Shorten repetitive code with a For Loop to increment variables. - python

I am drawing a blank, but I do think that writing this code manually seems excessive. Can I use a 'for loop' to shorten this and increment the variables below?
pos = 1000
m1.setAsHome() # set position as zero for all of the commands
m1.goTo(pos) # move to the original position oof the limit switch
while(m1.isBusy()):
continue
m1.free() # reset the motor
while(m2.isBusy()):
continue
m2.setAsHome() # set position as zero for all of the commands
m2.goTo(pos) # move to the original position oof the limit switch
while(m2.isBusy()):
continue
m2.free() # reset the motor
My hunch is something along the lines of this:
for i in range(4):
m = 0
print m[i].setAsHome()
of course this yields an error. Sorry for the newbie question, but I am sure there must be a way to shorten this. Furthermore, the code above keeps going to include 4 motors. Thank you.

You need a list of motors:
motors = [m1, m2, m3, m4]
So you can use a for loop:
for motor in motors:
motor.setAsHome()
motor.goTo(pos)

Related

How do I fix infinite loop bugs in Python?

I have a problem regarding a competition question I'm attempting to do. Here is the question (its a bit long)
""""
Welcome aboard, Captain! Today you are in charge of the first ever doughnut-shaped spaceship, The
Circular. There are N cabins arranged in a circle on the spaceship. They are numbered from 1 to N in
a clockwise direction around the ship. The ith and the (i + 1)th cabins are connected. So too are cabin
1 and cabin N.
Currently the ith cabin has Ai crewmates, however the spaceship cannot depart unless there are exactly
Bi crewmates in this cabin.
To achieve this, you have the power to pay crewmates to change cabins. You can pay a crewmate $1 to
move to an adjacent cabin. A crewmate can be asked to move multiple times, provided that you pay
them $1 each time.
What is the fewest dollars you must pay before you can depart? It is always be possible to depart.
""""
https://orac2.info/problem/aio22spaceship/ (the link to the intereactive Qs)
I searched the web and i found no solutions to the Q. My code seems to be infinite looping i guess but im not sure as i cant see what cases the sit uses to determine if my code is right.
Heres my code
#!/usr/bin/env python
import sys
sys.setrecursionlimit(1000000000)
#
# Solution Template for Spaceship Shuffle
#
# Australian Informatics Olympiad 2022
#
# This file is provided to assist with reading and writing of the input
# files for the problem. You may modify this file however you wish, or
# you may choose not to use this file at all.
#
# N is the number of cabins.
N = None
# A contains the initial number of crewmates in each cabin. Note that here the
# cabins are numbered starting from 0.
A = []
# B contains the desired number of crewmates in each cabin. Note that here the
# cabins are numbered starting from 0.
B = []
answer = 0
# Open the input and output files.
input_file = open("spacein.txt", "r")
output_file = open("spaceout.txt", "w")
# Read the value of N.
N = int(input_file.readline().strip())
# Read the values of A and B.
input_line = input_file.readline().strip()
A = list(map(int, input_line.split()))
input_line = input_file.readline().strip()
B = list(map(int, input_line.split()))
AM = A
#AM is my modifying set
# TODO: This is where you should compute your solution. Store the fewest
# dollars you must pay before you can depart into the variable
while AM != B:
#Check if the set is correct
#notfound is a testing variable to see if my code was looping due to input error
notfound = True
for i in range(N):
#Check which places needs people to be moved
while AM[i]>B[i]:
notfound = False
#RV and LV check the "neediness" for each half's people requirements. I check how many people
#are needed on one side compared to the other and subtract the "overflow of people"
RV = 0
LV = 0
for j in range(int(N/2-0.5)):
#The range thing makes sure that if N is odd, im splitting the middle but if N is even, i leave out the end pod
RV += B[(i+j+1)%N]-AM[(i+j+1)%N]
LV += B[(i-j-1)%N]-AM[(i-j-1)%N]
answer +=1
if RV>LV:
AM[i]+=-1
AM[(i+1)%N]+=1
else:
AM[i]+=-1
AM[(i-1)%N]+=1
print(AM,B)
if notfound:
break
print(answer)
# Write the answer to the output file.
output_file.write("%d\n" % (answer))
# Finally, close the input/output files.
input_file.close()
output_file.close()
please help i really neeed to know the answer, driving me mad ngl
Welp, there aren't any resources online and I've tried everything. I think the problem might be that because of my solving method, passengers may be flicked between two pods indefinitely. Not sure since i could make a case that demoed this.
also my post probably is messy since this is my first time posting

