m.car1.x[1] == 100
yield m.car1.v[0] == 0
yield m.car1.v[1] == 0
yield m.car1.time[0] == 0
m.car1_initcon = ConstraintList(rule=_init)
discretizer = TransformationFactory('dae.finite_difference')
discretizer.apply_to(m, nfe=15, scheme='BACKWARD')
solver = SolverFactory('ipopt')
solver.solve(m, tee=True)
print("final time = %6.2f" % (value(m.car1.tf)))