Count Vowels in String Python

I'm trying to count how many occurrences there are of specific characters in a string, but the output is wrong.

Here is my code:

inputString = str(input("Please type a sentence: "))
a = "a"
A = "A"
e = "e"
E = "E"
i = "i"
I = "I"
o = "o"
O = "O"
u = "u"
U = "U"
acount = 0
ecount = 0
icount = 0
ocount = 0
ucount = 0
if A or a in stri : acount = acount + 1
if E or e in stri : ecount = ecount + 1
if I or i in stri : icount = icount + 1
if o or O in stri : ocount = ocount + 1
if u or U in stri : ucount = ucount + 1
print(acount, ecount, icount, ocount, ucount)

If I enter the letter A the output would be: 1 1 1 1 1

5

29 Answers

What you want can be done quite simply like so:

>>> mystr = input("Please type a sentence: ")
Please type a sentence: abcdE
>>> print(*map(mystr.lower().count, "aeiou"))
1 1 0 0 0
>>>

In case you don't know them, here is a reference on map and one on the *.

def countvowels(string): num_vowels=0 for char in string: if char in "aeiouAEIOU": num_vowels = num_vowels+1 return num_vowels

(remember the spacing s)

>>> sentence = input("Sentence: ")
Sentence: this is a sentence
>>> counts = {i:0 for i in 'aeiouAEIOU'}
>>> for char in sentence:
... if char in counts:
... counts[char] += 1
...
>>> for k,v in counts.items():
... print(k, v)
...
a 1
e 3
u 0
U 0
O 0
i 2
E 0
o 0
A 0
I 0

1

data = str(input("Please type a sentence: "))
vowels = "aeiou"
for v in vowels: print(v, data.lower().count(v))

0

Use a Counter

>>> from collections import Counter
>>> c = Counter('gallahad')
>>> print c
Counter({'a': 3, 'l': 2, 'h': 1, 'g': 1, 'd': 1})
>>> c['a'] # count of "a" characters
3

Counter is only available in Python 2.7+. A solution that should work on Python 2.5 would utilize defaultdict

>>> from collections import defaultdict
>>> d = defaultdict(int)
>>> for c in s:
... d[c] = d[c] + 1
...
>>> print dict(d)
{'a': 3, 'h': 1, 'l': 2, 'g': 1, 'd': 1}

1

if A or a in stri means if A or (a in stri) which is if True or (a in stri) which is always True, and same for each of your if statements.

What you wanted to say is if A in stri or a in stri.

This is your mistake. Not the only one - you are not really counting vowels, since you only check if string contains them once.

The other issue is that your code is far from being the best way of doing it, please see, for example, this: Count vowels from raw input. You'll find a few nice solutions there, which can easily be adopted for your particular case. I think if you go in detail through the first answer, you'll be able to rewrite your code in a correct way.

For anyone who looking the most simple solution, here's the one

vowel = ['a', 'e', 'i', 'o', 'u']
Sentence = input("Enter a phrase: ")
count = 0
for letter in Sentence: if letter in vowel: count += 1
print(count)

2

>>> string = "aswdrtio"
>>> [string.lower().count(x) for x in "aeiou"]
[1, 0, 1, 1, 0]

1

Another solution with list comprehension:

vowels = ["a", "e", "i", "o", "u"]
def vowel_counter(str): return len([char for char in str if char in vowels])
print(vowel_counter("abracadabra"))
# 5

count = 0
string = raw_input("Type a sentence and I will count the vowels!").lower()
for char in string: if char in 'aeiou': count += 1
print count

2

I wrote a code used to count vowels. You may use this to count any character of your choosing. I hope this helps! (coded in Python 3.6.0)

while(True):
phrase = input('Enter phrase you wish to count vowels: ')
if phrase == 'end': #This will to be used to end the loop quit() #You may use break command if you don't wish to quit
lower = str.lower(phrase) #Will make string lower case
convert = list(lower) #Convert sting into a list
a = convert.count('a') #This will count letter for the letter a
e = convert.count('e')
i = convert.count('i')
o = convert.count('o')
u = convert.count('u')
vowel = a + e + i + o + u #Used to find total sum of vowels
print ('Total vowels = ', vowel)
print ('a = ', a)
print ('e = ', e)
print ('i = ', i)
print ('o = ', o)
print ('u = ', u)

Suppose,

S = "Combination"

import re
print re.findall('a|e|i|o|u', S)

Prints: ['o', 'i', 'a', 'i', 'o']

For your case in a sentence (Case1):

txt = "blah blah blah...."

import re
txt = re.sub('[\r\t\n\d\,\.\!\?\\\/\(\)\[\]\{\}]+', " ", txt)
txt = re.sub('\s{2,}', " ", txt)
txt = txt.strip()
words = txt.split(' ')
for w in words: print w, len(re.findall('a|e|i|o|u', w))

Case2

import re, from nltk.tokenize import word_tokenize
for w in work_tokenize(txt): print w, len(re.findall('a|e|i|o|u', w))

from collections import Counter
count = Counter()
inputString = str(input("Please type a sentence: "))
for i in inputString: if i in "aeiouAEIOU": count.update(i)
print(count)

sentence = input("Enter a sentence: ").upper()
#create two lists
vowels = ['A','E',"I", "O", "U"]
num = [0,0,0,0,0]
#loop through every char
for i in range(len(sentence)):
#for every char, loop through vowels for v in range(len(vowels)): #if char matches vowels, increase num if sentence[i] == vowels[v]: num[v] += 1
for i in range(len(vowels)): print(vowels[i],":", num[i])

count = 0
s = "azcbobobEgghakl"
s = s.lower()
for i in range(0, len(s)): if s[i] == 'a'or s[i] == 'e'or s[i] == 'i'or s[i] == 'o'or s[i] == 'u': count += 1
print("Number of vowels: "+str(count))

1

This works for me and also counts the consonants as well (think of it as a bonus) however, if you really don't want the consonant count all you have to do is delete the last for loop and the last variable at the top.

