Create a race in Python? - python

I'm very new to python 3. As an assignment, I am supposed to make a race in python basic Syntax and without importing any other additional function, where 3 mice are racing with different odds of winning.
from random import randint
import time
def race():
z = '----{,_,">' # The mouse.
j = ' '
print('\t'*13, '|') # The finish line. Very sophisticated.
while len(z) < 50:
time.sleep(1)
#or k in range(3):
x = randint(1, 6)
j = ' ' * x
z = j + z
print("\r" + str(z), end="") # For clearing printed text. I prefer not to import os.
As you might see, I've managed to make one; Two to go. But I've scratching my head for three hours on how to do it, that is how to make them go simultaneously ... Any help will be appreciated.

As you have seen, unfortunately the "\r" character only erases the current line, but it does not erase the others, making it impossible to create the program with this strategy. The only way to do this is to clear the console using multiple print("\n"). See this example:
from random import randint
import time
def clear():
print("\n"*80)
def race(mice = 2):
mice = ['----{,_,">' for i in range(mice)]
finished = [False,]
while not all(finished):
time.sleep(1)
clear()
print('\t' * 13, '|') # The finish line. Very sophisticated.
for i in range(len(mice)):
x = randint(1, 6)
spaces = ' ' * x
mice[i] = spaces + mice[i]
print(mice[i])
finished = map(lambda mouse: False if len(mouse) < 50 else True, mice)
race(3)

Make a function that takes a string (mouse) as an argument; adds a random number of spaces to it and returns the new string.
Make three mice.
For each mouse call the function and assign the result to that mouse.
repeat until one mouse wins - (string is a certain length).

Related

ArcGIS:Python - Adding Commas to a String

In ArcGIS I have intersected a large number of zonal polygons with another set and recorded the original zone IDs and the data they are connected with. However the strings that are created are one long list of numbers ranging from 11 to 77 (each ID is 11 characters long). I am looking to add a "," between each one making, it easier to read and export later as a .csv file. To do this I wrote this code:
def StringSplit(StrO,X):
StrN = StrO #Recording original string
StrLen = len(StrN)
BStr = StrLen/X #How many segments are inside of one string
StrC = BStr - 1 #How many times it should loop
if StrC > 0:
while StrC > 1:
StrN = StrN[ :((X * StrC) + 1)] + "," + StrN[(X * StrC): ]
StrC = StrC - 1
while StrC == 1:
StrN = StrN[:X+1] + "," + StrN[(X*StrC):]
StrC = 0
while StrC == 0:
return StrN
else:
return StrN
The main issue is how it has to step through multiple rows (76) with various lengths (11 -> 77). I got the last parts to work, just not the internal loop as it returns an error or incorrect outputs for strings longer than 22 characters.
Thus right now:
1. 01234567890 returns 01234567890
2. 0123456789001234567890 returns 01234567890,01234567890
3. 012345678900123456789001234567890 returns either: Error or ,, or even ,,01234567890
I know it is probably something pretty simple I am missing, but I can't seem remember what it is...
It can be easily done by regex.
those ........... are 11 dots for give split for every 11th char.
you can use pandas to create csv from the array output
Code:
import re
x = re.findall('...........', '01234567890012345678900123456789001234567890')
print(x)
myString = ",".join(x)
print(myString)
output:
['01234567890', '01234567890', '01234567890', '01234567890']
01234567890,01234567890,01234567890,01234567890
for the sake of simplicity you can do this
code:
x = ",".join(re.findall('...........', '01234567890012345678900123456789001234567890'))
print(x)
Don't make the loops by yourself, use python libraries or builtins, it will be easier. For example :
def StringSplit(StrO,X):
substring_starts = range(0, len(StrO), X)
substrings = (StrO[start:start + X] for start in substring_starts)
return ','.join(substrings)
string = '1234567890ABCDE'
print(StringSplit(string, 5))
# '12345,67890,ABCDE'

Solution works for sample data but online judge gives errors?

