Deleting consonants from a string in Python

Question

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')

Answer 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

Answer 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