Her is the python code:

data = input('Please give me a string: ')
data = data.lower()
vowels = ['a','e','i','o','u']
consonants = ['b','c','d','f','g','h','j','k','l','m','n','p','q','r','s','t','v','w','x','y','z']
vowelCount = 0
consonantCount = 0
for string in data: for i in vowels: if string == i: vowelCount += 1 for i in consonants: if string == i: consonantCount += 1
print('Your string contains %s vowels and %s consonants.' %(vowelCount, consonantCount))

Simplest Answer:
inputString = str(input("Please type a sentence: "))
vowel_count = 0
inputString =inputString.lower()
vowel_count+=inputString.count("a")
vowel_count+=inputString.count("e")
vowel_count+=inputString.count("i")
vowel_count+=inputString.count("o")
vowel_count+=inputString.count("u")
print(vowel_count)

1

from collections import defaultdict
def count_vowels(word): vowels = 'aeiouAEIOU' count = defaultdict(int) # init counter for char in word: if char in vowels: count[char] += 1 return count

A pythonic way to count vowels in a word, not like in java or c++, actually no need to preprocess the word string, no need for str.strip() or str.lower(). But if you'd like to count vowels case-insensitively, then before go into the for-loop, use str.lower().

vowels = ["a","e","i","o","u"]
def checkForVowels(some_string): #will save all counted vowel variables as key/value amountOfVowels = {} for i in vowels: # check for lower vowel variables if i in some_string: amountOfVowels[i] = some_string.count(i) #check for upper vowel variables elif i.upper() in some_string: amountOfVowels[i.upper()] = some_string.count(i.upper()) return amountOfVowels
print(checkForVowels("sOmE string"))

You can test this code here :

So have fun hope helped a lil bit.

...

vowels = "aioue"
text = input("Please enter your text: ")
count = 0
for i in text: if i in vowels: count += 1
print("There are", count, "vowels in your text")

...

1

def vowels(): numOfVowels=0 user=input("enter the sentence: ") for vowel in user: if vowel in "aeiouAEIOU": numOfVowels=numOfVowels+1 return numOfVowels
print("The number of vowels are: "+str(vowels()))

1

You could use regex and dict comprehension:

import re
s = "aeiouuaaieeeeeeee"

The regex function findall() returns a list containing all matches

Here x is the key and the lenght of the list returned by the regex is the count of each vowel in this string, note that regex will find any character you introduce into the "aeiou" string.

foo = {x: len(re.findall(f"{x}", s)) for x in "aeiou"}
print(foo)

returns:

{'a': 3, 'e': 9, 'i': 2, 'o': 1, 'u': 2}

string1='I love my India'
vowel='aeiou'
for i in vowel: print i + "->" + str(string1.count(i))

1

This is a simple one don't feel it complex search for ternary for loop in python you will get it.

print(sum([1 for ele in input() if ele in "aeiouAEIOU"]))

1

def vowel_count(string): string = string.lower() count = 0 vowel_found = False for char in string: if char in 'aeiou': #checking if char is a vowel count += 1 vowel_found = True if vowel_found == False: print(f"There are no vowels in the string: {string}") return count
string = "helloworld"
result = vowel_count(string) #calling function
print("No of vowels are: ", result)

def count_vowel(): cnt = 0 s = 'abcdiasdeokiomnguu' s_len = len(s) s_len = s_len - 1 while s_len >= 0: if s[s_len] in ('aeiou'): cnt += 1 s_len -= 1 print 'numofVowels: ' + str(cnt) return cnt
def main(): print(count_vowel())
main()

1

count = 0
name=raw_input("Enter your name:")
for letter in name: if(letter in ['A','E','I','O','U','a','e','i','o','u']): count=count + 1
print "You have", count, "vowels in your name."

1

 1 #!/usr/bin/python 2 3 a = raw_input('Enter the statement: ') 4 5 ########### To count number of words in the statement ########## 6 7 words = len(a.split(' ')) 8 print 'Number of words in the statement are: %r' %words 9 10 ########### To count vowels in the statement ########## 11 12 print '\n' "Below is the vowel's count in the statement" '\n' 13 vowels = 'aeiou' 14 15 for key in vowels: 16 print key, '=', a.lower().count(key) 17 

def check_vowel(char): chars = char.lower() list = [] list2 = [] for i in range(0, len(chars)): if(chars[i]!=' '): if(chars[i]=='a' or chars[i]=='e' or chars[i]=='i' or chars[i]=='o' or chars[i]=='u'): list.append(chars[i]) else: list2.append(chars[i]) return list, list2
char = input("Enter your string:")
list,list2 = check_vowel(char)
if len(list)==1: print("Vowel is:", len(list), list)
if len(list)>1: print("Vowels are:", len(list), list)
if len(list2)==1: print("Constant is:", len(list2), list2)
if len(list2)>1: print("Constants are:", len(list2), list2)