运算符$ c>运算符$ c>运算符 $ c> == 运算符 - 在运算符允许您检查特定项目是否在序列/集合中。

  1 in [1,2,3]＃True 
 1 in [2,3,4]＃False 
 'a'in ['a'，'e'，'i'，'o'，'u']＃True 
'a'in'aeiou'＃也是True 
 $ p> 



其他一些注释：
 

        $ c>，这是一种专门为是该组项目的项目X部分类型的操作而设计的数据类型。
 code> vowels = set（['a'，'e'，'i'，'o'，'u']）

  *  dict  s也有效，在  
 

在字符串上迭代
 

字符串是Python中的序列类型，这意味着你不需要去获取长度然后使用索引的所有努力 - 你可以只是遍历字符串，你会得到每个字符轮流：
 

例如：
 用于my_string中的字符：
元音：
＃... 
 
 

使用字符串初始化集合
 

上面，您可能已经注意到，创建一个具有预设值的集合（至少在Python 2.x中）涉及使用列表。这是因为 set（）类型构造函数接受一系列项。你还可能注意到，在上一节中，我提到字符串是Python中的序列 - 字符序列。
 

这意味着如果你想要一个设置字符，你实际上可以将这些字符的字符串传递给 set（）构造函数 - 您不需要有一个列表，字符串。换句话说，以下两行是等效的：
  set_from_string = set（'aeiou'）
 set_from_list = set （['a'，'e'，'i'，'o'，'u']）
 
整洁，是吗？ :)不过请注意，如果您想要设置一组字符串，而不是一组字符，这也可能会使您感到困惑。例如，以下两行不是相同：
  set_with_one_string = set（['cat ']）
 set_with_three_characters = set（'cat'）
 
一个元素：
 'cat'in set_with_one_string＃True 
'c'in set_with_one_string＃False 
 
后者是一个包含三个元素（每个元素都是一个字符）的集合：
 'c'in set_with_three_characters`＃True 
'cat'in set_with_three_characters＃False 
 
 

区分大小写
 

比较字符区分大小写。 'a'=='A'为False， $ 'A'in'aeiou'。为了解决这个问题，你可以转换你的输入，以匹配你正在比较的情况：
  lowercase_string = input_string。 lower（）
 
 
I have a homework question which asks to read a string through raw input and count how many vowels are in the string. This is what I have so far but I have encountered a problem:
def vowels():
    vowels = ["a","e","i","o","u"]
    count = 0
    string = raw_input ("Enter a string: ")
    for i in range(0, len(string)):
        if string[i] == vowels[i]:
            count = count+1
    print count

vowels()
It counts the vowels fine, but due to if string[i] == vowels[i]:, it will only count one vowel once as i keeps increasing in the range. How can I change this code to check the inputted string for vowels without encountering this problem?

in operator

You probably want to use the in operator instead of the == operator - the in operator lets you check to see if a particular item is in a sequence/set.
1 in [1,2,3] # True
1 in [2,3,4] # False
'a' in ['a','e','i','o','u'] # True
'a' in 'aeiou' # Also True


Some other comments:

Sets

The in operator is most efficient when used with a set, which is a data type specifically designed to be quick for "is item X part of this set of items" kind of operations.*
vowels = set(['a','e','i','o','u'])
*dicts are also efficient with in, which checks to see if a key exists in the dict.

Iterating on strings

A string is a sequence type in Python, which means that you don't need to go to all of the effort of getting the length and then using indices - you can just iterate over the string and you'll get each character in turn:

E.g.:
for character in my_string:
    if character in vowels:
        # ...


Initializing a set with a string

Above, you may have noticed that creating a set with pre-set values (at least in Python 2.x) involves using a list. This is because the set() type constructor takes a sequence of items. You may also notice that in the previous section, I mentioned that strings are sequences in Python - sequences of characters.

What this means is that if you want a set of characters, you can actually just pass a string of those characters to the set() constructor - you don't need to have a list one single-character strings. In other words, the following two lines are equivalent:
set_from_string = set('aeiou')
set_from_list = set(['a','e','i','o','u'])
Neat, huh? :) Do note, however, that this can also bite you if you're trying to make a set of strings, rather than a set of characters. For instance, the following two lines are not the same:
set_with_one_string = set(['cat'])
set_with_three_characters = set('cat')
The former is a set with one element:
'cat' in set_with_one_string # True
'c' in set_with_one_string # False
Whereas the latter is a set with three elements (each one a character):
'c' in set_with_three_characters` # True
'cat' in set_with_three_characters # False


Case sensitivity

Comparing characters is case sensitive. 'a' == 'A' is False, as is 'A' in 'aeiou'. To get around this, you can transform your input to match the case of what you're comparing against:
lowercase_string = input_string.lower()