<p>如果您希望使设置比完整的64键字典更简洁,您可以首先设置一个列表(或元组)字典,每个氨基酸一个,然后构建基本->;沿着以下几行查找表:</p>
<pre><code>from itertools import product
bases = 'UCAG'
amino_codes = dict(Asp=('GAU','GAC'),
Glu=('GAA', 'GAC'),
Leu=['CU'+b for b in bases] + ['UUA','UUG'],
Ser=['UC'+b for b in bases], # list comprehensions just to show the concept
Stop=('UAA,UAG,UGA') # You do the rest
)
bases_to_amino = {}
for triplet in product(bases, repeat=3):
base_key = ''.join(triplet)
amino = next((a for (a, pat) in amino_codes.items() if base_key in pat), 'missing')
bases_to_amino[base_key] = amino
from pprint import pprint
pprint(bases_to_amino)
{'AAA': 'missing',
'AAC': 'missing',
'AAG': 'missing',
'AAU': 'missing',
'ACA': 'missing',
'ACC': 'missing',
'ACG': 'missing',
'ACU': 'missing',
'AGA': 'missing',
'AGC': 'missing',
'AGG': 'missing',
'AGU': 'missing',
'AUA': 'missing',
'AUC': 'missing',
'AUG': 'missing',
'AUU': 'missing',
'CAA': 'missing',
'CAC': 'missing',
'CAG': 'missing',
'CAU': 'missing',
'CCA': 'missing',
'CCC': 'missing',
'CCG': 'missing',
'CCU': 'missing',
'CGA': 'missing',
'CGC': 'missing',
'CGG': 'missing',
'CGU': 'missing',
'CUA': 'Leu',
'CUC': 'Leu',
'CUG': 'Leu',
'CUU': 'Leu',
'GAA': 'Glu',
'GAC': 'Asp',
'GAG': 'missing',
'GAU': 'Asp',
'GCA': 'missing',
'GCC': 'missing',
'GCG': 'missing',
'GCU': 'missing',
'GGA': 'missing',
'GGC': 'missing',
'GGG': 'missing',
'GGU': 'missing',
'GUA': 'missing',
'GUC': 'missing',
'GUG': 'missing',
'GUU': 'missing',
'UAA': 'Stop',
'UAC': 'missing',
'UAG': 'Stop',
'UAU': 'missing',
'UCA': 'Ser',
'UCC': 'Ser',
'UCG': 'Ser',
'UCU': 'Ser',
'UGA': 'Stop',
'UGC': 'missing',
'UGG': 'missing',
'UGU': 'missing',
'UUA': 'Leu',
'UUC': 'missing',
'UUG': 'Leu',
'UUU': 'missing'}
</code></pre>
<p>在这种情况下,您不会从练习中得到太多,但是对于需要处理64个以上排列的类似情况,这些概念可能很有价值</p>
<p>旁注:</p>
<ul>
<li>纯粹主义者可能不喜欢我在<code>amino_codes</code>中混合元组和列表的方式。我承认这是一个较轻的罪行</李>
<li><code>amino = next(blah)</code>用于在找到匹配项后停止查找</李>
<li>您可以使用嵌套生成器进一步压缩最后5行代码。做这件事的智慧在于品味</li>
</ul>
<p></p>
<pre><code>find_amino = lambda t: next((a for (a, pat) in amino_codes.items() if t in pat), 'missing')
bases_to_amino = dict((bk, find_amino(bk)) for bk in (''.join(t) for t in product(bases, repeat=3)))
</code></pre>