Visualize Pathfinding Algorithm

I was doing a pathfinding visualizer in pygame and I pretty much finished but there's still one thing that I do not like about the algorithm part of it and it's the fact that when you press the visualize algorithm button it shows you the shortest path in yellow and all of the nodes the algorithm has visited ever in light blue but it shows you instantaneously and I want it to color the nodes accordingly step by step to actually reach the effect of visualizing (like in here https://clementmihailescu.github.io/Pathfinding-Visualizer/#), I tried to write some code in the function that seemed like it would have worked as intended but it didn't, here is the code:
# Breadth First Search Algorithm
def bfs(graph, start, goal):
explored = []
# Queue for traversing the
# graph in the BFS
queue = [[start]]
# If the desired node is
# reached
if start == goal:
return
# Loop to traverse the graph
# with the help of the queue
while queue:
path = queue.pop(0)
node = path[-1]
y, x = node
# Codition to check if the
# current node is not visited
if node not in explored and nodes_rows[x][y].color is not BLACK:
nodes_rows[x][y].color = LIGHT_BLUE
neighbours = graph[node]
# Loop to iterate over the
# neighbours of the node
for neighbour in neighbours:
new_path = list(path)
new_path.append(neighbour)
queue.append(new_path)
# Condition to check if the
# neighbour node is the goal
if neighbour == goal:
new_path.remove(start)
new_path.remove(goal)
return new_path
explored.append(node)
return None
The nodes_rows[x][y].color == color_name is the code that is responsible for coloring nodes on the grid which is represented by a dictionary(I provided it so it's gonna be easier for you to understand how coloring works in general in my program). The problem with that implementation is when I do add the coloring part at if statement to color all the neighbors it does it instantly on the grid without showing a kind of an animation that shows the coloring process node by node, my question is can I do it so it colors them each iteration rather than all at once by adding something to this code and not writing a new one and if I do need to write a new one that what is the instructions how can I do so?
Here is what I mean by coloring all at once like it does for now:
https://cdn.discordapp.com/attachments/772816508015083552/832303260911272046/PowerPoint_-_1_2021-04-15_20-13-35_Trim.mp4
Edit:
try:
while True:
if not ticks or pygame.time.get_ticks() - ticks >= 500:
ticks = pygame.time.get_ticks()
nodes = next(algorithm)
if nodes_rows[nodes[-1][1]][nodes[-1][0]].color != BLUE:
nodes_rows[nodes[-1][1]][nodes[-1][0]].color = LIGHT_BLUE
pygame.display.update()
except StopIteration:
pass
Tried doing it with yield and if I print it it does yield a list every half a second with a new explored node at the end like intended but it updates it all at once after waiting total amount of ticks I tried playing with the indent of display.update() but didn't work either I don't even know what to do at this point
Thanks to everyone contributing to help <3
Per the comments above, here is a simple example of a generator to help you grasp the idea.
def counter(x):
for i in range(x):
yield i
c = counter(3)
In: next(c)
Out: 0
In: next(c)
Out: 1
In: next(c)
Out: 2
What's happening is that every time you call next, the function will continue to run until it reaches the next yield, at which point it will return the yielded value.
It will also remember where it left off, so the next time you call next, the function will continue where it left off.
In your application, this could be used to yield the list of explored locations, then draw those locations in whatever color you like, call next(bfs) to step forward and yield the next list of explored locations, and so on until of course you find the solution and run out of items to yield.
One more example, perhaps a little more closely related to what you are trying to do:
def make_path():
path = []
i = 0
while True:
path.append(i)
i += 1
yield path
c = make_path()
for _ in range(6):
print(next(c))
Out: [0]
[0, 1]
[0, 1, 2]
[0, 1, 2, 3]
[0, 1, 2, 3, 4]
[0, 1, 2, 3, 4, 5]
i think the problem is that you do not update the canvas in your while queue loop.
the program will execute your bfs algorithm and then it will update the canvas.
i honestly don't use pygame very regularly, but i think to force the canvas to repaint you need to stick a pygame.display.update() inside your while loop.
Pygame is not my forte but, I think there are some missing things in your project:
1.- You don't really control when colors are shown on your screen. This can be achieved with this:
pygame.display.update()
Documentation: https://devdocs.io/pygame/ref/display#pygame.display.update
Usually when working with graphics, you add a color, and when then next screen painting update comes you see it, what you must do is to force the painting so you can animate safely.
2.- Coloring animations are not being done by default.
I couldn't help noticing this in your initial question: "it does it instantly on the grid without showing a kind of an animation that shows the coloring process node by node". If you want an animation like the example linked above (Pathfinding visualizer), I'm afraid you have to do it manually. Try creating a test with a white square and start painting growing circles inside interpolating colors, until you reach the walls and paint the full square. This should be similar to what are you trying to accomplish.
3.- Perhaps you should add some stops when you are animating your canvas.
The refresh rate of your hardware is way too faster than your human eye.
To properly see a coloring animation you should try adding execution stops, when interpolating colors.
nodes_rows[x][y].color = Color(r, g, b) # Start with a color and interpolate incrementing or decrementing r, g and b until your reach a final color.
pygame.display.update() # Update your screen
pygame.time.wait(1000) # In milliseconds, add some time wait so screen can be painted, if time is long enough perhaps screen updates meanwhile and you do not need to update the screen.
4.- (Optional) Parallel painting: Using threads.
If this does not work right for you, add threads.
As I stated before, pygame is not my forte, but consider adding threads to animate each square color. I've found a post here explaining this.
using threading in pygame
I hope this helps you and have a happy coding experience! :D

