python: program hanging! (print issue maybe?) - python

So, I'm quite nooby at python. I decided to make a program that makes prime numbers. I know there's probably a function built in that does this but I decided to do it myself.
number = 1
numlist = list()
for x in range (0, 1000):
numlist.append("")
print "Created list entry " + str(x)
while True:
number = number + 1
if number % 2 != 0:
numscrollerA = 1
numscrollerB = 1
while numscrollerA <= number:
if float(number) / float(numscrollerA) == float(int(number)):
numlist[numscrollerA] = "true"
if float(number) / float(numscrollerA) != float(int(number)):
numlist[numscrollerA] = "false"
numscrollerA = numscrollerA + 1
while numscrollerB <= number:
if numscrollerB != 1 and numscroller != number and numlist[numscrollerB] == "true":
primestatus = "false"
else:
primestatus = "true"
if primestatus == "true":
print number
I get "Created list entry x" 1000 times as I should. Then the program just hangs.

while numscrollerB <= number:
if numscrollerB != 1 and numscroller != number and numlist[numscrollerB] == "true":
primestatus = "false"
else:
primestatus = "true"
You don't increase numscrollerB in this loop, so it runs infinitedly. Anyway, You should rather use 'for loop':
for numscrollerB in range(1, number+1):
pass # do something

Your code is very unpythonic. Typical of a newcomer experienced in a different style of coding.
Your list is uneccessary.
In python you could create the list like this
def check_even(val):
#this contains your logic
return val % 2 == 0
evenslist = [check_even(i) for i in xrange(1, 1001)]
print numlist

Related

How to solve Luhn algoritm

there is a lot of information about how to write Luhn algortim. I'm trying it too and I think that I'am very close to succes but I have some mistake in my code and dont know where. The test card is VALID card but my algorithm says otherwise. Don't you know why? Thx for help
test = "5573497266530355"
kazde_druhe = []
ostatni = []
for i in test:
if int(i) % 2 == 0:
double_digit = int(i) * 2
if double_digit > 9:
p = double_digit - 9
kazde_druhe.append(p)
else:
kazde_druhe.append(double_digit)
else:
ostatni.append(int(i))
o = sum(ostatni)
k = sum(kazde_druhe)
total = o+k
if total % 10 == 0:
print(f"Your card is valid ")
else:
print(f"Your card is invalid ")
Finally! Thank you all for your help. Now it is working :-)
test = "5573497266530355" kazde_druhe = [] ostatni = []
for index, digit in enumerate(test):
if index % 2 == 0:
double_digit = int(digit) * 2
print(double_digit)
if double_digit > 9:
double_digit = double_digit - 9
kazde_druhe.append(double_digit)
else:
kazde_druhe.append(double_digit)
else:
ostatni.append(int(digit))
o = sum(ostatni)
k = sum(kazde_druhe)
total = o+k if total % 10 == 0:
print(f"Your card is valid ")
else:
print(f"Your card is invalid ")
From this description
2. With the payload, start from the rightmost digit. Moving left, double the value of every second digit (including the rightmost digit).
You have to check the digit position, not the number itself.
Change to this:
for i in range(len(test)):
if i % 2 == 0:
This code works. :)
I fixed you code as much as i could.
test = "5573497266530355"
#test = "3379513561108795"
nums = []
for i in range(len(test)):
if (i % 2) == 0:
num = int(test[i]) * 2
if num > 9:
num -= 9
nums.append(num)
else:
nums.append(int(test[i]))
print(nums)
print((sum(nums) % 10) == 0)
I found where your code went wrong.
On the line:
for i in test:
if int(i) % 2 == 0:
It should be:
for i in range(len(test)):
if i % 2 == 0:
You should not be using the element of the string you should be using the index of the element.

How to show that using 'cache' in a Python program is more efficient?

The program uses two ways to calculate the Collatz Conjecture sequence, I want to show that the method using cache, rather than starting from scratch is more efficient - uses fewer processes. I have tried using the time module and it keeps giving inconsistent results - sometimes the cache method appears to be slower than calculating the sequence from scratch.
cache = {}
def in_cache(n):
if n in cache.keys():
return True
else:
return False
def calculate_collatz(n):
count = 0
collatz_sequence = "N: " + str(n)
while n != 1:
if n % 2 == 0:
n = n // 2
else:
n = (3 * n) + 1
count += 1
collatz_sequence += " -> {}".format(n)
return (collatz_sequence, count)
def collatz(n):
if in_cache(n):
print("Using cache...")
return cache[n]
else:
print("Calculating from scratch...")
result = calculate_collatz(n)
cache[n] = result
return result
choice = ""
while choice != 'Q':
#print('The cache: ', cache)
choice = input("Enter an integer to calculate the Collatz Conjecture sequence. Enter Q to quit\n")
try:
n = int(choice)
print(collatz(n))
except:
print("You did not enter a whole number")

