I am a beginner in the field of python coding.I am stuck with this code which I wrote as a solution for the problem -"ambiguous permutations" on codechef.It shows time limit exceeded when I run it on the website.Please help me with this issue.The code is given below:
type=[]
while(True):
n = int(raw_input())
def inverse_calculation(listhere=[]):
listtwo =[]
for i in range(1,n+1):
a = listhere.index(i)
listtwo.append((int(a)+1))
if (listtwo == listhere):
return type.append('ambiguous')
else:
return type.append('not ambiguous')
if (n!=0):
list1 = []
list1 = [int(x) for x in raw_input().split()]
inverse_calculation(list1)
else:
break
for element in type:
print(element)
Here in the code your n value will always be not equal to zero ,that is, your if condition will always be true, come what may, because you are not decrementing or altering the value of n in you entire program. And since your while condition is always true you are bound to get this error.
So my suggestion is you try to alter the value of n in your program.
Related
So I am rather new to programming and just recently started with Classes and we are supposed to make a phonebook that can be loaded in seperate text files.
I however keep running into the problem in this section that when I get into the for-loop. It hits a brick wall on
if storage[2] == permaStorage[i].number:
And tells me "IndexError: list index out of range". I am almost certain it is due to permaStorage starts out empty, but even when I attempt to fill it with temporary instances of Phonebook it tells me it out of range. The main reason it is there is to check if a phone number already exists within the permaStorage.
Anyone got a good tip on how to solve this or work around it?
(Sorry if the text is badly written. Just joined this site and not sure on the style)
class Phonebook():
def __init__(self):
self.name = ''
self.number = ''
def Add(name1, number1):
y = Phonebook()
y.name = name1
y.number = number1
return y
def Main():
permaStorage = []
while True:
print " add name number\n lookup name\n alias name newname\n change name number\n save filename\n load filename\n quit\n"
choices = raw_input ("What would you like to do?: ")
storage = choices.split(" ")
if storage[0] == "add":
for i in range(0, len(permaStorage)+1):
if storage[2] == permaStorage[i].number:
print "This number already exists. No two people can have the same phonenumber!\n"
break
if i == len(permaStorage):
print "hej"
try:
tempbox = Add(storage[1], storage[2])
permaStorage.append(tempbox)
except:
raw_input ("Remember to write name and phonenumber! Press any key to continue \n")
I think problem is that permaStorage is empty list and then u try to:
for i in range(0, len(permaStorage)+1):
if storage[2] == permaStorage[i].number:
will cause an error because permaStorage has 0 items but u trying to get first (i=0, permaStorage[0]) item.
I think you should replace second if clause with first one:
for i in range(0, len(permaStorage)+1):
if i == len(permaStorage):
print "hej"
try:
tempbox = Add(storage[1], storage[2])
permaStorage.append(tempbox)
if storage[2] == permaStorage[i].number:
print "This number already exists. No two people can have the same phonenumber!\n"
break
So in this case if perStorage is blank you will append some value and next if clause will be ok.
Indexing starts at zero in python. Hence, a list of length 5 has the last element index as 4 starting from 0. Change range to range(0, len(permastorage))
You should iterate upto the last element of the list, not beyond.
Try -
for i in range(0, len(permaStorage)):
The list of numbers produced in range() is from the start, but not including the end, so range(3) == [0, 1, 2].
So if your list x has length 10, range(0, len(x)) will give you 0 through 9, which is the correct indices of the elements of your list.
Adding 1 to len(x) will produce the range 0 through 10, and when you try to access x[10], it will fail.
doc = "unsorted.txt"
out_fil = "H:\Grade 11\Intro to Computer Science\sorted.txt" # Used in the Windows variation of the program
out_file = "/Applications/Banter" # Used in the Mac variation of the program
import time
def main():
order = False
blank = []
passcount = 0
starttime = time.time()
numlist = CreateList(doc)
while not order:
passcount = passcount + 1
switch = False
switchcount = 0
print "1" # These are test prints to I used to find problems
for x in range (len(numlist)):
print "2" # These are test prints to I used to find problems
if numlist[x] > numlist[x+1]:
temp = numlist[x+1]
numlist[x+1] = numlist[x]
numlist[x] = temp
switchcount = switchcount + 1
switch = True
print "9" # These are test prints to I used to find problems
elif switch == 0:
order = True
CreateFile(numlist)
endtime = time.time()
print "This list required",endtime-starttime,"seconds to sort."
print "This list required",switchcount,"switches to sort."
print "This list required",passcount,"passes to sort."
def CreateList(doc):
sort = open(doc,"r")
numlist = sort.readlines()
sort.close()
for x in range (len(numlist)):
numlist[x] = int(numlist[x].strip())
return numlist
def CreateFile(numlist):
sort = open(doc,"w")
sort.write(str(numlist)+"\n")
sort.close()
return numlist
def List(numlist):
print numlist
main()
The main purpose of my program is to sort a list of integers from a file in order using the bubble sort method, and then put that list into a new file. I'm also detailing the amount of time it takes to perform this as well as the number of passes and switches within the program that it took to sort it completely.
