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Take some inputs until 0 encountered, print each word in lowercase and break the loop

Write a program to input the lines of text, print them to the screen after converting them to lowercase. The last line prints the number of input lines.
The program stops when it encounters a line with only zero.
Requirements: use infinite loop and break . statement
here's my code but it's saying Input 9 is empty
a = 1
l = 0
while a != 0:
a = input()
a.lower()
print(a)
l+=1
example inputs
TbH
Rn
ngL
bRb
0

I'd suggest this, it's a combination of these 2 comments, sorry
count = 0
while True:
a = input()
if a == '0':
break
else:
print(a.lower())
count += 1
print(count)

This may accomplish what you are trying to achieve:
def infinite_loop():
while True:
user_input = input('enter a value:\n> ')
if user_input == '0':
break
else:
print(user_input.lower())
if __name__ == '__main__':
infinite_loop()

There a few errors in your original code, here is some suggestions and fixes. This post is try to follow your code as much as it can, and point the changes needed.
Please see the comments and ask if you have any questions.
count = 0 # to count the lines
while w != '0': # input's string
w = input().strip() # get rid of \n space
word = w.lower() # convert to lower case
print(word)
count += 1
# while-loop stops once see '0'
Outputs: (while running)
ABBA
abba
Misssissippi
misssissippi
To-Do
to-do
0
0

How do I count the number of capital letters using a while loop

s = input()
i = 0
while i < len(s) and (s[i]) < "A" or "Z" < s([i]):
print(i)
I keep getting this wrong and I'm not sure what to do. I Do not want to use a for loop just a while loop. Thank you

You can do it by many ways.
If I were you I will do it using isupper() and sum() generator,
s = input("Type something...")
print(sum(1 for c in s if c.isupper()))
Using while as you asked,
s = input("Type something...")
i = 0
capital = 0
while i < len(s):
if s[i].isupper():
capital+=1
i = i + 1
print(capital)

Your while loop is currently written such that it will terminate at the first lowercase letter. You need the loop to go over the entire string, but keep count of uppercase letters as it goes.
s = input()
i = 0
c = 0
while i < len(s):
if "A" <= s[i] <= "Z":
c = c + 1 # c only goes up on capital letters
i = i + 1 # i goes up on every letter
print(i, c)
print(f"Capital letters: {c}")
An easier method is to use the sum function along with isupper:
s = input()
print(f"Capital letters: {sum(c.isupper() for c in s)}")

You are using the while for both the limit and the counting which won't work.
You have to use the while for the limit and an if for the counting:
s = input()
i = 0
count = 0
while i < len(s):
print(i)
if "A" <= s[i] <= "Z":
count += 1
i = i + 1
print(f'Capitals in "{s}" = {count}')
However, this code is very complicated and better is the answer from #AlwaysSunny or the comment from #Samwise

def Capital(In):
return sum([l.isupper() for l in In])
print(Capital(input()))

Try use this:
text = input()
count=0
for letter in text:
if letter == letter.upper():
count+=1
print(letter, end=" ")
print(count-1)
I hope I have explained clearly)

I wished to check if an input is in the string and then print the output

I intended to let the program check if the input matches with any character in a str and then print out the result, the player input and the underscores in the correct places. This is my test code so far:
astring = "apple"
bstring = "_ " * 5
print(bstring)
my_input = input("enter a letter")
for i, n in enumerate(astring):
if my_input == i:
bstring[n] = my_input
else:
i = i + 1
print(bstring)
However, only the underscores are being printed out. Can anyone help me?

In your loop, you should be checking to see if the letter at your current index of your string is the same as the letter at the current index of your input string, to do this you can use:
if i < len(my_input) and my_input[i] == n:
Also, strings in Python are immutable, and so you can't change them via index. Instead, use an array of _, so that you can change what is at a particular index. Then, at the end, join each element in your list by a space.
Lastly, there is no need to increment i, as this is done for you by your for loop:
astring='apple'
bstring=['_']*len(astring)
print(bstring)
my_input = input('enter a letter')
for i,n in enumerate(astring):
if i < len(my_input) and my_input[i] == n:
bstring[i] = n
print(' '.join(bstring))

for i,n in enumerate(astring):
'i' is the index, 'n' is the character. You have it the other way around in 'if'.

hope it will help you
astring='apple'
bstring=["_" for i in range(len(astring))]
print(bstring)
my_input=input('enter a letter')
for i,n in enumerate(astring):
if my_input==n:
bstring[i]=my_input
else:
i=i+1
print(*bstring)