This is the problem I am trying to solve:
B: The Foxen's Treasure
There are N (1 ≤ N ≤ 4) Foxen guarding a certain valuable treasure,
which you'd love to get your hands on. The problem is, the Foxen
certainly aren't about to allow that - at least, not while they're
awake.
Fortunately, through careful observation, you've seen that each Fox
has a regular sleep cycle. In particular, the ith Fox stays awake for
Ai (1 ≤ Ai ≤ 23) hours, then sleeps for Si (1 ≤ Si ≤ 23) hours,
repeating this pattern indefinitely (2 ≤ Ai + Si ≤ 24). At the start
of your treasure-nabbing attempt, the ith Fox is
exactly Oi (0 ≤ Oi < Ai + Si) hours into its cycle.
There are T (1 ≤ T ≤ 20) scenarios as described above. For each one,
you'd like to determine how soon all of the Foxen will be
simultaneously asleep, allowing you to grab their treasure, or if this
will simply never happen.
Input
Line 1: 1 integer, T
For each scenario:
Line 1: 1 integer, N
Next N lines: 3 integers, Ai, Si, and Oi, for i = 1..N
Output
For each scenario:
Line 1: 1 integer, the minimum number of hours after the start to
wait until all of the Foxen are asleep during the same hour. If this
will never happen, output the string "Foxen are too powerful" (without
quotes) instead.
Sample Input
2
2
2 1 2
2 2 1
3
1 1 0
1 1 0
1 1 1
Sample Output
6
Foxen are too powerful
My Solution works as expected when I input the given sample case and get expected output. But when I submit the code to online judge it gives clipped error. Now there is no detail of the error which makes it difficult to find what the problem is.
Here is the solution which I have worked so far:
# ai is awake hours
# si is sleep hours.
# ai + si <= 24.
# False == sleep. True == awake.
datasets = int(raw_input());
foxen = [];
number_of_foxen = 0;
foxes = [];
class fox:
def __init__(self, a, s, i):
self.awake = a;
self.sleep = s;
self.current = i;
awake = 0;
sleep = 0;
current = 0;
def next(self):
if ( self.sleep + self.awake-1 > self.current ) :
self.current = self.current+1;
else:
self.current = 0;
return self.current;
def check(self):
if(self.current>=self.awake):
return False;
return True;
def printdata(self):
print "awake="+str(self.awake)+" sleep="+str(self.sleep)+" current="+str(self.current);
#return "awake="+str(self.awake)+" sleep="+str(self.sleep)+" current="+str(self.current);
for i in range(0, datasets):
number_of_foxen = int(raw_input());
for j in range(0, number_of_foxen):
foxen.append(raw_input());
x = foxen[j].split();
a = fox(int(x[0]), int(x[1]), int(x[2]));
foxes.append(a);
solution = False;
for j in range(0, 48):
#print "hour number = " + str(j);
#for k in range(0, len(foxes)):
#print "fox number="+ str(k)+" "+ foxes[k].printdata()+str(foxes[k].check());
count = 0 ;
for k in range(0, len(foxes)):
if(foxes[k].check()==False):
count+=1;
#print "count = "+str(count);
#print len(foxes);
if( (int(count) == int(len(foxes))) and (solution == False) ):
#print "this runs now *************";
solution = True;
number = j;
for k in range(0, len(foxes)):
foxes[k].next();
if(solution==True):
print number;
else:
print "Foxen are too powerful";
#print "Foxen are too powerful";
foxen = [];
number_of_foxen = 0;
foxes = [];
The biggest problem with your code is that it is unreadable. Indeed, it looks like it was written with little concept of Python's strengths. Here is my suggestion:
#!/usr/bin/env python3
"""
The Foxen's Treasure puzzle from http://wcipeg.com/problem/acmtryouts1b
"""
from sys import stdin
from itertools import cycle
from euclid import lcm
debug = True # set to False before submission to mechanical judge
class Fox:
"""A Fox cointains its defining integers and other derived
bindings such as its cycle and schedule."""
def __init__(self, trio):
(self.awake_count, self.sleep_count, self.skip_count) = trio
self.cycle = 'a' * self.awake_count + 's' * self.sleep_count
self.schedule = cycle(self.cycle)
if debug: print('<Fox: {}> cycle {}'.format(trio, self.cycle))
# handle skips by discarding the first elements
for _ in range(self.skip_count):
next(self.schedule)
def find_all_sleeping(foxes):
"""Return an hour number if all foxes are sleeping at that hour."""
# only examine the LCM of all fox periods. If not there it will never be.
lcm_period = 1
for fox in foxes:
lcm_period = lcm(lcm_period, len(fox.cycle))
for hour in range(lcm_period):
states = [next(fox.schedule) for fox in foxes]
if debug: print('{:2d} {}'.format(hour, ' '.join(states)))
if 'a' not in states:
return hour
return None
def read_trials(fp):
"""Reads the entire input at once. Returns a list of trials.
Each trial is a list of Fox."""
trials = list()
trial_count = int(fp.readline())
for trial in range(trial_count):
if debug: print('--Read trial {}'.format(trial))
foxes = list()
fox_count = int(fp.readline())
for _ in range(fox_count):
fox = Fox([int(x) for x in fp.readline().split()])
foxes.append(fox)
trials.append(foxes)
return trials
