在字符串 Python 中计算元音

Question

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

Answer 1

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 * .

Answer 2

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)

Answer 3

>>> 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

Answer 4

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

Answer 5

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}

Answer 6

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)

Answer 7

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.

Answer 8

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

Answer 9

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

Answer 10

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)

Answer 11

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)

Answer 12

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))

Answer 13

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

Answer 14

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])

Answer 15

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))

Answer 16

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))

Answer 17

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)

Answer 18

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() .

Answer 19

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 : https://repl.it/repls/BlueSlateblueDecagons

So have fun hope helped a lil bit.

Answer 20

...

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")

...

Answer 21

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()))

Answer 22

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}

Answer 23

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)

Answer 24

这是一个简单的，不要觉得在python中搜索三元for循环很复杂，你会得到它。

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

Answer 25

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)

Answer 26

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()

Answer 27

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."

Answer 28

  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 

Answer 29

string1='I love my India'

vowel='aeiou'

for i in vowel:
  print i + "->" + str(string1.count(i))

Answer 30

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)