Now, the problem I'm having is that the list index falls out of range because it's comparing x and x+1 of my numlist. But, I need to compare x+1 in order to sort the two integers beside each others within the list. Is there any way I can fix the program so that it'll compare all the integers in the list and not try to compare the space that isn't in the list because of the x+1?
You could make your loop in this way:
for x in range ( len(numlist) -1 ):
Though stackoverflow is a great resource for answers, giving them up for homework assignments feels like cheating. I'll meet you half way: Make your loop run for one less iteration.
OK so let's say I have a situation where I have a bunch of objects in different classifications and I need to know the total possible combinations of these objects so I end up with an input that looks like this
{'raw':[{'AH':['P','C','R','Q','L']},
{'BG':['M','A','S','B','F']},
{'KH':['E','V','G','N','Y']},
{'KH':['E','V','G','N','Y']},
{'NM':['1','2','3','4','5']}]}
Where the keys AH, BG, KH, NM constitute groups, the values are list that hold individual objects and a finished group would constitute one member of each list, in this example KH is listed twice so each finished group would have 2 members of KH in it. I build something that handles this, it looks like this.
class Builder():
def __init__(self, data):
self.raw = data['raw']
node = []
for item in self.raw:
for k in item.keys():
node.append({k:0})
logger.debug('node: %s' % node)
#Parse out groups#
self.groups = []
increment = -2
while True:
try:
assert self.raw[increment].values()[0][node[increment][node[increment].keys()[0]]]
increment = -2
for x in self.raw[-1].values()[0]:
group = []
for k in range(0,len(node[:-1])):
position = node[k].keys()[0]
player = self.raw[k].values()[0][node[k][node[k].keys()[0]]]
group.append({position:player})
group.append({self.raw[-1].keys()[0]:x})
if self.repeatRemovals(group):
self.groups.append(group)
node[increment][node[increment].keys()[0]]+=1
except IndexError:
node[increment][node[increment].keys()[0]] = 0
increment-=1
try:
node[increment][node[increment].keys()[0]]+=1
except IndexError:
break
for group in self.groups:
logger.debug(group)
def repeatRemovals(self, group):
for x in range(0, len(group)):
for y in range(0, len(group)):
if group[x].values()[0] == group[y].values()[0] and x != y:
return False
return True
if __name__ == '__main__':
groups = Builder({'raw':[{'AH':['P','C','R','Q','L']},
{'BG':['M','A','S','B','F']},
{'KH':['E','V','G','N','Y']},
{'KH':['E','V','G','N','Y']},
{'NM':['1','2','3','4','5']}]})
logger.debug("Total groups: %d" % len(groups.groups))
The output of running this should clearly state my intended goal, if I have failed to do so in text. My concern is the time it takes to handle large classification of objects, when a classification has some 40 something objects in it, it is in the matrix three times and there are 7 other classifications with comparable object sizes. I think the numpy library could help me, but I am new to scientific programming and am not sure how to go about it, or if it would be worth it, could anyone provide some insight? Thank you.
Try this:
Remove duplicated values
Calculate all possibilities using permutation and factorial
Like that:
https://www.youtube.com/watch?v=Oc50d2GqXx0
I need to get user input to generate a list of 8 numbers, but when they input a number that is already in the list print and error . Without using the in function to determine if its in the list. Here's what I have so far.
def main():
myList = range(9)
a= True
for i in myList:
while a == True:
usrNum = int(input("Please enter a number: "))
if usrNum != myList[i]:
myList.append(usrNum)
print(i)
main()
Error for above code,
Scripts/untitled4.py", line 18, in main
myList.append(usrNum)
AttributeError: 'range' object has no attribute 'append'
The issue seems to be your way of generating myList. If you generate it with myList = [range(9)] you'll get:
[[0, 1, 2, 3, 4, 5, 6, 7, 8]]
Try using simply:
myList = range(9)
Also, you need to change myList.append[usrNum] with myList.append(usrNum) or you'll get a:
TypeError: 'builtin_function_or_method' object has no attribute '__getitem__'
You could also use wim's suggestion instead of the != operator:
if myList.__contains__(usrNum):
myList.append(usrNum)
There are two ways you can go about this:
Loop through the list to check each element.