for trial, foxes in enumerate(read_trials(stdin)):
if debug: print('--Run trial {}'.format(trial))
hour = find_all_sleeping(foxes)
if hour is None:
print('Foxen are too powerful')
else:
print(hour)
I suspect that the first concern is that it looks much longer than the OP; that is true, but if you take out the debugging code which shows how things are happening, and the docstrings that explain why it is doing things, it's actually a few line shorter than the OP.
The main loop of the OP is too long to understand without significant study, and a whole bunch of bad variable names makes that even harder. In contrast, there are places here where a value is given a name only to make the code more explicit about what an input line means. You'll find a number of
for _ in range(trial)
to show that the loop value is not used. This is a frequent idiom when dealing with fixed format input.
The Fox representation keeps the inner workings in the problem space. As noted in the exercise page, it makes more sense to look at things as a concurrence between sequences:
--Read trial 0
<Fox: [2, 1, 2]> cycle aas
<Fox: [2, 2, 1]> cycle aass
the offsets skip_count are not shown here, but they are clear in the trial run.
The input from the datafile is all kept inside read_trials() instead of scattered through the code. This confines the mess to one place rather than distributing it through the code. We know from the puzzle instructions that the datafile will not be large enough to care about. read_trials(fp) also takes a file-like object which allows it to read from an actual file, a StringIO buffer, or the standard input.
Once the Fox schedule generator is initialized, itertools.cycle will give an unending supply of the next letter in the sequence; it does the wrap-around for you.
It is worth noting that the primary data structure trials is a plain old list because it doesn't need anything more than that.
I've gotten a little weary of bad code being answered with worse code. Sure, this could be considered way more than the needs of an electronic judge where only the output matters. Contrariwise, I'm still puzzled by bits like (solution == False), a main loop that is 42 lines long and split between the top and bottom of the file, variables like i and j which convey no intent, the memory burden of False == awake (or did I mix them up?), dead code, no-op code, `range(0, n) and a whole bunch of magic numbers throughout.
Sure, you can code like the code doesn't matter, but if you are teaching yourself to code it is good to practice good practice. Yeah, you might never look at this piece of code again, but if you ain't gonna learn it now, then when?
In case you feel it a cheat to have imported lcm() there's no reason to write it a second time, so I referenced a homebrew package of which the relevant lines are:
def gcd(a, b):
"""Return the Greatest Common Divisor of a and b."""
while b:
a, b = b, a % b
return a
def lcm(a, b):
"""Return the Least Common Multiple of a and b."""
return abs(a * b) // gcd(a, b)
Jorge was correct in his comment, there doesn't appear to be any problem with your algorithm other than the arbitrary 48 hour cuttoff.
However:
1) your print statements do not use the correct syntax for Python 3+. For example, your final print statement print "Foxen are too powerful"; must be changed to work in Python 3, try print ('Foxen are too powerful') instead.
2) I'm seeing some odd C/MatLab-like syntax as well, lines being ended by a semicolon, and double brackets surrounding conditions in your if statements. This probably isn't a problem, but depending on how picky the system you are submitting the answer to is, you may want to clean it up a little.
3) Definitely increase the cutoff time for your search. I'd recommend a reasonably large value, on the order of 10,000 hours, just to be sure that it won't be a factor.
I've taken the liberty of making all of the above changes so I'm posting the resultant code now:
# ai is awake hours
# si is sleep hours.
# ai + si <= 24.
# False == sleep. True == awake.
datasets = int(raw_input())
foxen = []
number_of_foxen = 0
foxes = []
class fox:
def __init__(self, a, s, i):
self.awake = a
self.sleep = s
self.current = i
awake = 0
sleep = 0
current = 0
def next(self):
if ( self.sleep + self.awake-1 > self.current ):
self.current = self.current+1
else:
self.current = 0
return self.current
def check(self):
if(self.current>=self.awake):
return False
return True
def printdata(self):
print ("awake="+str(self.awake)+" sleep="+str(self.sleep)+" current="+str(self.current))
#return ("awake="+str(self.awake)+" sleep="+str(self.sleep)+" current="+str(self.current))
for i in range(0, datasets):
number_of_foxen = int(raw_input())
for j in range(0, number_of_foxen):
foxen.append(raw_input())
x = foxen[j].split()
a = fox(int(x[0]), int(x[1]), int(x[2]))
foxes.append(a)
solution = False
for j in range(0, 10000):
#print ("hour number = " + str(j))
