从Python中的字符串中删除辅音

Question

这是我的代码.我不确定我是否需要一个计数器才能工作.答案应该是'iiii'.

def eliminate_consonants(x):
        vowels= ['a','e','i','o','u']
        vowels_found = 0
        for char in x:
            if char == vowels:
                print(char)

eliminate_consonants('mississippi')

Answer 1

更正您的代码

这条线if char == vowels:是错的.它必须是if char in vowels:.这是因为您需要检查元音列表中是否存在该特定字符.除此之外,您需要print(char,end = '')(在python3中)将输出打印为iiii一行.

最终的计划就像

def eliminate_consonants(x):
        vowels= ['a','e','i','o','u']
        for char in x:
            if char in vowels:
                print(char,end = "")

eliminate_consonants('mississippi')

输出将是

iiii

其他方式包括

• 使用in字符串

  def eliminate_consonants(x):
      for char in x:
          if char in 'aeiou':
              print(char,end = "")
  

  看起来很简单,语句if char in 'aeiou'检查char字符串中是否存在aeiou.

• 列表理解

   ''.join([c for c in x if c in 'aeiou'])
  

  此列表推导将返回一个列表,该列表仅在角色所在时才包含字符 aeiou

• 生成器表达式

  ''.join(c for c in x if c in 'aeiou')
  

  这个gen exp将返回一个生成器,而不是仅在角色所在的情况下返回字符 aeiou

• 常用表达

  您可以使用re.findall仅发现字符串中的元音.代码

  re.findall(r'[aeiou]',"mississippi")
  

  将返回在字符串中找到的元音列表,即['i', 'i', 'i', 'i'].所以现在我们可以使用str.join然后再使用了

  ''.join(re.findall(r'[aeiou]',"mississippi"))
  

• str.translate 和 maketrans

  对于这种技术,您需要存储一个地图,该地图将每个非元音与一个None类型相匹配.为此您可以使用string.ascii_lowecase.制作地图的代码是

  str.maketrans({i:None for i in string.ascii_lowercase if i not in "aeiou"})
  

  这将返回映射.将它存储在变量中(这里m是地图)

  "mississippi".translate(m)
  

  这将删除aeiou字符串中的所有非字符.

• 运用 dict.fromkeys

  你可以dict.fromkeys一起使用sys.maxunicode.但请记住import sys!

  dict.fromkeys(i for i in range(sys.maxunicode+1) if chr(i) not in 'aeiou')
  

  现在用str.translate.

  'mississippi'.translate(m)
  

• 运用 bytearray

  正如JFSebastian在下面的评论中所提到的,您可以使用创建小写辅音的bytearray

  non_vowels = bytearray(set(range(0x100)) - set(b'aeiou'))
  

  使用这个,我们可以翻译这个词,

  'mississippi'.encode('ascii', 'ignore').translate(None, non_vowels)
  

  将返回b'iiii'.这可以str通过使用decodeie 轻松转换b'iiii'.decode("ascii").

• 运用 bytes

  bytes返回一个bytes对象,是不可变的版本bytearray.(具体是Python 3)

  non_vowels = bytes(set(range(0x100)) - set(b'aeiou'))
  

  使用这个,我们可以翻译这个词,

  'mississippi'.encode('ascii', 'ignore').translate(None, non_vowels)
  

  将返回b'iiii'.这可以str通过使用decodeie 轻松转换b'iiii'.decode("ascii").

时间比较

Python 3

python3 -m timeit -s "text = 'mississippi'*100; non_vowels = bytes(set(range(0x100)) - set(b'aeiou'))" "text.encode('ascii', 'ignore').translate(None, non_vowels).decode('ascii')"
100000 loops, best of 3: 2.88 usec per loop
python3 -m timeit -s "text = 'mississippi'*100; non_vowels = bytearray(set(range(0x100)) - set(b'aeiou'))" "text.encode('ascii', 'ignore').translate(None, non_vowels).decode('ascii')"
100000 loops, best of 3: 3.06 usec per loop
python3 -m timeit -s "text = 'mississippi'*100;d=dict.fromkeys(i for i in range(127) if chr(i) not in 'aeiou')" "text.translate(d)"
10000 loops, best of 3: 71.3 usec per loop
python3 -m timeit -s "import string; import sys; text='mississippi'*100; m = dict.fromkeys(i for i in range(sys.maxunicode+1) if chr(i) not in 'aeiou')" "text.translate(m)"
10000 loops, best of 3: 71.6 usec per loop
python3 -m timeit -s "text = 'mississippi'*100" "''.join(c for c in text if c in 'aeiou')"
10000 loops, best of 3: 60.1 usec per loop
python3 -m timeit -s "text = 'mississippi'*100" "''.join([c for c in text if c in 'aeiou'])"
10000 loops, best of 3: 53.2 usec per loop
python3 -m timeit -s "import re;text = 'mississippi'*100; p=re.compile(r'[aeiou]')" "''.join(p.findall(text))"
10000 loops, best of 3: 57 usec per loop

排序顺序的时间

translate (bytes)    |  2.88
translate (bytearray)|  3.06
List Comprehension   | 53.2
Regular expressions  | 57.0
Generator exp        | 60.1
dict.fromkeys        | 71.3
translate (unicode)  | 71.6

正如您所看到的,使用最终方法bytes是最快的.

Python 3.5

python3.5 -m timeit -s "text = 'mississippi'*100; non_vowels = bytes(set(range(0x100)) - set(b'aeiou'))" "text.encode('ascii', 'ignore').translate(None, non_vowels).decode('ascii')"
100000 loops, best of 3: 4.17 usec per loop
python3.5 -m timeit -s "text = 'mississippi'*100; non_vowels = bytearray(set(range(0x100)) - set(b'aeiou'))" "text.encode('ascii', 'ignore').translate(None, non_vowels).decode('ascii')"
100000 loops, best of 3: 4.21 usec per loop
python3.5 -m timeit -s "text = 'mississippi'*100;d=dict.fromkeys(i for i in range(127) if chr(i) not in 'aeiou')" "text.translate(d)"
100000 loops, best of 3: 2.39 usec per loop
python3.5 -m timeit -s "import string; import sys; text='mississippi'*100; m = dict.fromkeys(i for i in range(sys.maxunicode+1) if chr(i) not in 'aeiou')" "text.translate(m)"
100000 loops, best of 3: 2.33 usec per loop
python3.5 -m timeit -s "text = 'mississippi'*100" "''.join(c for c in text if c in 'aeiou')"
10000 loops, best of 3: 97.1 usec per loop
python3.5 -m timeit -s "text = 'mississippi'*100" "''.join([c for c in text if c in 'aeiou'])"
10000 loops, best of 3: 86.6 usec per loop
python3.5 -m timeit -s "import re;text = 'mississippi'*100; p=re.compile(r'[aeiou]')" "''.join(p.findall(text))"
10000 loops, best of 3: 74.3 usec per loop

排序顺序的时间

translate (unicode)  |  2.33
dict.fromkeys        |  2.39
translate (bytes)    |  4.17
translate (bytearray)|  4.21
List Comprehension   | 86.6
Regular expressions  | 74.3
Generator exp        | 97.1

Answer 2

你可以尝试这样的pythonic方式，

In [1]: s = 'mississippi'
In [3]: [char for char in s if char in 'aeiou']
Out[3]: ['i', 'i', 'i', 'i']

功能;

In [4]: def eliminate_consonants(x):
   ...:     return ''.join(char for char in x if char in 'aeiou')
   ...: 

In [5]: print(eliminate_consonants('mississippi'))
iiii