How to fix a String index out of range exception in Python

There is some issue with my python code. I am making a program that finds the occurrences of the letter A in a word and if that letter is found and the next letter is not the letter A the A is swapped with the next letter.
As an example TAN being TNA but WHOA staying as WHOA
AARDVARK being ARADVRAK
The issue is when I input ABRACADABRA I get a string index out of range exception. Before I had that exception I had the word that prints it as BRACADABRIi'm not sure why if I have to add another loop in my program.
If you guys also have anymore efficient way to run the code then the way I have please let me know!
def scrambleWord(userInput):
count = 0
scramble = ''
while count < len(userInput):
if userInput[count] =='A' and userInput[count+1] != 'A':
scramble+= userInput[count+1] + userInput[count]
count+=2
elif userInput[count] != 'A':
scramble += userInput[count]
count+=1
if count < len(userInput):
scramble += userInput(len(userInput)-1)
return scramble
#if a is found switch the next letter index with a's index
def main():
userInput = input("Enter a word: ")
finish = scrambleWord(userInput.upper())
print(finish)
main()

When you get to the end of the string and it is an 'A' your program is then asking for the next character which is off the end of the string.
Change the loop so it doesn't include the last character:
while count < len(userInput)-1:
if ...

You can modify your code as below:
def scrambleWord(userInput):
count = 0
scramble = ''
while count < len(userInput):
if count < len(userInput)-1 and userInput[count] =='A' and userInput[count+1] != 'A':
scramble+= userInput[count+1] + userInput[count]
count+=2
else:
scramble += userInput[count]
count+=1
return scramble
You are not checking the condition (count < len(userInput)-1) when logic tries to check for A's occurrence and swap with next letter. It throws string index out of range exception.

The issue arises in your code when last character in input is 'A'.
This is because your first if in the loop tries to access 'count + 1' character during last iteration.
And since there's no character at that position, you get index error.
The simplest solution would be to make a separate if condition for the same.
Updated snippet for while loop might look like this -
# while start
while count < len_: # len_ is length of input
if count + 1 >= len_:
break # break outta loop, copy last character
current = inp[count]
next_ = inp[count + 1]
if current == 'A':
op += ( next_ + current) # op is result
count += 1
else:
op += current
# increment counter by 1
count += 1
# rest of the code after while is same
Another small issue in your code is while copying last character ( after loop ends ), you should use [ ] instead of ( ) to refer last character in input string.

Just for fun :
from functools import reduce
def main():
word = input("Enter a word: ").lower()
scramble = reduce((lambda x,y : x[:-1]+y+'A' \
if (x[-1]=='a' and y!=x[-1]) \
else x+y),word)
print(scramble.upper())
main()

how to check repeating Vowels

I'm pretty new to python and I'm having trouble with my
if then else statements and I only get is "no repeating vowels" which mean my rep_vowel is still returning 0
so the program rules are as follows.
if no vowel appears next to itself (e.g. hello), then print:
no vowel repeats
if exactly one vowel is repeated in sequence at least once (e.g. committee) then print a message that indicates which vowel repeats:
only vowel e repeats
if more than one vowel repeats (e.g. green door) then print:
more than one vowel repeats
ignore upper case - lower case differences: assume all the input is always lowercase
answer = input("Enter a string: ")
rep_vowel = 0
i = 0
length_Answer = len(answer)
next_string = 1
curChar = answer[0+rep_vowel]
for i in range(0,length_Answer):
if answer[0 + i] in ["a","e","i","o","u"]:
i =+ 1
next_string = answer[0+i+i]
if next_string == answer:
rep_vowel =+ 1
if rep_vowel == 0:
print("no repeating vowles")
elif rep_vowel > 1:
print("more than 1 repeating vowels")
else:
print ("the letter "+ str(curChar) +" repeats")

You have a few mistakes so i'll try to address several of them:
You do a lot of [0 + something] indexing, which is useless, since 0 + something always equals to something, so yo should just do indexing with [something]
Changing the value of i with i += 1 is bad because you are already increasing it as part of the loop
All you have to do to find a match is simply match the current letter to the next one, if both are the same and they are also vowels, you've found a match.
You are initializing unnecessary variables such as i = 0 only to have them overridden in the next lines
Adding all of those together:
answer = input("Enter a string: ")
vowels = "aeiou"
repeats = [] # this list will hold all repeats of vowels
for i in range(len(answer) - 1): # i'll explain the -1 part at the end
if answer[i] in vowels and answer[i] == answer[i + 1]:
repeats.append(answer[i])
if len(repeats) == 0:
print("no repeating vowles")
elif len(repeats) > 1:
print("more than 1 repeating vowels")
else:
print("the letter " + repeats[0] + " repeats")
This still doesn't take every possible input into account, but it should get you started on a final solution (or perhaps that's enough). For example, input of teest will give the correct result but the input of teeest doesn't (depends on your definition of correct).
About the len(answer-1) range, that's only to make sure we don't go out of bounds when doing answer[i + 1], so we're stopping on the next to last letter instead.

