About the Average Word Length problem

Question

I am struggling with this problem since yesterday. I did what i can do, but I couldn't get 5/5. No matter what i did, I always get 3/5.
can you see what I'm missing in this code?
# as a remainder, the average word length problem is: you have an input with multiple words, and the output should be the average of letters per word rounded to the nearest whole number.

And Here's the code:

import string
punc= string.punctuation

test= input()

phrase=" "
count_letter =0
count_word =0
for i in range(len(test)):
phrase +=str(test[i])
if phrase[i] == " ":
count_word +=1
if phrase[i] != " " and (phrase[i] not in punc): # not counting spaces and punctuations
count_letter +=1
print(round((count_letter +1)/count_word))

Answer 1

Try using:

(count_letter//count_word)+1

Also, im really glad i read your question. i had no idea there was a string library. I used regex to solve this.

Answer 2

Your code adapted.
I bypassed the word =" ", and iterated directly from the input.
Also a minor change to word_count =

This now works.
Thanks heaps for sharing your interesting concept

https://code.sololearn.com/cdm958D8hhbU/?ref=app

Answer 3

https://code.sololearn.com/c3tFNP5i63Vp

Answer 4

import string
punc= string.punctuation

test= input()

phrase=" "
count_letter =0
count_word =0
for i in range(len(test)):
phrase +=str(test[i])
print("character: " + phrase[i] + " ~ ")
if phrase[i] == " ":
count_word +=1
print("Adding a Word")
elif phrase[i] != " " and (phrase[i] not in punc): # not counting spaces and punctuations
count_letter +=1
print("Adding a Letter")
print(round(((count_letter)//count_word)+1))

Answer 5

I am also playing with the code and getting some discrepancies when I introduce an extra " " space into the sentence.
The code counts an extra word each time it sees chr(32), " ".

Answer 6

Try using math.ceil() instead of round()

Answer 7

also you can’t use the code i gave you for rounding because if you floor a number like 3.0 it becomes 3.0. And then you add 1, it becomes 4.

math.ceil() is the right choice. But thats not the issue.

——————-Spoiler ahead:

The issue starts all the way at the beginning of your code with the line phrase = “ “

because then you do:
phrase += str(test[i])
That means phrase is now “ word”
Which means phrase[i] = “ “
And your if statement says
if phrase[i] = “ “ then add a word!!!

To solve your issue, all you have to do is is change:
phrase = “ “
to:
phrase = “”

And you’re done. Lol, coding can be so cruel. All this over a damn whitespace.

Answer 8

you have to remove the floor and replace it with ceil and get rid of the + 1

right now you are doing this:

13//3 = 3.0 + 1 + 1 = 5
i just copied yours and removed that and it worked.

edit: nope. i was using the old one you wrote. Davids code is good for learning more methods, but i think you should try to make your code work. It’s almost there. I’ll post mine as well above yours. Your missing two conditional statements.

Answer 9

text = input()

def longPromText():
countWord = len(text.split())
countLetters = len(text.replace(' ',''))

print(countLetters / countWord)

longPromText()

Answer 10

Aymane Boukrouh has hit the nail on the head.
The task requires you to "round up" to the nearest whole number.

Answer 11

It did surely made a difference. I'm now getting 4/5, but now a new problem came out. The (count_letter) part, when doing
((count_letter +1)/count_word) test 5 fails. And when instead doing (count_letter/count_word) test 1 fails 🤷‍♂️.

Answer 12

Ivan thank you, but it didn't fix the second problem.

Answer 13

To round up to the next number you can use ceil() in the math module.

from math import ceil
print(ceil(4.1)) # output: 5

Answer 14

Ivan Thanks, I don't know if it does work because I copied the code David Ashton wrote to test it and it worked. But it is a completely different approach.

Answer 15

words = input()

len_words= len(words) - words.count(" ")
tot = len_words / (words.count(" ")+1)
print(tot)

Answer 16

Try my code for #Python

import string
import math
punc= string.punctuation

test= input()

phrase= ""
count_letter =-1
count_word =1
for i in range(len(test)):
phrase +=str(test[i])
if phrase[i] == " ":
count_word +=1
if phrase[i] != " " and (phrase[i] not in punc):
count_letter +=1
print(math.ceil((count_letter//count_word)) +1)

Answer 17

ok so i figured out your problem. First use the code that shows you the failing test case like the one i gave you. Here’s a copy of your code that i changed a little. Run it on your end and you will see what is happening that is causing you to round incorrectly.

Answer 18

'''#My Answer:
import re,math as m

x = list(map(lambda y: len(y), ''.join(re.findall('[A-Za-z ]',input())).split(' ')))

print(m.ceil((sum(x)/len(x))))

'''

import string
import math
punc= string.punctuation

test= input()

phrase= ""
count_letter =0
count_word =0
for i in range(len(test)):
phrase +=str(test[i])
print(phrase + "-----------")
if phrase[i] == " " or phrase[i] in punc:
count_word +=1
print("Add a word")
if phrase[i] != " " and (phrase[i] not in punc):
count_letter +=1
print("Add a letter")
# you need to check to see if the before a punctuation there is another punctuation. like “Hello there...”
print(math.ceil(count_letter/count_word))

Answer 19

Try my code:
Average word length:

import string
import math
punc= string.punctuation

test= input()

phrase= ""
count_letter =-1
count_word =1
for i in range(len(test)):
phrase +=str(test[i])
if phrase[i] == " ":
count_word +=1
if phrase[i] != " " and (phrase[i] not in punc):
count_letter +=1
print(math.ceil((count_letter//count_word)) +1)

Answer 20

word = input()
str = word.split()
import re, math
mystr = re.sub(r"[^A-Za-z]", "", word)
print (int(math.ceil(len(mystr)/len(str))))