Optimize For loop for speed python

Like pointed out, let me explain a bit of the code. I have 2 for loops because i have a board of a game called Tablut. I have to check from the pawn position (row_index and column_index) if there is an obstacle on the left and on the right. So, the first loop take the row_index of the pawn, then check for every row on the right side and stop when it finds an obstacle. Then do the same for the left side. If i use a single loop i don't have a way to stop searching in one direction if found an obstacle and start looking in the other, because the break will stop the entire for. The loop are used to generate all the possible actions that a move can do inside the board in that turn. Action(Position(....)) is one action that the pawn can make.
So, basically, i check in the right and left direction, if the next tile is empty i can move there, so i create a new action and add it to a list of all the possible moves, if there is an obstacle, since i can't jump it, i stop the loop.
i have this for loop and i would like to optimize it for speed.
for row in range(row_index - 1, -1, -1):
if bitboard_util.get_bit(obstacle_bitboard, row, column_index) != 1:
action = Action(Position(row_index, column_index), Position(row, column_index), color)
all_available_moves_for_pawn.append(action)
else:
break
for row in range(row_index + 1, 9):
if bitboard_util.get_bit(obstacle_bitboard, row, column_index) != 1:
action = Action(Position(row_index, column_index), Position(row, column_index), color)
all_available_moves_for_pawn.append(action)
else:
break
You could optimize this a little by preparing a structure based on your obstacle bits that will give the range of available "moves" for each position. You can prepare that structure once and reuse it multiple times to obtain the move lists directly for all column_index/row_index afterward.
from itertools import accumulate
rowCount = 9
colCount = 9
obstacleRanges = []
for col in range(colCount):
obstacles = [-1] + [ row*(bitboard_util.get_bit(obstacle_bitboard, row, col) == 1) for row in range(rowCount) ] + [rowCount]
prevObstacle = [*accumulate(obstacles,max)]
nextObstacle = [*accumulate(obstacles[::-1],lambda p,o:[p,o][o>0])][::-1]
obstacleRanges.append([*zip(prevObstacle[1:-1],nextObstacle[1:-1])])
obstacleRanges will contain tuples with the position (row) of the previous and next obstacles for every position on the board. This is computed once for the whole board and allows you to find the range of moves directly without additional calls to get_bit
usage:
prevObstacle,nextObstacle = obstacleRanges[column_index,row_index]
for moveRow in range(prevObstacle+1, nextObstacle):
action = Action(Position(row_index, column_index), Position(moveRow, column_index), color)
all_available_moves_for_pawn.append(action)
Note that this may or may not improve performance depending on the distribution and number of you pawns and also on how many times you query the actions with the same obstacle layout. For example, if your pawns ARE the obstacles and they are moving, it would be preferable to update the structure with the effect of pawn movements rather than recalculate it completely.