The in operator is effectively doing:
for each value in the list:
if the value is what you're looking for
return True
if you reach the end of the list:
return False
If you can add that check into your code, you'll have your problem solved.
Use an alternate way of tracking which elements have been added
Options include a dict, or bits of an int.
For example, create checks = {}. When you add an value to the list, set checks[usrNum] = True. Then checks.get(usrNum, False) will return a boolean indicating whether the number already exists. You can simplify that with a collections.DefaultDict, but I suspect that may be more advanced than you're ready for.
The first is probably the result your instructor is after, so I'll give you a simple version to work with and massage to fit your needs.
myList = []
while True:
usrNum = int(input())
found = False
for v in myList:
if usrNum == v:
found = True
if not found:
myList.append(usrNum)
else:
#number was already in the list, panic!
Most instructors will be more impressed, and hence award better grades, if you can figure out how to do something like method 2, however.
You could do something like this, modify as needed (not sure when/if you want to break when the user enters a number that is already in the list, etc.)
This prompts for user input until they enter an item that already exists in the list, then it prints a message to the user, and stops execution.
def main():
mylist = range(9)
while True:
usrNum = int(input("Please enter a number: "))
if existsinlist(mylist, usrNum):
print("{} is already in the list {}".format(usrNum, mylist))
break
else:
mylist.append(usrNum)
def existsinlist(lst, itm):
for i in lst:
if itm == i:
return True
return False
Perhaps the point of this homework assignment is to help you understand how an operator like in is more efficient to read (and write, and compile) than the explicit loop that I used in the existsinlist function.
Not sure if list-comperehension would be allowable in this case, but you also could've done something like this, without relying on the existsinlist helper function:
def main():
mylist = range(9)
while True:
usrNum = int(input("Please enter a number: "))
if [i for i in mylist if i == usrNum]:
print("{} is already in the list {}".format(usrNum, mylist))
break
else:
mylist.append(usrNum)
In this case, the result of the list-comprehension can be evaluated for truthiness:
An empty list like [] results if no matching value exists, and this will be considered False
A non-empty list will result if at least one matching value exists, and this will be considered True
Yet another option which short-circuits and may be preferable:
if any(usrNum == i for i in mylist)
I'm newbie in Python and I'm struggling in create a list of sums generated by a for loop.
I got an school assignment where my program have to simulate the scores of a class of blind students in a multiple choice test.
def blindwalk(): # Generates the blind answers in a test with 21 questions
import random
resp = []
gab = ["a","b","c","d"]
for n in range(0,21):
resp.append(random.choice(gab))
return(resp)
def gabarite(): # Generates the official answer key of the tests
import random
answ_gab = []
gab = ["a","b","c","d"]
for n in range(0,21):
answ_gab.append(random.choice(gab))
return(answ_gab)
def class_tests(A): # A is the number of students
alumni = []
A = int(A)
for a in range(0,A):
alumni.append(blindwalk())
return alumni
def class_total(A): # A is the number of students
A = int(A)
official_gab = gabarite()
tests = class_tests(A)
total_score = []*0
for a in range(0,A):
for n in range(0,21):
if tests[a][n] == official_gab[n]:
total_score[a].add(1)
return total_score
When I run the class_total() function, I get this error:
total_score[a].add(1)
IndexError: list index out of range
Question is: How I valuate the scores of each student and create a list with them, because this is what I want to do with the class_total() function.
I also tried
if tests[a][n] == official_gab[n]:
total_score[a] += 1
But I got the same error, so I think I don't fully understand how lists work in Python yet.
Thanks!
(Also, I'm not a English native-speaker, so please tell me if I couldn't be clear enough)
This line:
total_score = []*0
And in fact, any of the following lines:
total_score = []*30
total_score = []*3000
total_score = []*300000000
Cause total_score to be instantiated as an empty list. It doesn't even have a 0th index, in this case! If you'd like to initiate every value to x in a list of length l , the syntax would look more like:
my_list = [x]*l
Alternatively, instead of thinking about the size before-hand, you can use .append instead of trying to access a particular index, as in:
my_list = []
my_list.append(200)
# my_list is now [200], my_list[0] is now 200
my_list.append(300)
# my_list is now [200,300], my_list[0] is still 200 and my_list[1] is now 300