在字符串 Python 中计算元音
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Count Vowels in String Python
提问by user2975192
I'm trying to count how many occurrences there are of specific characters in a string, but the output is wrong.
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。
Here is my code:
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)
If I enter the letter Athe output would be: 1 1 1 1 1
如果我输入这封信A，输出将是：1 1 1 1 1
回答by user2975192
回答by Prashant Kumar
Use a Counter
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
Counteris only available in Python 2.7+. A solution that should work on Python 2.5 would utilize defaultdict
Counter仅在 Python 2.7+ 中可用。应该在 Python 2.5 上工作的解决方案将利用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}
回答by sashkello
if A or a in strimeans if A or (a in stri)which is if True or (a in stri)which is always True, and same for each of your ifstatements.
if A or a in stri意味着if A or (a in stri)which is if True or (a in stri)which always True，并且对于您的每个if语句都是相同的。
What you wanted to say is if A in stri or a in stri.
你想说的是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.
这是你的错误。不是唯一的 - 您并没有真正计算元音，因为您只检查字符串是否包含它们一次。
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.
另一个问题是，您的代码远不是最好的方法，例如，请参阅：从原始输入计算元音。您会在那里找到一些不错的解决方案，这些解决方案可以很容易地用于您的特定情况。我认为如果您详细了解第一个答案，您将能够以正确的方式重写您的代码。
回答by inspectorG4dget
>>> 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
回答by kyle k
data = str(input("Please type a sentence: "))
vowels = "aeiou"
for v in vowels:
print(v, data.lower().count(v))
回答by hanish
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)
（记住间距 s）
回答by avina k
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()
回答by Simanto Bagchi
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."
回答by timgeb
For brevity and readability, use a dictionary comprehension.
为了简洁和可读性，请使用字典理解。
>>> 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.
或者，您可以考虑命名元组。
>>> 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)
回答by user6143812
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