Firstly, you have to indent your code.
to say if (condition) then do print('hello') you write it this way:
if condition:
print('hello')
Secondly, you are using i =+ 1 which is the same as i=1
I think you meant i +=1 which is i = i+1
Finally, I suggest this code:
answer = input("Enter a string: ")
vowel_repeated_count = 0
length_Answer = len(answer)
i=0
while (i <length_Answer-1):
#we check if it's a vowel
if answer[i] in ["a","e","i","o","u"]:
#we check if it's followed by the same vowel
if answer[i+1] == answer[i]:
#increment the vowel_repeated_count
vowel_repeated_count +=1
#we save the vowel for the display
vowel = answer[i]
#we skip the other same repeated vowels
#example: abceeed, we skip the third e
while (answer[i] == vowel and i < length_Answer-1):
i +=1
#we add this incrementation because we're in a while loop
i +=1
if vowel_repeated_count == 0:
print("no repeating vowles")
elif vowel_repeated_count == 1:
print("the letter "+ str(vowel) +" repeats")
else:
print ("more than 1 repeating vowels")

You have some logical errors. It's time consuming to edit that. You can try this, I have modified your code. Hope it will work for you. I have commented beside every important line.
answer = input("Enter a string: ")
is_found = {} #a dictionary that will hold information about how many times a vowel found,initially all are 0
is_found["a"]=0
is_found["e"] = 0
is_found['i']=0
is_found['o']=0
is_found['u']=0
vowels =["a","e","i","o","u"]
for i in range(0,len(answer)):
if answer[i] in vowels:
is_found[answer[i]] = is_found[answer[i]]+1 # if a vowel found then increase its counter
repeated=0 #let 0 repeated vowel
previously_repeated=False #to trace whether there is a previously repeated character found
curChar=None
for key,value in is_found.items(): #iterate over dictionary
if previously_repeated and value>1: #if a vowel found and previously we have another repeated vowel.
repeated=2
elif previously_repeated==False and value>1: # we don't have previously repeated vowel but current vowel is repeated
curChar=key
previously_repeated=True
repeated=1
if repeated== 0:
print("no repeating vowles")
elif repeated> 1:
print("more than 1 repeating vowels")
else:
print ("the letter "+ str(curChar) +" repeats")

There is no need to increment your counter i. In your for loop, it will increment itself each time it goes through the for loop. Also, you need a variable to keep track of how many times the vowel repeats.
answer = input("Enter a string: ")
rep_vowel = 0
length_Answer = len(answer)
vowelList=["a","e","i","o","u"]
vowelRepeated = []
#this will go from i=0 to length_Answer-1
for i in range(length_Answer):
if (answer[i] in vowelList) and (answer[i+1] in vowelList):
if (answer[i] == answer[i+1]):
vowelRepeated.append(answer[i])
repVowel += 1
if rep_vowel==0:
print("no repeating vowels")
elif rep_vowel==1:
print("only one vowel repeated:")
print(vowelRepeated)
else:
print("multiple vowels repeated:")
print(vowelRepeated)

for such counting, I will prefer to use a dictionary to keep the counting number. Your code has been modified for your reference
answer = input("Enter a string: ")
length_Answer = len(answer)
count = dict()
for i in range(length_Answer):
if answer[i] in ["a","e","i","o","u"]:
if answer[i+1] == answer[i]:
if answer[i] in count:
count[answer[i]] += 1
else:
count[answer[i]] = 1
rep_vowel = len(count)
if rep_vowel == 0:
print("no repeating vowles")
elif rep_vowel > 1:
print("more than 1 repeating vowels")
else:
for k in count:
vowel = k
print("the letter " + vowel + " repeats")

You have a few issues with your solution :
1) You never use curChar, i'm guessing you wanted to enter the next_string value into it after the '==' statement.
2) You compare your next_string to answer, this will always be a false statement.
3) Also no need to use [0+i], [i] is good enough
Basically what you want to do is this flow :
1) Read current char
2) Compare to next char
3) If equal put into a different variable
4) If happens again raise a flag
Optional solution :
vowel_list = ["a","e","i","o","u"]
recuring_vowel_boolean_list = [answer[index]==answer[index+1] and answer[index] in vowel_list for index in range(len(answer)-1)]
if not any(recuring_vowel_boolean_list ):
print("no repeating vowels")
elif (recuring_vowel_boolean_list.count(True) > 1):
print("More then 1 repeating vowels")
else:
print("The letter {} repeats".format(answer[recuring_vowel_boolean_list.index(True)]))