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ollegr [7]
2 years ago
7

I WILL MARK BRAINLEST

Computers and Technology
1 answer:
mezya [45]2 years ago
5 0

Answer: I think its the last one, there is often more than one right choice :)

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Program:
Mkey [24]

Answer:

See explaination

Explanation:

#method to print menu & handle user choice

def print_menu(input_str):

#printing menu

print('MENU')

print('c - Number of non-whitespace characters')

print('w - Number of words')

print('f - Fix capitalization')

print('r - Replace punctuation')

print('s - Shorten spaces')

print('q - Quit\n')

#getting choice

choice=input('Choose an option: ').lower()

#identifying choice

if choice=='c':

#displaying number of non white space chars in input_str

print('Number of non-whitespace characters:',get_num_of_non_WS_characters(input_str))

elif choice=='w':

#displaying number of words in input_str

print('Number of words:',get_num_of_words(input_str))

elif choice=='f':

#fixing capitalization and getting updated string and count of values capitalized

input_str,count=fix_capilization(input_str)

#displaying results

print('Number of letters capitalized:',count)

print('Edited text:',input_str)

elif choice=='r':

#replacing punctuation, displaying updated text

input_str=replace_punctuation(input_str)

print('Edited text:', input_str)

elif choice=='s':

#shortening spaces and displaying updated text

input_str = shorten_space(input_str)

print('Edited text:', input_str)

#returning choice and input_str

return choice,input_str

#returns the number of non white space chars in input_str

def get_num_of_non_WS_characters(input_str):

count=0

#looping through each character in input_str

for i in input_str:

if not i.isspace():

#i is not a space

count+=1

return count

#returns the number of words in input_str

def get_num_of_words(input_str):

#splitting words into list of tokens by space

words=input_str.split(' ')

count=0

#counting all non empty strings in words list

for i in words:

if len(i)>0:

count+=1

return count

#method to fix capitalization and return updated string and count of letters updated

def fix_capilization(input_str):

count=0

beginning=True #starting letter should be capitalized

result=''

for i in input_str:

if beginning and i.isalpha():

#start of a sentence and i is alphabetic

if i.islower():

#converting i to upper case, incrementing count

i=i.upper()

count+=1

result+=i

#not start of a sentence

beginning=False

elif i in '?.!':

#i is either ? or . or !, next letter should be capitalized

beginning=True

result+=i

else:

#any other character

result+=i

return result,count

#method to replace exclamation and semicolons with period and comma respectively

def replace_punctuation(input_str,exclamationCount =0,semicolonCount=0):

result=''

for i in input_str:

if i=='!':

i='.'

exclamationCount+=1

elif i==';':

i=','

semicolonCount+=1

result+=i

print('Punctuation replaced')

#displaying replaced values counts

print('exclamationCount:',exclamationCount)

print('semicolonCount:',semicolonCount)

return result

#removes all double or more spaces in input_str

def shorten_space(input_str):

input_str=input_str.strip()

result=''

prev=None

for i in input_str:

if prev==None:

result+=i

elif i==' ':

if prev != ' ':

result+=i

else:

result+=i

prev=i

return result

if __name__ == '__main__':

#getting input, printing it

input_str=input('Enter a sample text:\n')

print('\nYou entered:',input_str)

choice=' '

#looping until choice becomes q

while choice!='q':

choice,input_str=print_menu(input_str)

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n200080 [17]

There are 0.5 strides in one girth
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hope it helps
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Question 3 / 5
Andrej [43]

Answer:

False

Explanation:

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