CODEWITHSUNDEEP
pythondictionarylist-comprehensionfrequencyletter
user3375672
user3375672

Reputation: 3768

python position frequency dictionary of letters in words

To efficiently get the frequencies of letters (given alphabet ABC in a dictionary in a string code I can make a function a-la (Python 3) :

def freq(code):
   return{n: code.count(n)/float(len(code)) for n in 'ABC'}

Then

code='ABBBC'   
freq(code)

Gives me 

{'A': 0.2, 'C': 0.2, 'B': 0.6}

But how can I get the frequencies for each position along a list of strings of unequal lengths ? For instance mcode=['AAB', 'AA', 'ABC', ''] should give me a nested structure like a list of dict (where each dict is the frequency per position):

[{'A': 1.0, 'C': 0.0, 'B': 0.0}, 
 {'A': 0.66, 'C': 0.0, 'B': 0.33},
 {'A': 0.0, 'C': 0.5, 'B': 0.5}]

I cannot figure out how to do the frequencies per position across all strings, and wrap this in a list comprehension. Inspired by other SO for word counts e.g. the well discussed post Python: count frequency of words in a list I believed maybe the Counter module from collections might be a help.

Understand it like this - write the mcode strings on separate lines:

AAB
AA
ABC

Then what I need is the column-wise frequencies (AAA, AAB, BC) of the alphabet ABC in a list of dict where each list element is the frequencies of ABC per columns.

Upvotes: 1

Views: 1275

Answers (3)

Eric Duminil
Eric Duminil

Reputation: 54303

Your code isn't efficient at all :

  • You first need to define which letters you'd like to count
  • You need to parse the string for each distinct letter

You could just use Counter:

import itertools
from collections import Counter
mcode=['AAB', 'AA', 'ABC', '']
all_letters = set(''.join(mcode))

def freq(code):
  code = [letter for letter in code if letter is not None]
  n = len(code)
  counter = Counter(code)
  return {letter: counter[letter]/n for letter in all_letters}

print([freq(x) for x in itertools.zip_longest(*mcode)])
# [{'A': 1.0, 'C': 0.0, 'B': 0.0}, {'A': 0.6666666666666666, 'C': 0.0, 'B': 0.3333333333333333}, {'A': 0.0, 'C': 0.5, 'B': 0.5}]

For Python2, you could use itertools.izip_longest.

Upvotes: 0

Feodoran
Feodoran

Reputation: 1822

A much shorter solution:

from itertools import zip_longest

def freq(code):
    l = len(code) - code.count(None)
    return {n: code.count(n)/l for n in 'ABC'}

mcode=['AAB', 'AA', 'ABC', '']
results = [ freq(code) for code in zip_longest(*mcode) ]
print(results)

Upvotes: 1

Heiko Oberdiek
Heiko Oberdiek

Reputation: 1708

Example, the steps are shortly explained in comments. Counter of module collections is not used, because the mapping for a position also contains characters, that are not present at this position and the order of frequencies does not seem to matter. 

def freq(*words):
    # All dictionaries contain all characters as keys, even
    # if a characters is not present at a position.
    # Create a sorted list of characters in chars.
    chars = set()
    for word in words:
        chars |= set(word)

    chars = sorted(chars)

    # Get the number of positions.
    max_position = max(len(word) for word in words)

    # Initialize the result list of dictionaries.
    result = [
        dict((char, 0) for char in chars)
        for position in range(max_position)
    ]

    # Count characters.
    for word in words:
        for position in range(len(word)):
            result[position][word[position]] += 1

    # Change to frequencies
    for position in range(max_position):
        count = sum(result[position].values())
        for char in chars:
            result[position][char] /= count  # float(count) for Python 2

    return result


# Testing
from pprint import pprint
mcode = ['AAB', 'AA', 'ABC', '']
pprint(freq(*mcode))

Result (Python 3):

[{'A': 1.0, 'B': 0.0, 'C': 0.0},
 {'A': 0.6666666666666666, 'B': 0.3333333333333333, 'C': 0.0},
 {'A': 0.0, 'B': 0.5, 'C': 0.5}]

In Python 3.6, the dictionaries are even sorted; earlier versions can use OrderedDict from collections instead of dict.

Upvotes: 1

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