Normalising Short Tandem Repeat Units

def circular_permuted1(seq):
    all_possible = []
    for i in range(len(seq)):
        permuted_seq = (seq + seq)[i:i+len(seq)]
        all_possible.append(permuted_seq)
    return(all_possible)


from collections import deque

def circular_permuted2(seq):
    all_possible = []
    d = deque(a)
    for i in range(len(a)):
        d.rotate()
        all_possible.append(''.join(list(d)))
    return(all_possible)


import toolz

def circular_permuted3(seq):
    all_possible = []
    for permuted_seq in toolz.sliding_window(len(seq), seq+seq[:-1]):
        all_possible.append(''.join(permuted_seq))
    return(all_possible)


def circular_permuted4(seq):
    n = len(seq)
    return([''.join([seq[i - j] for i in range(n)]) for j in range(n)])


def circular_permuted5(seq):
    n = len(seq)
    return([(seq + seq)[i:i+len(seq)] for i in range(n)])


def circular_permuted6(x):
    return([x[i:]+x[:i] for i in range(len(x))])


a = "ATCG"


%%timeit -r10
circular_permuted1(a)

100000 loops, best of 10: 2.09 µs per loop



%%timeit -r10
circular_permuted2(a)

100000 loops, best of 10: 3.62 µs per loop



%%timeit -r10
circular_permuted3(a)

The slowest run took 4.22 times longer than the fastest. This could mean that an intermediate result is being cached 
100000 loops, best of 10: 3.65 µs per loop



%%timeit -r10
circular_permuted4(a)

100000 loops, best of 10: 5.8 µs per loop



%%timeit -r10
circular_permuted5(a)

100000 loops, best of 10: 2.01 µs per loop



%%timeit -r10
circular_permuted6(a) # Winner!

1000000 loops, best of 10: 1.87 µs per loop

from Bio.Seq import Seq
from Bio.Alphabet import generic_dna

def self_and_rev_complement1(in_dna):
    all_possible = [in_dna]
    
    # Get reverse complement
    dna = Seq(in_dna, generic_dna)
    rev_complement = str(dna.reverse_complement())
    all_possible.append(rev_complement)
    return(all_possible)

from skbio.sequence import DNASequence

def self_and_rev_complement2(in_dna):
    all_possible = [in_dna]
    
    # Get reverse complement
    dna = DNASequence(in_dna)
    rev_complement = str(dna.reverse_complement())
    all_possible.append(rev_complement)
    return(all_possible)


%%timeit -r10
self_and_rev_complement1(a)

The slowest run took 6.29 times longer than the fastest. This could mean that an intermediate result is being cached 
100000 loops, best of 10: 5.99 µs per loop



%%timeit -r10
self_and_rev_complement2(a)

The slowest run took 4.44 times longer than the fastest. This could mean that an intermediate result is being cached 
100000 loops, best of 10: 10.7 µs per loop



self_and_rev_complement2(a)




['ATCG', 'CGAT']

from Bio.Seq import Seq
from Bio.Alphabet import generic_dna

def normalise_str(in_dna):
    """Find all possible eqivalent STR sequences. 
    And return the first alphabetically.
    
    For example, TA = AT. But would return AT.
    """
    all_possible = [in_dna]
    
    # Get reverse complement
    dna = Seq(in_dna, generic_dna)
    rev_complement = str(dna.reverse_complement())
    all_possible.append(rev_complement)
    
    # Permute
    for seq in [in_dna, rev_complement]:
        for permuted_seq in circular_permuted1(seq): # Switch to faster permutation (6)
            all_possible.append(permuted_seq)

    # Sort and take the first
    all_possible.sort()
    return(all_possible[0])


from skbio.sequence import DNASequence

def normalise_str2(in_dna):
    """Find all possible eqivalent STR sequences. 
    And return the first alphabetically.
    
    For example, TA = AT. But would return AT.
    """
    all_possible = [in_dna]
    
    # Get reverse complement
    dna = DNASequence(in_dna)
    rev_complement = str(dna.reverse_complement())
    all_possible.append(rev_complement)
    
    # Permute
    for seq in [in_dna, rev_complement]:
        for permuted_seq in circular_permuted1(seq): # Switch to faster permutation (6)
            all_possible.append(permuted_seq)

    # Sort and take the first
    all_possible.sort()
    return(all_possible[0])


%%timeit -r10
normalise_str(a)

100000 loops, best of 10: 12.9 µs per loop



%%timeit -r10
normalise_str2(a)

10000 loops, best of 10: 17.4 µs per loop