How can I extract three characters at a time from a Python string?

Question

I want to write a function that takes a long string of characters (a protein sequence like 'UGGUGUUAUUAAUGGUUU') and extracts three characters at a time from it (i.e. the codons). It can either return each set of three characters one after another, or a list containing all the sets of three characters. Either way would work. But I'm having some trouble figuring out exactly how to do this cleanly.

Here's what I have so far:

def get_codon_list(codon_string):
    codon_start = 0
    codon_length = 3
    codon_end = 3
    codon_list = []
    for x in range(len(codon_string) // codon_length):
        codon_list.append(codon_string[codon_start:codon_end])
        codon_start += codon_length
        codon_end += codon_length
    return codon_list

It works to return a list of the codons, but it seems very inefficient. I don't like using hard-coded numbers and incrementing variables like that if there is a better way. I also don't like using for loops that don't actually use the variable in the loop. It doesn't seem like a proper use of it.

Any suggestions for how to improve this, either with a specific function/module, or just a better Pythonic technique?

Thanks!

Related is this link: stackoverflow.com/questions/22571259/… This method uses text wrapping — Robert McClardy
– Robert McClardy, Commented Feb 15, 2020 at 23:45
Does this answer your question? Split string every nth character? — AMC
– AMC, Commented Feb 16, 2020 at 0:17

abc · Accepted Answer · 2020-02-15 23:41:35Z

4

You can use a list comprehension and get a slice of length 3 from the string at each time.

>>> s="UGGUGUUAUUAAUGGUUU"
>>> res = [s[i:i+3] for i in range(0,len(s),3)]
>>> res
['UGG', 'UGU', 'UAU', 'UAA', 'UGG', 'UUU']

answered Feb 15, 2020 at 23:41

abc

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Tomerikoo · Accepted Answer · 2020-02-15 23:53:55Z

3

You can simply use the step argument of the range function to avoid maintaining the variables:

def get_codon_list(codon_string):
    codon_length = 3
    codon_list = []

    for codon_start in range(0, len(codon_string), codon_length):
        codon_end = codon_start + codon_length
        codon_list.append(codon_string[codon_start:codon_end])

    return codon_list

And then it can become as a list-comprehension:

def get_codon_list(codon_string):
    codon_length = 3

    codon_list = [codon_string[x:x+codon_length] for x in range(0, len(codon_string), codon_length)]

    return codon_list

edited Feb 15, 2020 at 23:53

answered Feb 15, 2020 at 23:42

Tomerikoo

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Comments

Randy · Accepted Answer · 2020-02-15 23:41:16Z

2

The itertools grouper recipe is perfect for that (https://docs.python.org/3/library/itertools.html#itertools-recipes):

In [1]: from itertools import zip_longest

In [2]: def grouper(iterable, n, fillvalue=None):
   ...:     "Collect data into fixed-length chunks or blocks"
   ...:     # grouper('ABCDEFG', 3, 'x') --> ABC DEF Gxx"
   ...:     args = [iter(iterable)] * n
   ...:     return zip_longest(*args, fillvalue=fillvalue)
   ...:

In [3]: list(grouper('UGGUGUUAUUAAUGGUUU', 3))
Out[3]:
[('U', 'G', 'G'),
 ('U', 'G', 'U'),
 ('U', 'A', 'U'),
 ('U', 'A', 'A'),
 ('U', 'G', 'G'),
 ('U', 'U', 'U')]

answered Feb 15, 2020 at 23:41

Randy

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1 Comment

aneroid Over a year ago

Maybe finish that with [''.join(tup) for tup in grouper('UGGUGUUAUUAAUGGUUU', 3)].

C-RAD · Accepted Answer · 2020-02-15 23:46:04Z

0

You might want to use a while loop here and increment the index by 3 each iteration, printing the next three letters, and exiting when the inedex is within 3 of the length

answered Feb 15, 2020 at 23:46

C-RAD

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Comments

jsuisnul · Accepted Answer · 2020-02-15 23:57:31Z

0

With regular expression :

import re

def get_codon_list(codon_string):    
    return list(re.findall(r"(\w{3})", codon_string))

answered Feb 15, 2020 at 23:57

jsuisnul

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