how to set the orientation of the stimulus for each trial in psychopy

I am fairly new to the python language and psychopy. I am practicing it by creating dummy experiments. Here, I am trying to create an experiment about bayesian brain. Non-vertical lines will be presented to the participant while no respond is expected from the participants, just exposure. Then for the last trial (it stays on the monitor for longer period of time to be responded), it is expected from the participant to judge whether the last line trial is vertical or not? (after exposing to non-vertical lines, I am expecting to see a change in perception of verticality).
However, there are so many things that I couldn't learn from the web. I am pretty sure you guys can help me easily.
My primary problem is; how to set up the orientation of the line? I found out the stim.ori but not sure how to use it on 'line' stimuli. Below I've attached the codes that I made so far. Also, I have added some extra questions with #.
I tried to be clear as much as I can. Sorry for my bad english.
Thank you!
from psychopy import visual, core, event #import some libraries from PsychoPy
import random
#create a window
mywin = visual.Window([800,600],monitor="testMonitor", units="deg")
#stimuli
lineo = visual.Line(mywin, start=(-5, -1), end=(-5, 1))
fixation = visual.GratingStim(mywin, size=0.2, pos=[0,0], color = 'black')
#draw the stimuli and update the window
n = 5 # trial number
i = 0
while i < n:
#fixation
fixation.draw()
mywin.flip()
presses = event.waitKeys(1)
# stimulus
orientationlist = [20,30,40,50,60] # I want to draw the orientation info from this list
x = random.choice(orientationlist)
lineo.ori((x)) #
lineo.draw()
mywin.flip()
presses= event.waitKeys(2)
i +=1
if i == 5: # how do I change the number into the length of the trial; len(int(n) didnt work.
lineo.draw()
mywin.flip()
presses = event.waitKeys(4)
#quiting
# I dont know how to command psychopy for quiting the
# experiment when 'escape' is pressed.
#cleanup
mywin.close()
core.quit()
There's a few things that you would want to do differently. I've updated your code and marked changes with the comment "CHANGE". Changing the orientation of a stimulus is pretty consistent in psychopy, so it's no different for Line than any other visual stimulus type.
from psychopy import visual, core, event #import some libraries from PsychoPy
import random
#create a window
mywin = visual.Window([800,600],monitor="testMonitor", units="deg")
#stimuli
lineo = visual.Line(mywin, start=(-5, -1), end=(-5, 1))
fixation = visual.GratingStim(mywin, size=0.2, pos=[0,0], color = 'black')
orientationlist = [20,30,40,50,60] # CHANGED. No need to redefine on every iteration of the while-loop.
#draw the stimuli and update the window
n = 5 # trial number
for i in range(n): # CHANGED. This is much neater in your case than a while loop. No need to "manually" increment i.
#fixation
fixation.draw()
mywin.flip()
event.waitKeys(1) # CHANGED. No need to assign output to anything if it isn't used.
# stimulus
lineo.ori = random.choice(orientationlist) # CHANGED. Alternative: lineo.setOri(random.choice(orientationlist)).
lineo.draw()
mywin.flip()
event.waitKeys(2)
# At this point, all the stimuli have been shown. So no need to do an if-statement within the loop. The following code will run at the appropriate time
lineo.draw()
mywin.flip()
event.waitKeys(keyList=['escape']) # CHANGED. Listen for escape, do not assign to variable
# CHANGED. No need to call core.quit() or myWin.close() here since python automatically cleans everything up on script end.