simple string program doesent work, python

A friend of mine told me that she needs help with some homework, I owe her a favor so I said fine, why not. she needed help with a program that checks a sequence, if the sequence is made of the same 2 chars one after the other it will print "yes" (for example "ABABABAB" or "3$3$3$3:)
The program works fine with even length strings (for example "abab") but not with odd length one ("ububu")
I made the code messy and "bad" in purpose, computers is her worst subject so I don't want it to look obvious that someone else wrote the code
the code -
def main():
StringInput = input('your string here - ')
GoodOrBad = True
L1 = StringInput[0]
L2 = StringInput[1]
i = 0
while i <= len(StringInput):
if i % 2 == 0:
if StringInput[i] == L1:
i = i + 1
else:
GoodOrBad = False
break
if i % 2 != 0:
if StringInput[i] == L2:
i = i + 1
else:
GoodOrBad = False
break
if GoodOrBad == True:
print("yes")
elif GoodOrBad != True:
print("no")
main()
I hope someone will spot the problem, thanks you if you read everything :)
How about (assuming s is your string):
len(set(s[::2]))==1 & len(set(s[1::2]))==1
It checks that there is 1 char in the even locations, and 1 char in the odd locations.
a) Showing your friend bad and messy code makes her hardly a better programmer. I suggest that you explain to her in a way that she can improve her programming skills.
b) If you check for the character at the even position and find that it is good, you increment i. After that, you check if i is odd (which it is, since you found a valid character at the even position), you check if the character is valid. Instead of checking for odd position, an else should do the trick.
You can do this using two methods->
O(n)-
def main():
StringInput = input('your string here - ')
GoodOrBad = True
L1 = StringInput[0]
L2 = StringInput[1]
i = 2
while i < len(StringInput):
l=StringInput[i]
if(l==StringInput[i-2]):
GoodOrBad=True
else:
GoodOrBad=False
i+=1
if GoodOrBad == True:
print("yes")
elif GoodOrBad == False:
print("no")
main()
Another method->
O(1)-
def main():
StringInput = input('your string here - ')
GoodOrBad = True
L1 = set(StringInput[0::2])
L2 = set(StringInput[1::2])
if len(L1)==len(L2):
print("yes")
else:
print("no")
main()
There is a lot in this that I would change, but I am just showing the minimal changes to get it to work. There are 2 issues.
You have an off by one error in the code:
i = 0
while i <= len(StringInput):
# in the loop you index into StringInput
StringInput[i]
Say you have 5 characters in StringInput. Because your while loop is going from i = 0 to i < = len(StringInput), it is going to go through the values [0, 1, 2, 3, 4, 5]. That last index is a problem since it is off the end off StringInput.
It will throw a 'string index out of range' exception.
You need to use:
while i < len(StringInput)
You also need to change the second if to an elif (actually it could just be an else, but...) so you do not try to test both in the same pass of the loop. If you go into the second if after the last char has been tested in the first if it will go out of range again.
elif i % 2 != 0:
So the corrected code would be:
def main():
StringInput = input('your string here - ')
GoodOrBad = True
L1 = StringInput[0]
L2 = StringInput[1]
i = 0
while i < len(StringInput):
if i % 2 == 0:
if StringInput[i] == L1:
i = i + 1
else:
GoodOrBad = False
break
elif i % 2 != 0:
if StringInput[i] == L2:
i = i + 1
else:
GoodOrBad = False
break
if GoodOrBad == True:
print("yes")
elif GoodOrBad != True:
print("no")
main()
def main():
StringInput = input('your string here - ')
MaxLength = len(StringInput) // 2 + (len(StringInput) % 2 > 0)
start = StringInput[:2]
chained = start * MaxLength
GoodOrBad = chained[:len(StringInput)] == StringInput
if GoodOrBad == True:
print("yes")
elif GoodOrBad != True:
print("no")
I believe this does what you want. You can make it messier if this isn't bad enough.

String Conversion Error "String index out of range" in python

Hello I'm asking about the conversion issue with strings and integers.
I have a piece code that requires the conversion between strings and integers but I just can't get it to work, I wonder if anyone could help me. I receive String index out of range error.
Here is the code
def ISBN(bc):
total = 0
up = 0
down = 11
for x in range(10):
sbc = str(bc)
ibc = int(sbc[up])
total += (ibc * down)
#total += (int(sbc[up])*down)
up += 1
down -+ 1
mod = (total % 12)
if mod == 10:
total = "x"
print ("The ISBN book code is: " + bc + total)
w = 0
while w == 0:
a = int(input("Please input the 10 digit book number:\n"))
b = str(a)
if len(b) == 9:
ISBN(a)
else:
print ("Sorry book code not 10 digits long")
restart = input("Would you like to use the book code changer again?\n")
restart = restart.lower
if restart == "yes" or restart == "y":
print ("--------------------------------------------------\n")
elif restart == "no" or restart == "n":
print ("Thank you for using the ISBN book code changer\n")
w = 1
if len(b) == 9:
This should be 10.
mod = (total % 12)
This should be 11.

determinating if the input is even or odd numbers

Hello I am trying to write a program in python that asks the user to input a set of numbers of 1's and 0's and I want the program to tell me if I have and even number of zeros or an odd number of zeros or no zero's at all. Thanks for your help!!
forstate = "start"
curstate = "start"
trans = "none"
value = 0
print "Former state....:", forstate
print "Transition....:", trans
print "Current state....", curstate
while curstate != "You hav and even number of zeros":
trans = raw_input("Input a 1 or a 0: ")
if trans == "0" and value <2:
value = value + 1
forstate = curstate
elif trans == "1" and value < 2:
value = value + 0
forstate = curstate
curstate = str(value) + " zeros"
if value >= 2:
curstate = "You have and even number of zeros"
print "former state ...:", forstate
print "Transition .....:", trans
print "Current state....", curstate
Looks like you're trying to do a finite state machine?
try:
inp = raw_input
except NameError:
inp = input
def getInt(msg):
while True:
try:
return int(inp(msg))
except ValueError:
pass
START, ODD, EVEN = range(3)
state_next = [ODD, EVEN, ODD]
state_str = ['no zeros yet', 'an odd number of zeros', 'an even number of zeros']
state = START
while True:
num = getInt('Enter a number (-1 to exit)')
if num==-1:
break
elif num==0:
state = state_next[state]
print 'I have seen {0}.'.format(state_str[state])
Edit:
try:
inp = raw_input
except NameError:
inp = input
START, ODD, EVEN = range(3)
state_next = [ODD, EVEN, ODD]
state_str = ['no zeros yet', 'an odd number of zeros', 'an even number of zeros']
def reduce_fn(state, ch):
return state_next[state] if ch=='0' else state
state = reduce(reduce_fn, inp('Enter at own risk: '), START)
print "I have seen " + state_str[state]
It sounds like homework, or worse an interview questions, but this will get you started.
def homework(s):
counter = 0
if '0' in s:
for i in s:
if i == '0':
counter = counter + 1
return counter
don't forget this part over here
def odd_or_even_or_none(num):
if num == 0:
return 'This string contains no zeros'
if num % 2 == 0
return 'This string contains an even amount of zeros'
else:
return 'This string contains an odd amount of zeros'
if you call homework and give it a string of numbers it will give you back the number of 0
homework('101110101')
now that you know how many 0s you need to call odd_or_even_or_none with that number
odd_or_even_or_none(23)
so the solution looks like this
txt = input('Feed me numbers: ')
counter = str( homework(txt) )
print odd_or_even_or_none(counter)
try:
inp = raw_input
except NameError:
inp = input
zeros = sum(ch=='0' for ch in inp('Can I take your order? '))
if not zeros:
print "none"
elif zeros%2:
print "odd"
else:
print "even"
The simple solution to your problem is just to count the zeros, then print a suitable message. num_zeros = input_stream.count('0')
If you're going to build a finite state machine to learn how to write one, then you'll learn more writing a generic FSM and using it to solve your particular problem. Here's my attempt - note that all the logic for counting the zeros is encoded in the states and their transitions.
class FSMState(object):
def __init__(self, description):
self.transition = {}
self.description = description
def set_transitions(self, on_zero, on_one):
self.transition['0'] = on_zero
self.transition['1'] = on_one
def run_machine(state, input_stream):
"""Put the input_stream through the FSM given."""
for x in input_stream:
state = state.transition[x]
return state
# Create the states of the machine.
NO_ZEROS = FSMState('No zeros')
EVEN_ZEROS = FSMState('An even number of zeros')
ODD_ZEROS = FSMState('An odd number of zeros')
# Set up transitions for each state
NO_ZEROS.set_transitions(ODD_ZEROS, NO_ZEROS)
EVEN_ZEROS.set_transitions(ODD_ZEROS, EVEN_ZEROS)
ODD_ZEROS.set_transitions(EVEN_ZEROS, ODD_ZEROS)
result = run_machine(NO_ZEROS, '01011001010')
print result.description

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