Deleting consonants from a string in Python

问题

Here is my code. I'm not exactly sure if I need a counter for this to work. The answer should be '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')

回答1:

Correcting your code

The line if char == vowels: is wrong. It has to be if char in vowels:. This is because you need to check if that particular character is present in the list of vowels. Apart from that you need to print(char,end = '') (in python3) to print the output as iiii all in one line.

The final program will be like

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

eliminate_consonants('mississippi')

And the output will be

iiii

---

Other ways include

• Using in a string

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

  As simple as it looks, the statement if char in 'aeiou' checks if char is present in the string aeiou.

• A list comprehension

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

  This list comprehension will return a list that will contain the characters only if the character is in aeiou

• A generator expression

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

  This gen exp will return a generator than will return the characters only if the character is in aeiou

• Regular Expressions

  You can use re.findall to discover only the vowels in your string. The code

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

  will return a list of vowels found in the string i.e. ['i', 'i', 'i', 'i']. So now we can use str.join and then use

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

• str.translate and maketrans

  For this technique you will need to store a map which matches each of the non vowels to a None type. For this you can use string.ascii_lowecase. The code to make the map is

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

  this will return the mapping. Do store it in a variable (here m for map)

  "mississippi".translate(m)
  

  This will remove all the non aeiou characters from the string.

• Using dict.fromkeys

  You can use dict.fromkeys along with sys.maxunicode. But remember to import sys first!

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

  and now use str.translate.

  'mississippi'.translate(m)
  

• Using bytearray

  As mentioned by J.F.Sebastian in the comments below, you can create a bytearray of lower case consonants by using

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

  Using this we can translate the word ,

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

  which will return b'iiii'. This can easily be converted to str by using decode i.e. b'iiii'.decode("ascii").

• Using bytes

  bytes returns an bytes object and is the immutable version of bytearray. (It is Python 3 specific)

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

  Using this we can translate the word ,

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

  which will return b'iiii'. This can easily be converted to str by using decode i.e. b'iiii'.decode("ascii").

---

Timing comparison

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

The timings in sorted order

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

As you can see the final method using bytes is the fastest.

---

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

The timings in sorted order

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

回答2:

You can try pythonic way like this,

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

Function;

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

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

回答3:

== tests for equality. You are looking to see if any of the characters exist in the string that are in your list 'vowels'. To do that, you can simply use in such as below.

Additionally, I see you have a 'vowels_found' variable but are not utilizing it. Below one example how you can solve this:

def eliminate_consonants(x):
    vowels= ['a','e','i','o','u']
    vowels_found = 0
    for char in x:
        if char in vowels:
            print(char)
            vowels_found += 1

    print "There are", vowels_found, "vowels in", x

eliminate_consonants('mississippi')

Your output would then be:

i
i
i
i
There are 4 vowels in mississippi

来源：https://stackoverflow.com/questions/29998052/deleting-consonants-from-a-string-in-python