[英]calculate all possible combinations of RNA codons for a protein sequence
我有一個蛋白質序列:
sequence_protein = 'IEEATHMTPCYELHGLRWVQIQDYAINVMQCL'
以及每種蛋白質的 tRNA 密碼子表:
codon_table = {
'A': ('GCT', 'GCC', 'GCA', 'GCG'),
'C': ('TGT', 'TGC'),
'D': ('GAT', 'GAC'),
'E': ('GAA', 'GAG'),
'F': ('TTT', 'TTC'),
'G': ('GGT', 'GGC', 'GGA', 'GGG'),
'H': ('CAT', 'CAC'),
'I': ('ATT', 'ATC', 'ATA'),
'K': ('AAA', 'AAG'),
'L': ('TTA', 'TTG', 'CTT', 'CTC', 'CTA', 'CTG'),
'M': ('ATG',),
'N': ('AAT', 'AAC'),
'P': ('CCT', 'CCC', 'CCA', 'CCG'),
'Q': ('CAA', 'CAG'),
'R': ('CGT', 'CGC', 'CGA', 'CGG', 'AGA', 'AGG'),
'S': ('TCT', 'TCC', 'TCA', 'TCG', 'AGT', 'AGC'),
'T': ('ACT', 'ACC', 'ACA', 'ACG'),
'V': ('GTT', 'GTC', 'GTA', 'GTG'),
'W': ('TGG',),
'Y': ('TAT', 'TAC'),}
然后我寫了一個 function 它將給出一個元組,其中包含每種蛋白質的可能密碼子:
tRNA = []
for i in sequence_protein:
for residue in i:
tRNA.append(codon_table[residue])
這給了這個 output:
[('ATT', 'ATC', 'ATA'),
('GAA', 'GAG'),
('GAA', 'GAG'),
('GCT', 'GCC', 'GCA', 'GCG'),
('ACT', 'ACC', 'ACA', 'ACG'),
('CAT', 'CAC'),
('ATG',),
('ACT', 'ACC', 'ACA', 'ACG'),
('CCT', 'CCC', 'CCA', 'CCG'),
('TGT', 'TGC'),
('TAT', 'TAC'),
('GAA', 'GAG'),
('TTA', 'TTG', 'CTT', 'CTC', 'CTA', 'CTG'),
('CAT', 'CAC'),
('GGT', 'GGC', 'GGA', 'GGG'),
('TTA', 'TTG', 'CTT', 'CTC', 'CTA', 'CTG'),
('CGT', 'CGC', 'CGA', 'CGG', 'AGA', 'AGG'),
('TGG',),
('GTT', 'GTC', 'GTA', 'GTG'),
('CAA', 'CAG'),
('ATT', 'ATC', 'ATA'),
('CAA', 'CAG'),
('GAT', 'GAC'),
('TAT', 'TAC'),
('GCT', 'GCC', 'GCA', 'GCG'),
('ATT', 'ATC', 'ATA'),
('AAT', 'AAC'),
('GTT', 'GTC', 'GTA', 'GTG'),
('ATG',),
('CAA', 'CAG'),
('TGT', 'TGC'),
('TTA', 'TTG', 'CTT', 'CTC', 'CTA', 'CTG')]
有沒有辦法計算序列的所有可能的密碼子組合(基本上計算元組中所有單獨元素的乘積)? 並且還要計算在不首先生成序列的情況下會有多少產品?
我嘗試使用產品 function 但這使我的筆記本崩潰:s
combs = []
for a in product(*tRNA):
combs.append(a)
print(a)
sequence_protein = 'IEEATHMTPCYELHGLRWVQIQDYAINVMQCL'
total_number_combinations = np.prod([ len(codon_table[aa]) for aa in sequence_protein ])
最優雅的是itertools:
from itertools import product
tRNA = [codon_table[aa] for aa in sequence_protein]
for i in product(*tRNA):
#...do whatever you have to do with these combinations.
但您可以使用自定義 function。 只需使用yield
這樣您就不會一次生成所有序列並避免 memory 問題。
import itertools
list_codons = [('ATT', 'ATC', 'ATA'),
('GAA', 'GAG'),
('GAA', 'GAG'),
('GCT', 'GCC', 'GCA', 'GCG'),
('ACT', 'ACC', 'ACA', 'ACG'),
('CAT', 'CAC'),
('ATG',),
('ACT', 'ACC', 'ACA', 'ACG'),
('CCT', 'CCC', 'CCA', 'CCG'),
('TGT', 'TGC'),
('TAT', 'TAC'),
('GAA', 'GAG'),
('TTA', 'TTG', 'CTT', 'CTC', 'CTA', 'CTG'),
('CAT', 'CAC'),
('GGT', 'GGC', 'GGA', 'GGG'),
('TTA', 'TTG', 'CTT', 'CTC', 'CTA', 'CTG'),
('CGT', 'CGC', 'CGA', 'CGG', 'AGA', 'AGG'),
('TGG',),
('GTT', 'GTC', 'GTA', 'GTG'),
('CAA', 'CAG'),
('ATT', 'ATC', 'ATA'),
('CAA', 'CAG'),
('GAT', 'GAC'),
('TAT', 'TAC'),
('GCT', 'GCC', 'GCA', 'GCG'),
('ATT', 'ATC', 'ATA'),
('AAT', 'AAC'),
('GTT', 'GTC', 'GTA', 'GTG'),
('ATG',),
('CAA', 'CAG'),
('TGT', 'TGC'),
('TTA', 'TTG', 'CTT', 'CTC', 'CTA', 'CTG')]
counter = 0; max_proc = 1000000; list_seq = []
for x in itertools.product(*list_codons):
counter += 1
if counter % max_proc == 0:
#Do your stuff by slice and clear the list
list_seq = []
list_seq.append(x)
print (counter)
print (x)
就是這樣,沒有更多的RAM問題
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