Question

#Write a function called average_word_length that takes as
#input a string called my_string, and returns as output the
#average length of the words in the string.

In writing this function, note the following:
#
# - You should account for consecutive spaces. A string like
# "Hi Lucy" is two words with an average length of 3.0
# characters.
# - You should not assume the string starts with a letter.
# A string like " David" has one word with an average
# length of 5.0 characters.
# - You should not count punctuation marks toward the
# length of a word. A string like "Hi, Lucy" has two
# words with an average length of 3.0 characters: the ,
# after "Hi" does not count as a character in the word.
# The only punctuation marks you need to handle are
# these: . , ! ?
# - You may assume the only characters in the string are
# letters, the punctuation marks listed above, and spaces.
# - If my_string is not a string, you should instead return
# the string, "Not a string".
# - If there are no words in my_string, you should instead
# return the string, "No words". This could happen for
# strings like "" (an empty string) and ".,!?" (a string
# of only punctuation marks). You may assume that any
# of these punctuation marks will always be followed by
# at least one space.
#
#Here are a few hints that might help you:
#
# - You can peak at the first character in my_string with
# my_string[0]. If my_string is "Hi, Lucy", then the value
# of my_string[0] is "H". You don't have to do this, but
# you can if you want.
# - There are lots of ways you can do this. If you're
# stuck, try taking a step back and thinking about the
# problem from a fresh perspective.
# - If you're still stuck, try counting words and letters
# separately, and worrying about average length only
# after both have been counted.
# - The word count should equal the number of letters that
# come immediately after a space or the start of the
# string. The character count should simply equal the
# number of characters besides spaces and punctuation
# marks. The average word length should be character
# count divided by word count.

#Write your function here!
average_word_length(my_string)

#When your function works, the following code should
#output:
#2.0
#3.0
#4.0
#Not a string
#No words

print(average_word_length("Hi"))
print(average_word_length("Hi, Lucy"))
print(average_word_length(" What big spaces you have!"))
print(average_word_length(True))
print(average_word_length("?!?!?! ... !"))

Answer 1

# do comment if any problem arises

# Code

# this function returns true if given string contains alphabets

def has_alpha(words):

    for char in words:

        if char.isalpha():

            return True

    return False

def average_word_length(words):

    # if input is not a string

    if not isinstance(words, str):

        return "Not a string"

    # if words doesnot contains alphabets

    if not has_alpha(words):

        return "No words"

    # remove punctuations

    words = words.replace('.', '')

    words = words.replace(',', '')

    words = words.replace('!', '')

    # strip

    words=words.lstrip()

    words=words.rstrip()

    # split words by spaces

    words = words.split(' ')

    length = 0

    # calculate total length of words

    for word in words:

        length += len(word)

    # return average length

    return length/len(words)

print(average_word_length("Hi"))

print(average_word_length("Hi, Lucy"))

print(average_word_length(" What big spaces you have!"))

print(average_word_length(True))

print(average_word_length("?!?!?! ... !"))

Screenshot of code:

Output: