Question

假设s是一串小写字符。

编写一个程序，计算字符串s中包含的元音数量。有效的元音包括：'a'，'e'，'i'，'o'和'u'。例如，如果s = 'azcbobobegghakl'，您的程序应打印：

元音数量：5

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)

谁能告诉我它有什么问题？

Answer 1

有几个问题。首先，您对vowels的分配并不符合您的想法：

>>> vowels = 'a' or 'e' or 'i' or 'o' or 'u'
>>> vowels
'a'

Python懒惰地评估or;只要任何谓词评估True，就会返回它。非空序列（包括""以外的字符串评估True，因此会立即返回'a'。

其次，当您迭代s时，无论如何都会忽略该分配：

>>> for vowels in "foo":
    print(vowels)


f
o
o

for x in y:依次将可迭代y中的每个项目分配给名称x，因此以前分配给x的所有内容都无法通过该名称访问。

我认为你想要的是：

count = 0
vowels = set("aeiou")
for letter in s:
    if letter in vowels:
        count += 1

Answer 2

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

Answer 3

使用自己的循环。

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

您也可以使用字符串格式：

print ('Number of vowels: {} '.format(count))

Answer 4

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)

以上代码适用于初学者

Answer 5

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

Answer 6

您可以使用此代码查找计数

{{1}}

Answer 7

for i in s:
    if i in vowels:
        num_vowels += 1
print (num_vowels)

Answer 8

str = 'aeioubbaeiouggaeiouss'
vow = set("aeiouAEIOU")
count = 0

for alpha in str:
    if alpha in vow:
        count += 1
print(count)

计数将给出元音进入字符串的总次数。

Answer 9

这是一个利用Procfile运算符和web循环的简单解决方案：

in

为什么for？要使其与任何包含大写字母的元音的字符串一起使用。

Answer 10

你需要一个数组。这是一个元素序列，在你的情况下是字符。这是一种在Python中定义数组的方法：

vowels  = ['a','e','i','o','u']

Answer 11

//Using while loop:- 
%%time
s = 'services'
m = list(s)
count = 0
while m:
    d = m.pop()
    if (d is 'a') or (d is 'e') or (d is 'i') or (d is 'o') or (d is 'u'):
        count += 1

结果是： - CPU时间：用户0 ns，sys：0 ns，总计：0 ns 壁挂时间：18.8μs //使用for循环一行： -

%%time
vowels = ['a', 'e', 'i', 'o', 'u']
sum([s.count(elem) for elem in vowels])

结果是： - CPU时间：用户0 ns，sys：0 ns，总计：0 ns 壁时间：18.6μs

Answer 12

strings = raw_input('Enter a string: ')

vowels = 0

for char in strings:
    if(char ==('a') or char ==('e') or char ==('i') or char ==('o') or char ==('u')):
        vowels += 1

print('Number of vowels: ' + str(vowels))

Answer 13

count=0
for letter in s:

    if(letter=='a' or letter == 'e' or letter =='i' or letter =='o' or letter=='u'):
         count=count+1

print("Number of vowels:",count)

Answer 14

total = 0
for c in s:
    if c == 'a' or c == 'e' or c == 'i' or c == 'o' or c == 'u':
        total += 1
print "Number of vowels: " + str(total)

替代解决方案

num = 0
for letter in s:
  if letter in "aeiou":
    num+=1
print "Number of vowels:" + str(num)

Answer 15

用于计算字符串中的元音

s = "Some string here"

或

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

这将给出给定字符串中元音总数的输出

Answer 16

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

Answer 17

这是一个使用Counter的样本，对于较大的字符串来说，它比Sundar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'])

以下是比较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))

结果（平均六次运行），版本：这个，Sundar，简单总和：26 26 26 时间：392 494 2626
相对：80％100％532％

Answer 18

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

使用string.count()

Answer 19

print("\nThe count of vowels are:",sum([1 for i in input("\nEnter the string\n\n").lower() if i in ['a','e','i','o','u']]),"\n")

输入字符串

azcbobobegghakl

元音的数量是：5

print("\nThe count of vowels are:",sum([1 for i in input("\nEnter the string\n\n").lower() if i in ['a','e','i','o','u']]),"\n")

输入字符串

azCBOBOBegghakl

元音的数量是：5

Answer 20

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())])

语句set(vowels).intersection(c.keys())这将返回句子中出现的dictint元音

用于计算元音的Python代码

20 个答案: