Python: while should stop iterating if 1 of 2 conditions are reached? - python

I have set a very simple example here:
def sourcecode(a,b,c,alternatief=False):
volume_a, volume_b = 0, 0
while volume_a != c or volume_b != c:
print(volume_a,volume_b)
volume_a += 1
volume_b += 2
if volume_a == c:
return volume_a
elif volume_b == c:
return volume_b
print(sourcecode(7,3,5))
Of course this will be an infinite loop as always 1 of the two will never reach c.
But what I try to figure out is how to write my while statement to stop if one of two conditions are met.


It is a relatively simple tweak:
while volume_a != c and volume_b != c:
^
This requires both to be True to continue, so stops looping if either one becomes False.
With or:
True or True == True
True or False == True
False or True == True
False or False == False # need both to match to break out
With and:
True and True == True
True and False == False # break out here
False and True == False # or here
False and False == False


You need to use and, not or. When one of the condition is False, the loop will quit:
while volume_a != c and volume_b != c:
print(volume_a,volume_b)
volume_a += 1
volume_b += 2
If you're using or, it will stop only when both conditions are False. By using and, the loop will continue only if both conditions are True.
Hope this helps!


others response are good but you could simply use an infinite loop and returning from it
def sourcecode(a,b,c,alternatief=False):
volume_a, volume_b = 0, 0
while True:
if volume_a == c:
return volume_a
if volume_b == c:
return volume_b
print(volume_a,volume_b)
volume_a += 1
volume_b += 2
print(sourcecode(7,3,5))

Related

***Time limit exceeded*** error on python program

when i try to print this line:
print(perfect_square(0))
i should get True but instead i get a time limit exceeded error and i dont know how to fix it.
i tried chaging it to an elif statment instead of 2 separate if statements but i still get that error
This is my current code:
def perfect_square(n):
s = 1
while s != n:
if s*s == n:
return True
elif s == 0:
return True
else:
s +=1
return False
def perfect_cube(n):
s = 1
while s != n:
if s*s * s == n:
return True
elif s == 0:
return True
else:
s +=1
return False

Seems quite clear to me why the perfect_square(0) and perfect_cube(0) cases cause an infinite loop. You start s=1 and always increment it s+=1. It will never be equal to n=0 so you get an infinitely running program. Maybe try making checks for invalid values of n?
def perfect_cube(n):
if n < 1: return False
# ...

CS50P PSETS 2, Vanity plates

I'm having some issues with the psets 2 of cs50p, precisely I'm talking about the "Vanity Plates" problem, where I fulfilled all requests except one, which said:
“Numbers cannot be used in the middle of a plate; they must come at the end. For example, AAA222 would be an acceptable … vanity plate; AAA22A would not be acceptable. The first number used cannot be a ‘0’.” Can you help me? Thank's
this is the code I wrote so far:
def main():
plate = input("Plate: ")
if is_valid(plate):
print("Valid")
else:
print("Invalid")
def is_valid(s):
if s.isalnum() | s[:2].isalpha() | 2 < len(s) < 6 | :
else:
return False
main()

you have to consider all the cases one by one, this is how I solved it:
def main():
plate = input("Plate: ")
if is_valid(plate):
print("Valid")
else:
print("Invalid")
def is_valid(s):
if len(s) < 2 or len(s) > 6:
return False
elif not s[0].isalpha() or not s[1].isalpha():
return False
elif checkFirstZero(s):
return False
elif checkMiddleZero(s):
return False
elif last(s):
return False
elif worng(s):
return False
return True
def last(s):
isAlp = False
isNum = False
for w in s:
if not w.isalpha():
isNum = True
else:
if isNum:
return True
return False
def checkCuntuNNumber(s):
isFirstTry = True
isNum = False
for w in s:
if not w.isalpha():
if isFirstTry:
isNum = True
isFirstTry = False
if isNum and s[-1].isalpha():
return True
def checkMiddleZero(s):
isFirstTry = True
isNum = False
for w in s:
if not w.isalpha():
if isFirstTry:
isNum = True
isFirstTry = False
if isNum and s[-1].isalpha():
return True
else:
return False
def checkFirstZero(s):
for w in s:
if not w.isalpha():
if int(w) == 0:
return True
else:
return False
def worng(s):
for w in s:
if w in [" ", ".", ","]:
return True
return False
main()

This is how I did it. I am sure there is an easier way to do it out there but hopefully this helps :)
characters = ['A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z']
numbers = ['1','2','3','4','5','6','7','8','9','0']
def main ():
plate = (input ("Plate: ")).upper()
if is_valid(plate):
print ('Valid')
else:
print ('Invalid')
def is_valid (s):
#Check whether length is between 2 and 6 included
if len(s) < 2 or len(s) > 6:
return False
elif char_check(s):
return False
elif char_start(s):
return False
elif zero_check(s):
return False
elif alpha_follow_check (s):
return False
else:
return True
#Check for valid characters
def char_check(s):
for i in s:
if not (i in characters or i in numbers):
return True
#Check whether first two are letters
def char_start (s):
for i in s[:2]:
if not i in characters:
return True
#Check if zero is first number listed
def zero_check (plate_response):
length_string = len (plate_response)
letter_position = 0
number_present = 0
zero_position = None
if any (i in numbers for i in plate_response):
for i in plate_response [0:length_string]:
if i == '0':
zero_position = letter_position
break
letter_position = letter_position + 1
for i in plate_response [0:zero_position]:
if i in numbers:
number_present = 1
if number_present == 0:
return True
else:
return False
#Check alphabet follows numbers
def alpha_follow_check (plate_response):
length_string = len (plate_response)
letter_position = 0
number_position = None
if any (i in numbers for i in plate_response):
for i in plate_response [0:length_string]:
if i in numbers:
number_position = letter_position
break
letter_position = letter_position + 1
for i in plate_response [number_position:length_string]:
if i in characters:
return True
else:
return False
main ()