#for k in range(0, len(foxes)):
#print ("fox number="+ str(k)+" "+ foxes[k].printdata()+str(foxes[k].check()))
count = 0
for k in range(0, len(foxes)):
if(foxes[k].check()==False):
count+=1
#print ("count = "+str(count))
#print (len(foxes))
if (int(count) == int(len(foxes)) and (solution == False)):
#print ("this runs now *************")
solution = True
number = j
for k in range(0, len(foxes)):
foxes[k].next()
if(solution == True):
print (number)
else:
print ("Foxen are too powerful")
#print ("Foxen are too powerful")
foxen = []
number_of_foxen = 0
foxes = []
Enjoy and Good Luck!
Interesting problem, here's my code:
import sys
# Globals
debugLevel = 0
fileMode = True # True if loading data from a file.
# Constants
AWAKE = 0
ASLEEP = -1
def gcd(a, b):
"""Return greatest common divisor using Euclid's Algorithm."""
while b:
a, b = b, a % b
return a
def lcm(a, b):
"""Return lowest common multiple."""
return a * b // gcd(a, b)
def readData(f):
''' Read in the problem data and store in data structures
'''
numTrials = int(f.readline().strip())
if debugLevel >= 4:
print("Num trials: ", numTrials)
trialData = []
for _ in range(numTrials):
numFoxen = int(f.readline().strip())
allFoxenHoursInfo = []
for _ in range(numFoxen):
aFoxHoursInfo = f.readline().split()
aFoxHoursInfo = list(map(int, aFoxHoursInfo))
allFoxenHoursInfo.append(aFoxHoursInfo)
trialData.append((numFoxen, allFoxenHoursInfo))
if debugLevel >= 8:
print("Trial data\n", trialData)
return numTrials, trialData
def runTrials(trialData):
'''
Go through each lot of foxen, and their sleep/awake schedules and
See if there's a time that all of them will be asleep.
'''
global debugLevel
for trial in trialData:
numFoxen, allFoxenHoursInfo = trial
# Create a table of the status of each fox in each hour
row = [AWAKE] * (numFoxen+1)
row[0] = 0
hoursTable = [row]
# Cycle length for each fox is the number of hours they spend awake then asleep.
cycleLength = [0] * (numFoxen)
# This is the number of hours into the cycle each fox is at the start
startingPosInCycle= [0] * (numFoxen)
# Initialise the first row
for fox in range(numFoxen):
cycleLength[fox] = allFoxenHoursInfo[fox][0] + allFoxenHoursInfo[fox][1] # Time awake plus time asleep
startingPosInCycle[fox] = allFoxenHoursInfo[fox][2] # % cycleLength[fox]
if startingPosInCycle[fox] >= allFoxenHoursInfo[fox][0]:
hoursTable[0][fox+1] = ASLEEP
if debugLevel >= 4:
print("Initial table: ", hoursTable)
# lcm = lowest common multiple and it's implemented above.
# For this problem, we only need to look at the lcm of all the cycle lengths for the foxen.
numIterations = 1
for fox in range(numFoxen):
numIterations = lcm(numIterations, cycleLength[fox])
# Go around a loop adding a new row to the table for each new hour containing the updated
# statuses of each fox.
for hourNum in range(1, numIterations):
allFoxesSleeping = False
# Update our hours table by creating a new row and calculating the status of each fox
newRow = [AWAKE] * (numFoxen+1)
newRow[0] = hourNum
for fox in range(numFoxen):
currentPosInCycle = (startingPosInCycle[fox] + hourNum) % cycleLength[fox]
if currentPosInCycle >= allFoxenHoursInfo[fox][0]:
newRow[fox+1] = ASLEEP
hoursTable.append(newRow)
if debugLevel >= 4:
print("Hours table\n", hoursTable)
# See if all foxen are sleeping, if they are, success
numFoxesSleeping = hoursTable[hourNum].count(ASLEEP)
if numFoxesSleeping == numFoxen:
allFoxesSleeping = True
print(hourNum)
break
if not allFoxesSleeping:
print('Foxen are too powerful')
def main():
'''Reads, runs, and outputs problem specific data.'''
# Initialisation
#strDir = ".\\"
# if fileMode:
# dataSource = open(strDir + "DataFile.txt", 'r')
# else:
dataSource = sys.stdin
# Read in the input data.
numTrials, trialData = readData(dataSource)
# Run each trial, outputting the result of that trial
runTrials(trialData)
sys.stdout.flush()
# Cleanup
# if fileMode:
# dataSource.close()
if __name__ == '__main__':
main()
Unfortunately, it does not pass the judge either. I have no idea why. I get this output:
Test case #1: WA [0.178s, 3628K] (0/1) (Details)
Your Output (clipped)
6
Foxen are too powe
Final score: 0/1
Interesting problem. I have emailed the author, because there is an inconsistency in the problem definition and the sample input data. He says this: Oi (0 ≤ Oi < Ai + Si) but then gives 1 1 1 on the last line of sample input data. And 1 is not strictly less than 1+1.
So who knows what other data the judge might be using...
The bit at the bottom that is commented out re files, lets me work with files and an IPython console rather than a Python console and pasting the data in, which I find slow and annoying.
Also, it's a little strange I reckon to not be able to see the data the judge is using. Surely seeing the data you are working against would enable the problem to be run and debugged offline, then when it's going, a new online submit could be done.

