使用列表，字典和循环匹配密码子

Question

This is one of the more difficult problems I have faced so far so excuse my for not providing a substantial attempt at doing it.

I want a program that prints matching codons in the format:

AAA : TTT
GGG : CCC
TTT : AAA
CCC : GGG
    .
    .
    .

Here is what I did:

pairs = {'A':'T','C':'G','T':'A','G':'C'}
codonsA = ['AAG', 'TAC', 'CGG', 'GAT', 'TTG', 'GTG', 'CAT', 'GGC', 'ATT', 'TCT']
codonsB = ['TAA', 'CTA', 'AAC', 'TTC', 'AGA', 'CAC', 'CCG', 'ATG', 'GCC', 'GTA']
for A in codonsA:
    print A + ' :',
    for B in codonsB:
        print B,
    print

#OUTPUT:

AAG : TAA CTA AAC TTC AGA CAC CCG ATG GCC GTA
TAC : TAA CTA AAC TTC AGA CAC CCG ATG GCC GTA
CGG : TAA CTA AAC TTC AGA CAC CCG ATG GCC GTA
GAT : TAA CTA AAC TTC AGA CAC CCG ATG GCC GTA
TTG : TAA CTA AAC TTC AGA CAC CCG ATG GCC GTA
GTG : TAA CTA AAC TTC AGA CAC CCG ATG GCC GTA
CAT : TAA CTA AAC TTC AGA CAC CCG ATG GCC GTA
GGC : TAA CTA AAC TTC AGA CAC CCG ATG GCC GTA
ATT : TAA CTA AAC TTC AGA CAC CCG ATG GCC GTA
TCT : TAA CTA AAC TTC AGA CAC CCG ATG GCC GTA

What I need to do now is to get rid off 9 codons from B from each row on the right of the colon and leave only the one codon on the right that matches the codon from A on the left. How do I do that?

Answer 1

You don't even need that codonsB list. A codon maps onto its anticodon complement 1:1.

def anticodon(codon):
    """returns the anticodon complement for a given codon"""
    return ''.join(pairs[c] for c in codon)

anticodon('AAG')
Out[5]: 'TTC'

You're free to check if each element in codonsA has an anticodon in codonsB , if you need to make that check.

all(anticodon(c) in codonsB for c in codonsA)
Out[6]: True

And the output I think you were originally looking for:

for codon in codonsA:
    print '{} : {}'.format(codon,anticodon(codon))

AAG : TTC
TAC : ATG
CGG : GCC
GAT : CTA
TTG : AAC
GTG : CAC
CAT : GTA
GGC : CCG
ATT : TAA
TCT : AGA