16

我正在尝试计算字符串中特定字符出现的次数，但输出错误。

这是我的代码：

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)

如果我输入字母A，输出将是：1 1 1 1 1

4

29 回答 29

23

你想要的可以像这样简单地完成：

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

如果你不认识他们，这里有一个参考map和一个*。

于 2013-11-14T00:07:24.897 回答
15 
def countvowels(string):
    num_vowels=0
    for char in string:
        if char in "aeiouAEIOU":
           num_vowels = num_vowels+1
    return num_vowels

（记住间距 s）

于 2014-09-17T07:52:06.310 回答
10 
data = str(input("Please type a sentence: "))
vowels = "aeiou"
for v in vowels:
    print(v, data.lower().count(v))
于 2013-11-14T02:29:38.930 回答
10 
>>> 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
于 2013-11-14T00:07:58.633 回答
6

用一个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仅在 Python 2.7+ 中可用。应该适用于 Python 2.5 的解决方案将利用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}
于 2013-11-14T00:07:35.747 回答
5

if A or a in stri表示if A or (a in stri)which is if True or (a in stri)which is always True，并且对于您的每个if陈述都是相同的。

你想说的是if A in stri or a in stri。

这是你的错误。不是唯一的 - 你并没有真正计算元音，因为你只检查字符串是否包含它们一次。

另一个问题是您的代码远不是最好的方法，例如，请参阅：Count vowels from raw input。您会在那里找到一些不错的解决方案，可以轻松地针对您的特定情况采用这些解决方案。我认为，如果您详细阅读第一个答案，您将能够以正确的方式重写您的代码。

于 2013-11-14T00:07:41.460 回答
5

对于寻找最简单解决方案的任何人，这是一个

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)
于 2017-09-15T19:06:07.570 回答
3 
>>> string = "aswdrtio"
>>> [string.lower().count(x) for x in "aeiou"]
[1, 0, 1, 1, 0]
于 2017-11-24T22:26:34.870 回答
2

列表理解的另一个解决方案：

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
于 2020-11-08T09:54:21.290 回答
2 
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
于 2016-10-07T18:37:22.490 回答
1 
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)
于 2020-09-03T22:55:52.397 回答
1

我写了一个用于计算元音的代码。您可以使用它来计算您选择的任何字符。我希望这有帮助！（用 Python 3.6.0 编码）

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)
于 2017-01-13T19:24:36.443 回答
1

认为，

S = "组合"

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

打印：['o', 'i', 'a', 'i', 'o']

对于您的案例（案例1）：

txt = "等等等等...."

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))

案例2

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))
于 2018-09-11T10:27:19.320 回答
0 
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()))
于 2019-10-08T06:45:31.297 回答
0

这是一个简单的不觉得复杂的在python中搜索三元for循环你会得到它。

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

于 2021-03-26T12:10:22.357 回答
0 
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)
于 2018-03-03T06:21:31.420 回答
0

...

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")

...

于 2019-06-20T17:17:33.047 回答
0 
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))
于 2016-10-03T13:09:51.083 回答
0

您可以使用正则表达式和字典理解：

import re
s = "aeiouuaaieeeeeeee"

正则表达式函数 findall() 返回一个包含所有匹配项的列表

这里 x 是键，正则表达式返回的列表的长度是该字符串中每个元音的计数，请注意，正则表达式会找到您在“aeiou”字符串中引入的任何字符。

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

返回：

{'a': 3, 'e': 9, 'i': 2, 'o': 1, 'u': 2}
于 2020-06-15T19:21:35.497 回答
0 
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)
于 2021-05-10T18:29:36.050 回答
0 
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

一种计算单词中元音的pythonic方法，不像 in javaor c++，实际上不需要预处理单词字符串，不需要str.strip()or str.lower()。但是如果你想不区分大小写地计算元音，那么在进入 for 循环之前，使用str.lower().

于 2019-02-26T13:42:06.153 回答
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])
于 2016-07-11T04:12:17.553 回答
0 
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"))

您可以在此处测试此代码：https ://repl.it/repls/BlueSlateblueDecagons

所以玩得开心，希望对 lil 有所帮助。

于 2019-03-29T23:09:07.903 回答
0

这对我有用，并且也计算辅音（将其视为奖励）但是，如果您真的不想要辅音计数，您所要做的就是删除最后一个 for 循环和顶部的最后一个变量。

她是python代码：

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))
于 2017-09-27T18:03:30.243 回答
-1 
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()
于 2015-06-23T03:34:25.130 回答
-1 
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)
于 2021-06-15T09:19:32.473 回答
-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."
于 2015-09-03T08:17:19.010 回答
-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
于 2016-04-01T07:38:25.377 回答
-1 
string1='I love my India'

vowel='aeiou'

for i in vowel:
  print i + "->" + str(string1.count(i))
于 2017-05-28T16:39:30.977 回答