How do I debug a python multiple-sentence generator?

I made my own Python code like this:
import random
# nouns-----------------------------------------------------------------------------------------------------------------
nouns = 'noun1 noun2 noun3 noun4 noun5 noun6 noun7 noun8 noun9 noun10 noun11 noun12 noun13 noun14 noun15 noun16' .split()
def getRandomWord(nounList):
wordIndex = random.randint(0, len(nounList) - 1)
return nounList[wordIndex]
noun_var = getRandomWord(nouns)
# verbs-----------------------------------------------------------------------------------------------------------------
verbs = 'verb1 verb2 verb3 verb4 verb5 verb6 verb7 verb8 verb9 verb10 verb11 verb12 verb13 verb14 verb15 verb16' .split()
def getRandomWord(verbList):
wordIndex = random.randint(0, len(verbList) - 1)
return verbList[wordIndex]
verb_var = getRandomWord(verbs)
# infinitives-----------------------------------------------------------------------------------------------------------
infi = 'verb1 verb2 verb3 verb4 verb5 verb6 verb7 verb8 verb9 verb10 verb11 verb12 verb13 verb14 verb15 verb16' .split()
def getRandomWord(infiList):
wordIndex = random.randint(0, len(infiList) - 1)
return infiList[wordIndex]
infi_var = getRandomWord(infi)
# print strings
for i in range(1):
print (noun_var, verb_var, infi_var)
But whenever I wanted to change the 1 in "Line 25, Char 16" into an input variable letting you be able to make the choice of how many sentences wanted, it just repeats the same sentence again and again instead of generating new ones. Is there another way to make an app with Python with the same principles or is there a way to fix this?
This code is in Python-3.x
If you want to display a different sentence in each iteration of your loop, then generate a new sentence. Right now you're picking 3 random words, and then printing them over and over again.
for i in range(int(input())):
print (random.choice(nouns), random.choice(verbs), random.choice(infi))