Python recursion depth exceeded limit exceeded, and don't know how to delete recursion

Maybe the problem can be solved by deleting all those functions, can't it?
However, i really don't know what to do to get the source work.
By the way, it just simulates a horse in a chesstable, going around and around, randomly, trying to visit each square once; and I get a recursion depth exceeded error.
import random
def main():
global tries,moves
tries,moves=0,0
restart()
def restart():
global a,indexes,x,y
a=[[0 for y in range(8)] for x in range(8)] #Costrutto chic
indexes=[x for x in range(8)]
#Random part
x=random.randint(0,7)
y=random.randint(0,7)
a[x][y]=1
start()
def start():
global i,indexes,moves,tries
i=0
random.shuffle(indexes) #List filled with random numbers that i'll use as indexes
while i<=7:
if indexes[i]==0:
move(-2,-1)
elif indexes[i]==1:
move(-2,1)
elif indexes[i]==2:
move(-1,-2)
elif indexes[i]==3:
move(-1,2)
elif indexes[i]==4:
move(1,-2)
elif indexes[i]==5:
move(1,2)
elif indexes[i]==6:
move(2,-1)
elif indexes[i]==7:
move(2,1)
i+=1
for _ in a:
if 0 in _:
print "Wasted moves: %d"%(moves)
tries+=1
moves=0
restart()
print "Success obtained in %d tries"%(tries)
def move(column,row):
global x,y,a,moves
try: b=a[x+row][y+column]
except IndexError: return 0
if b==0 and 0<=x+row<=7 and 0<=y+column<=7:
x=x+row
y=y+column
a[x][y]=1
moves+=1
start()
else: return 0
try :main()
#except: print "I couldn't handle it" <-Row added to prevent python from returning a huge amount of errors
EDIT: This is the modified version, which still does not works, but at least it's an improvement:
import random
def move((column,row),x,y):
try: cell=squares_visited[x+row][y+column]
except IndexError: return x,y ## NONE TYPE OBJECT
if cell==0 and 0<=x+row<=7 and 0<=y+column<=7:
x+=row
y+=column
squares_visited[x][y]=1
return x,y ## NONE TYPE OBJECT
squares_visited=[[0] * 8 for _ in range(8)]
x=random.randint(0,7)
y=random.randint(0,7)
squares_visited[x][y]=1
moves=[(-2,-1),(-2,1),(-1,-2),(-1,2),(1,-2),(1,2),(2,-1),(2,1)]
indexes=list(range(8))
tries=0
print "The horse starts in position %d,%d"%(x,y)
while True:
random.shuffle(indexes)
for _ in indexes:
cells=move(moves[indexes[_]],x,y) ##Passing as arguments x and y looks weird
x=cells[0]
y=cells[1]
#If you out this for cicle, there are no legal moves anymore(due to full completion with 1, or to lack of legit moves)
for _ in squares_visited:
if 0 in _:
squares_visited=[[0] * 8 for _ in range(8)]
tries+=1
else:
print "You managed to do it in %d tries."%(tries)
This code has a lot of problems -- enough that it's worth going over in full:
import random
def main():
global tries,moves
The first of many examples of over-use of global variables. When possible, pass parameters; or create a class. This is a general strategy that will help you construct more comprehensible (and thus more debuggable) algorithms; and in a general sense, this is part of why your code fails -- not because of any particular bug, but because the complexity of your code makes it hard to find bugs.
tries,moves=0,0
restart()
def restart():
global a,indexes,x,y
Why do you name your board a? That's a terrible name! Use something descriptive like squares_visited.
a=[[0 for y in range(8)] for x in range(8)] #Costrutto chic
indexes=[x for x in range(8)]
Minor: in python 2, [x for x in range(8)] == range(8) -- they do exactly the same thing, so the list comprehension is unnecessary. In 3, it works a little differently, but if you want a list (rather than a range object) just pass it to list as in (list(range(8))).
#Random part
x=random.randint(0,7)
y=random.randint(0,7)
a[x][y]=1
start()
So my understanding of the code so far is that a is the board, x and y are the starting coordinates, and you've marked the first spot visited with a 1. So far so good. But then things start to get hairy, because you call start at the end of restart instead of calling it from a top-level control function. That's theoretically OK, but it makes the recursion more complicated than necessary; this is another part of your problem.
def start():
global i,indexes,moves,tries
Argh more globals...
i=0
random.shuffle(indexes) #List filled with random numbers that i'll use as indexes
while i<=7:
if indexes[i]==0:
move(-2,-1)
elif indexes[i]==1:
move(-2,1)
elif indexes[i]==2:
move(-1,-2)
elif indexes[i]==3:
move(-1,2)
elif indexes[i]==4:
move(1,-2)
elif indexes[i]==5:
move(1,2)
elif indexes[i]==6:
move(2,-1)
elif indexes[i]==7:
move(2,1)
i+=1
Ok, so what you're trying to do is go through each index in indexes in sequence. Why are you using while though? And why is i global?? I don't see it being used anywhere else. This is way overcomplicated. Just use a for loop to iterate over indexes directly, as in
