计算元音的 Python 代码
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Python code to count vowels
提问by user3766695
Assume sis a string of lower case characters.
假设s是一串小写字符。
Write a program that counts up the number of vowels contained in the string s. Valid vowels are: 'a', 'e', 'i', 'o', and 'u'. For example, if s = 'azcbobobegghakl', your program should print:
编写一个程序来计算字符串中包含的元音数量s。有效的元音是：'a'，'e'，'i'，'o'，和'u'。例如，如果s = 'azcbobobegghakl'，您的程序应该打印：
Number of vowels: 5
元音数： 5
I have this so far
到目前为止我有这个
count = 0
vowels = 'a' or 'e' or 'i' or 'o' or 'u'
for vowels in s:
count +=1
print ('Number of vowels: ' + count)
Can anyone tell me what is wrong with it?
谁能告诉我它有什么问题？
回答by Fredrik Pihl
As a start, try this:
首先，试试这个：
In [9]: V = ['a','e','i','o','u']
In [10]: s = 'azcbobobegghakl'
In [11]: sum([1 for i in s if i in V])
Out[11]: 5
回答by Padraic Cunningham
Using your own loop.
使用您自己的循环。
count = 0
vowels = ['a' , 'e' , 'i' ,'o' , 'u']
for char in s:
if char in vowels: # check if each char in your string is in your list of vowels
count += 1
print ('Number of vowels: ' + str(count)) # count is an integer so you need to cast it as a str
You can use string formatting also:
您也可以使用字符串格式：
print ('Number of vowels: {} '.format(count))
回答by jonrsharpe
A couple of problems. First, your assignment to vowelsdoesn't do what you think it does:
几个问题。首先，您的分配vowels没有做您认为的那样：
>>> vowels = 'a' or 'e' or 'i' or 'o' or 'u'
>>> vowels
'a'
Python evaluates orlazily; as soon as any of the predicates evaluates Trueit is returned. Non-empty sequences, including strings other than ""evaluate True, so 'a'is returned straight away.
Pythonor惰性求值；一旦任何谓词求值，True它就会被返回。非空序列，包括除""评估之外的字符串True，因此'a'会立即返回。
Second, when you iterate over s, you ignore that assignment anyway:
其次，当您迭代 时s，您无论如何都会忽略该分配：
>>> for vowels in "foo":
print(vowels)
f
o
o
for x in y:assigns each item in the iterable yto the name xin turn, so anything previously assigned to xis not longer accessible via that name.
for x in y:依次将 iterabley中的每个项目分配给名称x，因此以前分配给的任何内容x都不能再通过该名称访问。
I think what you want is:
我想你想要的是：
count = 0
vowels = set("aeiou")
for letter in s:
if letter in vowels:
count += 1
回答by sundar nataraj
A different implementation using counter
使用计数器的不同实现
from collections import Counter
s='azcbobobegghakl'
vowels="aeiou"
c=Counter(s)
print sum([c[i] for i in set(vowels).intersection(c.keys())])
the statement set(vowels).intersection(c.keys())this returns the dictint vowels present in the sentence
声明set(vowels).intersection(c.keys())这将返回dictint元音出现在句子
回答by Aashish P
This is also another solution,
这也是另一种解决方案，
In [12]: vowels = ['a', 'e', 'i', 'o', 'u']
In [13]: str = "azcbobobegghakl"
In [14]: sum([str.count(elem) for elem in vowels])
Out[14]: 5
Using string.count()
使用 string.count()
回答by Rahul Khatri
here is the simple one:
这是一个简单的：
count = 0 #initialize the count variable
def count_vowel(word): #define a function for counting the vowels
vowels = 'aeiouAEIOU' #A string containing all the vowels
for i in range(word): #traverse the string
if i in vowels: #check if the the character is contained in the vowel string
count = count + 1 #update the count
return count
回答by xxyzzy
Here is a sample that uses Counter and is more compact and even a little faster than Sundar's for larger strings:
这是一个使用 Counter 的示例，对于较大的字符串，它比 Sundar 的更紧凑，甚至更快一点：
from collections import Counter
cnt = Counter('this and that')
sum([cnt[x] for x in 'aeiou'])
Here is a time test to compare 3 approaches:
这是比较 3 种方法的时间测试：
import time
from collections import Counter
s = 'That that is is is not that that is not is not. This is the understanding of all who begin to think.'
c = Counter(s)
dt1 = dt2 = dt3 = dt4 = 0;
vowels = ['a' , 'e' , 'i' ,'o' , 'u']
for i in range(100000):
ms0 = time.time()*1000.0
s1 = sum([c[x] for x in 'aeiou'])
ms1 = time.time()*1000.0
dt1 += ms1 - ms0
for i in range(100000):
ms1 = time.time()*1000.0
s2 = sum([c[x] for x in set(vowels).intersection(c.keys())])
ms2 = time.time()*1000.0
dt2 += ms2 - ms1
for i in range(100000):
ms2 = time.time()*1000.0
s3 = 0
for char in s:
if char in vowels: # check if each char in your string is in your list of vowels
s3 += 1
ms3 = time.time()*1000.0
dt3 += ms3 - ms2
print('sums:', s1, s2, s3)
print('times:', dt1, dt2, dt3)
print('relative: {:.0%}{:.0%}{:.0%}'.format(dt1/dt2, dt2/dt2, dt3/dt2))
Results (average of six runs), versions: this, Sundar, simple
sums: 26 26 26
times: 392 494 2626
relative: 80% 100% 532%
结果（六次运行的平均值），版本：this，Sundar，简单和：26 26 26 次：392 494 2626
相对：80% 100% 532%
回答by iLabQC
x = len(s)
a = 0
c = 0
while (a < x):
if s[a] == 'a' or s[a] == 'e' or s[a] == 'i' or s[a] == 'o' or s[a] == 'u':
c += 1
a = a+1
print "Number of vowels: " + str(c)
The above code is for beginners
以上代码适合初学者
回答by Luke D
My solution:
我的解决方案：
s = 'aassfgia'
vowels = 0
for x in s:
if x == 'a' or x == 'e' or x == 'i' or x == 'o' or x == 'u':
vowels += 1
print x
print vowels
回答by vatsal
For counting vowels from string
用于计算字符串中的元音
s = "Some string here"
or
或者
s = intput(raw_input("Enter ur string"))
s1 = s.lower()
count = 0
vowels = set("aeiou")
for letter in s1:
if letter in vowels:
count += 1
print 'Number of vowels:' ,count
this will give u output for total count of vowels in given string
这将为您提供给定字符串中元音总数的输出
