Requires assistance with vowel checker program

Question

Write a python program that asks the user for a word, then calculates and prints the vowel worth of the entered word using these instructions: Suppose you calculate the vowel worth of a word based on the following instructions:

a   5 points
e   4 points
i   3 points
o   2 points
u   1 point.

My code:

word = str(input("Enter a word:"))

def vowel(Word):
    global word
    Word = word
    score = 1

    if "a" or "A" in word:
        score += 5
    elif "e" or "E" in word:
        score += 4
    elif "i" or "I" in word:
        score += 3
    elif "o" or "O" in word:
        score += 2
    elif "u" or "U" in word:
        score += 1

    print("Your word scored",score,"in the vowel checker")

print(vowel(word))

Edit: FOR LOOP

word = input("Enter a word:")

def vowel(Wo_rd): global word Wo_rd = word score = 0 for char in word.lower(): if char == 'a' or "A": score += 5 elif char == "e" or "E": score += 4 elif char == "i" or "I": score += 3 elif char == "o" or "O": score += 2 elif char == "u" or "U": score += 1 a = "Your word scored",score," in the word checker test" return a

print(vowel(word))

word = str(input("Enter a word:")) def vowel(Word): global word Word = word score = 1 if "a" or "A" in word: score += 5 elif "e" or "E" in word: score += 4 elif "i" or "I" in word: score += 3 elif "o" or "O" in word: score += 2 elif "u" or "U" in word: score += 1 print("Your word scored",score,"in the vowel checker") print(vowel(word)) — JackAnderson, Jan 16 '17 at 16:40
show in question your code and FULL error message (Traceback) — furas, Jan 16 '17 at 16:40
`if "a" in word or "A" in word:` or shorter `if "a" in word.lower():` — furas, Jan 16 '17 at 16:41
if a appears twice in a word are you supposed to add 10 to the score? — depperm, Jan 16 '17 at 16:42
There is no error message. Just when i enter the word "Hello" it outputs 16, which it shouldnt — JackAnderson, Jan 16 '17 at 16:42
`if` and `elif` are exclusive. If one is `True`, the others aren't checked. — roganjosh, Jan 16 '17 at 16:42
better use `for` to get char-by-char from word and check this char with vowel — furas, Jan 16 '17 at 16:43
if a appears twice in a word are you supposed to add 10 to the score? — JackAnderson, Jan 16 '17 at 16:43
for char in word.lower(): if char == 'a': i will try this. Thanks — JackAnderson, Jan 16 '17 at 16:46
Possible duplicate of [How do I test one variable against multiple values?](http://stackoverflow.com/questions/15112125/how-do-i-test-one-variable-against-multiple-values) — MooingRawr, Jan 16 '17 at 16:49

score 3 · Accepted Answer · edited Jan 16 '17 at 16:52

3

word = str(input("Enter a word:"))

def vowel(word):
    score = 1

    for character in word:
        character = character.lower()
        if character == 'a':
            score += 5
        elif character == 'e':
            score += 4
        elif character == 'i':
            score += 3
        elif character == 'o':
            score += 2
        elif character == 'u':
            score += 1

    print("Your word scored",score,"in the vowel checker")

vowel(word)

Things to note:

In Python, strings are iterable, so you can loop over each character.
Note that you don't have to (and shouldn't) use a global.
The use of character.lower() simplifies our conditionals.
Instead of printing the output in the vowel function, you could instead just return score and put the print statement in the last line.

P.S. Given the question, shouldn't a word's score start at 0, not 1?

edited Jan 16 '17 at 16:52

depperm

10,606
4
43
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answered Jan 16 '17 at 16:48

pbaranay

585
5
17

So does your example include upper and lower case letters , or does it only accept lower case letters, And yes you are correct in saying that the score should be 0 – JackAnderson Jan 16 '17 at 16:56
@JackAnderson My example works for both upper- and lower-case letters. `character = character.lower()` turns each letter into lowercase, which simplifies our conditionals. This allows us to have to check against "a", not "a" or "A" like in the original code. – pbaranay Jan 16 '17 at 16:58
1

It's more efficient to use `for character in word.lower():` because you only call `lower()` once that way. Of course there are other ways to improve efficiency too – Chris_Rands Jan 16 '17 at 17:05

score 1 · Answer 2 · answered Jan 16 '17 at 16:50

First if you are passing word to vowel I don't know why you use global. You are calling vowel inside a print so vowel should return a string instead of printing a string itself. Next by using a for loop you can check each character of the word and increment the score even if multiple occurances of a vowel appear.

word = str(input("Enter a word:"))

def vowel(word):
  score = 1
  for c in word:
    if "a" == c.lower():
        score += 5
    elif "e" == c.lower():
        score += 4
    elif "i" == c.lower():
        score += 3
    elif "o" == c.lower():
        score += 2
    elif "u" == c.lower():
        score += 1

  return "Your word scored "+str(score)+" in the vowel checker"

print(vowel(word))

score 0 · Answer 3 · answered Jul 17 '18 at 12:17

Time complexity is important

for character in word:
    character = character.lower()
    if  re.match("a",character):
        score += 5
    elif re.match("e",character):
        score += 4
    elif re.match("i",character):
        score += 3
    elif re.match("o",character):
        score += 2
    elif re.match("u",character):
        score += 1

print("Your word scored",score,"in the vowel checker")

Enter a word:hello

pbaranay code, Your word scored 7 in the vowel checker

--- 2.4024055004119873 seconds ---

My code,Your word scored 7 in the vowel checker

--- 0.004721164703369141 seconds ---

Requires assistance with vowel checker program

3 Answers3