for index in indexes:
if index==0:
...
Ok, now for the specific problems...
for _ in a:
if 0 in _:
print "Wasted moves: %d"%(moves)
tries+=1
moves=0
restart()
print "Success obtained in %d tries"%(tries)
I don't understand what you're trying to do here. It seems like you're calling restart every time you find a 0 (i.e. an unvisited spot) on your board. But restart resets all board values to 0, so unless there's some other way to fill the board with 1s before hitting this point, this will result in an infinite recursion. In fact, the mutual recursion between move and start might be able to achieve that in principle, but as it is, it's way too complex! The problem is that there's no clear recursion termination condition.
def move(column,row):
global x,y,a,moves
try: b=a[x+row][y+column]
except IndexError: return 0
if b==0 and 0<=x+row<=7 and 0<=y+column<=7:
x=x+row
y=y+column
a[x][y]=1
moves+=1
start()
else: return 0
Here, in principle, the idea seems to be that if your move hits a 1 or goes off the board, then the current branch of the recursion terminates. But because i and indexes are global above in start, when start is re-called, it re-shuffles indexes and resets i to 0! The result is sheer chaos! I can't even begin to comprehend how that will effect the recursion; it seems likely that because i gets reset at the beginning of start every time, and because every successful call of move results in a call of start, the while loop in start will never terminate!
I suppose it's possible that eventually this process will manage to visit every square, at which point things might work as expected, but as it is, this is too complex even to predict.
try :main()
#except: print "I couldn't handle it" <-Row added to prevent python from returning a huge amount of errors
Not sure what you mean by that last line, but it doesn't sound like a good sign -- you're papering over an error instead of finding the root cause.
I'm going to play with this code a bit and see if I can get it to behave marginally better by de-globalizing some of its state... will report back soon.
Update:
Ok I de-globalized indexes as described above. I then replaced the start/restart recursion with an infinite loop in restart, a return statement in start where the call to restart used to be, and a sys.exit() at the end of start (to break out of the infinite loop on success). The result behaves more as expected. This is still poor design but it works now, in the sense that it recursively tries a bunch of random paths until every local position has been visited.
Of course it still doesn't ever succeed; it just keeps looping. Actually finding a solution will probably require a lot more rethinking of this algorithm! But following my above suggestions should help some, at least.
start() and move() call each other, making it an indirect recursive call BUT the move() return startment get out of move() and notout of the recursion.
You see, when you are calling a function, that calls a function that calls a functions, it all goes in a stack that reference each calls. When you get a your final result, you are supposed to go backward, and unstack these function calls.
Here, you don't, you call move(), that calls start(), and it it returns something then you just keep going deeper in the stack.
Try to make an iterative version of your code. Recursion is hard, start with something easier.
If you do want to persist in the recursive version, make move() call itself, and then go backward in the stack from it self once it reach the out condition. It will be clearer than dealing with recursive calls from two functions.
BTW:
Avoid using global variables. Either pass the data as arguments, or use a class. I would use argument passing, it will force you to come up with a better algo that this one.
the while loop is not necessary. Replace it with a for loop on indexes
the huge if/elif statement is not necessary, replace it with a dictionary
You should end up with somthing like this:
for i in indexes:
move(*MOVES[i])
MOVES, being a dict of values of i associated with params for move().
you may want to use generators instead of your list comprehensions, but that would require some algo changes. It could be better for your memory footprint. At the very least, make this shorter:
[x for x in range(8)] can be written range(8)
[[0 for y in range(8)] for x in range(8)] can be written [[0] * 8 for x in range(8)]
range() can be replaced by xrange()

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