从原始输入计数元音 [英] Count vowels from raw input
问题描述
我有一个家庭作业问题，要求通过原始输入读取一个字符串，并计算字符串中有多少个元音。这是我到目前为止，但我遇到一个问题：
def vowels（）：
vowels = a，e，i，o，u]
count = 0
string = raw_input（输入字符串：） （0，len（string））：
if string [i] == vowels [i]：
count = count + 1
print count
vowels ）
它计算元音很好，但由于 == vowels [i]：，它只会计算一个元音一次，因为 i 在范围内不断增加。如何更改此代码以检查输入的元音字符串，而不会遇到此问题？
运算符运算符运算符 $ 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
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?
解决方案 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()
这篇关于从原始输入计数元音的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
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
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()
这篇关于从原始输入计数元音的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
