Converting string to binary representation in python

Question

I want to create binary values for words based on their content of vowels and consonants, where vowels receive a value of '0' and consonants get a value of '1'.

For example, 'haha' would be represented as 1010, hahaha as 101010.

common_words = ['haha', 'hahaha', 'aardvark', etc...]

dictify = {}

binary_value = []

#doesn't work
for word in common_words: 
    for x in word:
        if x=='a' or x=='e' or x=='i' or x=='o' or x=='u':
            binary_value.append(0)
            dictify[word]=binary_value
        else:
            binary_value.append(1)
            dictify[word]=binary_value

-With this I am getting too many binary digits in the resulting dictionary:

>>>dictify
{'aardvark': [0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 1,...}

desired output:

>>>dictify
{'haha': 1010,'hahaha': 101010, 'aardvark': 00111011}

I am thinking of a solution that doesn't involve a loop within a loop...

dictify = {w:"".join('0' if c in 'aeiouAEIOU' else '1' for c in w) for w in common_words} — mshsayem
– mshsayem, Commented Feb 17, 2014 at 2:40
Your desired output isn't really possible-- 00111011 won't work as an integer because there's no way to preserve the initial zeroes. You could use a string or a list. — DSM
– DSM, Commented Feb 17, 2014 at 2:41
Please post your actual code. The code you posted can't work each and binary_value are never set. — Chris Johnson
– Chris Johnson, Commented Feb 17, 2014 at 2:46

user2357112 · Accepted Answer · 2014-02-17 02:41:37Z

The code you've posted doesn't work because all words share the same binary_value list. (It also doesn't work because number_value and each are never defined, but we'll pretend those variables said binary_value and word instead.) Define a new list for each word:

for word in common_words:
    binary_value = []
    for x in word:
        if x=='a' or x=='e' or x=='i' or x=='o' or x=='u':
            binary_value.append(0)
            dictify[word]=binary_value
        else:
            binary_value.append(1)
            dictify[word]=binary_value

If you want the output to look like 00111011 rather than a list, you'll need to make a string. (You could make an int, but then it would look like 59 instead of 00111011. Python doesn't distinguish "this int is base 2" or "this int has 2 leading zeros".)

for word in common_words:
    binary_value = []
    for x in word:
        if x.lower() in 'aeiou':
            binary_value.append('0')
        else:
            binary_value.append('1')
    dictify[word] = ''.join(binary_value)

mshsayem · Accepted Answer · 2014-02-17 02:48:54Z

2

user2357112 explains your code. Here is just another way:

>>> common_words = ['haha', 'hahaha', 'aardvark']
>>> def binfy(w):
        return "".join('0' if c in 'aeiouAEIOU' else '1' for c in w)

>>> dictify = {w:binfy(w) for w in common_words}
>>> dictify
{'aardvark': '00111011', 'haha': '1010', 'hahaha': '101010'}

edited Feb 17, 2014 at 2:48

answered Feb 17, 2014 at 2:43

mshsayem

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Comments

kojiro · Accepted Answer · 2014-02-17 03:14:24Z

This seems like a job for translation tables. Assuming your input strings are all ASCII (and it seems likely or the definition of exactly what is a vowel gets fuzzy), you can define a translation table this way*:

# For simplicity's sake, I'm only using lowercase letters
from string import lowercase, maketrans
tt = maketrans(lowercase, '01110111011111011111011111')

With the above table, the problem becomes trivial:

>>> 'haha'.translate(tt)
'1010'
>>> 'hahaha'.translate(tt)
'101010'
>>> 'aardvark'.translate(tt)
'00111011'

Given this solution, you can build dictify very simply with a comprehension:

dictify = {word:word.translate(tt) for word in common_words} #python2.7
dictify = dict((word, word.translate(tt)) for word in common_words) # python 2.6 and earlier

*This can also be done with Python 3, but you have to use bytes instead of strings:

from string import ascii_lowercase
tt = b''.maketrans(bytes(ascii_lowercase, 'ascii'), b'01110111011111011111011111')
b'haha'.translate(tt)
...

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