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

A

1 1 1 1 1

stri

'aabccc'.count('c')

20 Answers
20

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 *.

map

*

>>> 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


counts = i:0 for i in 'aeiouAEIOU'

counts=.fromkeys('aeiouAEIOU',0)

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)

Use a Counter

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

Counter

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


d = defaultdict(int)

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


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.

if A or a in stri

if A or (a in stri)

if True or (a in stri)

True

if

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

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.

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


string.lower()

if char in "aeiou":

For brevity and readability, use a dictionary comprehension.

>>> inp = raw_input() # use input in Python3
hI therE stAckOverflow!
>>> search = inp.lower()
>>> v:search.count(v) for v in 'aeiou'
'a': 1, 'i': 1, 'e': 3, 'u': 0, 'o': 2


Alternatively, you can consider a named tuple.

>>> from collections import namedtuple
>>> vocals = 'aeiou'
>>> s = 'hI therE stAckOverflow!'.lower()
>>> namedtuple('Vowels', ' '.join(vocals))(*(s.count(v) for v in vocals))
Vowels(a=1, e=3, i=1, o=2, u=0)


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)


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)

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


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))


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

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)


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('[rtnd,.!?\/()[]]+', " ", txt)
txt = re.sub('s2,', " ", 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))


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()


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 #!/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


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