In python, how would I go about linking user input to a index in a list?

I would like to create a binary puzzle with python.
At the moment I already made a 6x6, 8x8 and 10x10 layout which is shown based on the difficulty that the players wishes to play. The purpose of the puzzle can be compared with a game of sudoku, you want to input either 0 or 1 on a given location by the player. Below you will find what I currently have for the layout.
if graad == 1:
easy = [['A', 'B', 'C', 'D', 'E'],
['_','_','_','_','_','_','_'],
[0,1,0,1,0,1,' |1'],
[1,0,1,0,1,0,' |2'],
[0,1,0,1,0,1,' |3'],
[1,0,1,0,1,0,' |4'],
[0,1,0,1,0,1,' |5'],
[1,0,1,0,1,0,' |6']]
i = 0
while i < len(easy):
j = 0
s = ""
while j < len(easy[i]):
s = s + str(easy[i][j]) + " "
j = j + 1
print (s)
i = i + 1
Now the problem that I am facing is, how can I let python know that when a player fills in position 3 on column C and row 5 with a 0 for example?
I was thinking of an IF statement that checks the input on either a A, B, C D, E... Row 1,2,3,4,5.. but that is going to be a lot of if statements.
Edit1: Ok so to clarify.I wanted to post a picture but need more posts.
For example, I have a game board of 6x6 cells. Some of them are filled with a 1 and some of them are filled with 0 and most of them are empty because the goal is to have it look in the end like my layout in the python code.(That's the solution). So you want the user to fill in those empty cells.
Now, let's say that the player wants to fill in A-1 with a 1, how will python know that input A-1 is linked to index [0][0] in the list?
A simple way to convert your letter indices to numbers is to use the ord() function, which returns the numerical code of a single character. Since you are using upper-case letters, with 'A' being the label for the column with index 0, you can do
column = ord(letter) - ord('A')
That will convert 'A' to 0, 'B' to 1, etc.
Here's a short example program vaguely based on the code on your question.
It accepts moves in the form A10 to set location A1 to '1', 'B30' to set location B3 to '0'. It accepts lower case letters, too, so 'd11' is the same as 'D11'. Hit Ctrl-C to exit.
Tested on Python 2.6.6, but it should work correctly on Python 3. (To run it on Python 2, change input() to raw_input()).
#! /usr/bin/env python
def print_grid(g):
gsize = len(g)
base = ord('A')
print(' '.join([chr(base + i) for i in range(gsize)]))
print((gsize * 2) * '-')
for i, row in enumerate(g, 1):
print(' '.join(row) + ' | ' + str(i))
print('\n')
def main():
gsize = 9
rowstr = gsize * '_'
grid = [list(rowstr) for i in range(gsize)]
print_grid(grid)
while True:
move = input('Enter move: ')
letter, number, bit = move.strip()
col = ord(letter.upper()) - ord('A')
row = int(number) - 1
grid[row][col] = bit
print_grid(grid)
if __name__ == "__main__":
main()
If you work with a pandas DataFrame to hold your correct answer of the game you can easily check things. The pandas package has a good documentation (and a lot of Q&A here on stackoverflow).
The setup of your correct answer:
import pandas as pd
data = [[0,1,0,1,0,1],
[1,0,1,0,1,0],
[0,1,0,1,0,1],
[1,0,1,0,1,0],
[0,1,0,1,0,1],
[1,0,1,0,1,0]]
easy = pd.DataFrame(data)
easy.columns = ['A','B','C','D','E','F']
print easy
The item at position 'A',0 (python starts to number from 0) is given by easy['A'][0]. For more information about indexing a pandas DataFrame object visit the documentation.
Another usefull thing, a DataFrame object is printable, making it unnecessary to write a print command yourself.
If using DataFrames is overkill for you, another option is to work with a 'translation' dictionary. This dictionary will use the letters for keys and the corresponding column number as a value.
>>> column = {'A':0, 'B':1, 'C':2, 'D':3, 'E':4, 'F':5}
>>> print column['A']
0

How to make a list of lists in python

I am trying to make a list of lists in python using random.random().
def takeStep(prevPosition, maxStep):
"""simulates taking a step between positive and negative maxStep, \
adds it to prevPosition and returns next position"""
nextPosition = prevPosition + (-maxStep + \
( maxStep - (-maxStep)) * random.random())
list500Steps = []
list1000Walks = []
for kk in range(0,1000):
list1000Walks.append(list500Steps)
for jj in range(0 , 500):
list500Steps.append(list500Steps[-1] + takeStep(0 , MAX_STEP_SIZE))
I know why this gives me what it does, just don't know how to do something about it. Please give the simplest answer, in new at this and don't know a lot yet.
for kk in xrange(0,1000):
list500steps = []
for jj in range(0,500):
list500steps.append(...)
list1000walks.append(list500steps)
Notice how I am creating an empty array (list500steps) each time in the first for loop? Then, after creating all the steps I append that array (Which is now NOT empty) to the array of walks.
import random
def takeStep(prevPosition, maxStep):
"""simulates taking a step between positive and negative maxStep, \
adds it to prevPosition and returns next position"""
nextPosition = prevPosition + (-maxStep + \
( maxStep - (-maxStep)) * random.random())
return nextPosition # You didn't have this, I'm not exactly sure what you were going for #but I think this is it
#Without this statement it will repeatedly crash
list500Steps = [0]
list1000Walks = [0]
#The zeros are place holders so for the for loop (jj) below. That way
#The first time it goes through the for loop it has something to index-
#during "list500Steps.append(list500Steps[-1] <-- that will draw an eror without anything
#in the loops. I don't know if that was your problem but it isn't good either way
for kk in range(0,1000):
list1000Walks.append(list500Steps)
for jj in range(0 , 500):
list500Steps.append(list500Steps[-1] + takeStep(0 , MAX_STEP_SIZE))
#I hope for MAX_STEP_SIZE you intend on (a) defining the variable (b) inputing in a number
You might want to print one of the lists to check the input.

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