idk if will help, but the part that i've had the most difficulty in this problem was: "Numbers cannot be used in the middle of a plate; they must come at the end, AAA22A would not be acceptable", then i learned that you can create a full list from the plate that the user inputed, and how to actually use it, with the:
ls = list(s)
for i in range(len(ls)):
After that, we check when the first number appears. "if == '0'" ,then returns False to the function.
After that, if the first number isn't a 0, the program checks if the next item in that list is letter, and, if it is, also return False.
i < len(ls) -1 => this part guarantee that the program will not run in the last item of the list
ls[i+1].isalpha() => and this part check that, if the item on the list was a number, and then the next item is a letter, it returns False
I hope it helps someone, i've spend a lot of time trying to figure it out what to do, and then reached this solution: "for i in range(len(ls))".
Now my code is complete and working.
My code:
def main():
plate = input("Plate: ")
if is_valid(plate):
print("Valid")
else:
print("Invalid")
def is_valid(s):
if not s.isalnum():
return False
elif len(s) < 4 or len(s) > 7:
return False
elif s[0].isdigit()or s[1].isdigit():
return False
elif s[-1].isalpha() or s[-2].isalpha():
return False
else:
ls = list(s)
for i in range(len(ls)):
if ls[i].isdigit():
if ls[i] == '0':
return False
elif i < len(ls) -1 and ls[i+1].isalpha():
return False
else:
return True
main()

Finding Cyclops Numbers in Python

I am working on a problem in which I need to return True or False after checking to see whether a number is a cyclops number or not. A cyclops number is made up of odd number of digits, consists of only one zero and that zero is located in the middle. Here's what I have so far:
def is_cyclops(n):
strNum = str(n)
for i, el in enumerate(strNum):
if(len(strNum) % 2 == 0):
return False
else:
# find middle number is zero
# no other zeros exist
# return True
is_cyclops(0) # True
is_cyclops(101) # True
is_cyclops(1056) # False
is_cyclops(675409820) # False
How can I find the median number (without using numpy) & ensure it is a zero, and it is the only zero that exists in that number?

This worked for me:
def is_cyclops(num: int) -> bool:
str_ = str(num)
if not len(str_) % 2:
return False
if not str_.count('0') == 1:
return False
mid_index = len(str_) // 2
if str_[mid_index] == '0':
return True
return False
print(
is_cyclops(0),
is_cyclops(101),
is_cyclops(1056),
is_cyclops(675409820)
)
Output:
True True False False

As it looks like you've had a good attempt here, I'll help out.
def is_cyclops(n):
strNum = str(n)
if(len(strNum) % 2 == 0):
return False
else:
middle_index = len(strNum)//2
if strNum[middle_index] != "0": return False # find middle number is zero
if strNum.count("0") > 1: return False # no other zeros exist
return True
is_cyclops(0) # True
is_cyclops(101) # True
is_cyclops(1056) # False
is_cyclops(675409820) # False

Return true doesn't work but return false does

How come
def allEven(n):
for c in n:
if c % 2 != 0:
return False
return True
works but
def allEven(n):
for c in n:
if c % 2 == 0:
return True
return False
doesn't?
With the second one, when I type allEven([8, 0, -1, 4, -6, 10]), it says it's True.

In your second method, you return True once you find an even c in n which is not what your method supposed to do: return true IFF all cs are even.

Return Statement terminate loop and end program. If c % 2 == 0 is True, it Return True and terminate program not check all values.
Try This
def allEven(n):
for c in n:
if c % 2 == 0:
continue
else:
return False
return True

In a nested if loop, how to return true for multiple matching conditions?

In the below program, even though all the if conditions are matching, it returns true just once. How do i make it return true and print as many times as the conditions match?
lotto_numbers = [1,1,1]
fireball_number = 1
user_input1 = user_input2 = user_input3 = 1
def fbcheck():
if lotto_numbers == [user_input1,user_input2,fireball_number]:
return True
elif lotto_numbers == [fireball_number, user_input2, user_input3]:
return True
elif lotto_numbers == [user_input1, fireball_number, user_input3]:
return True
else:
return False
if (fbcheck() == True):
print ('you won')

You can use all:
def fbcheck():
user_data = [user_input1,user_input2,fireball_number]
lotto_numbers = [1,1,1]
print([a==b for a, b in zip(lotto_numbers, user_data)])
return all(a==b for a, b in zip(lotto_numbers, user_data))
print(fbcheck())
Output:
[True, True, True]
True

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