如何将序列代码字符串作为字典返回，它告诉我氨基酸发生的频率？

如何解决如何将序列代码字符串作为字典返回，它告诉我氨基酸发生的频率？

基本上，我正在尝试编写一个程序，该程序运行一个字符串，并返回所有可能的氨基酸以及它们出现的频率。我制作了这个程序，它给了我名字而不是数字。有人可以帮我吗？

DNA_Codons = {
# U
'UUU': 'Phenylalanin','UCU': 'Serin','UAU': 'Tyrosin','UGU': 'Cystein',# UxU
'UUC': 'Phenylalanin','UCC': 'Serin','UAC': 'Tyrosin','UGC': 'Cystein',# UxC
'UUA': 'Leucin','UCA': 'Serin','UAA': '---','UGA': '---',# UxA
'UUG': 'Leucin','UCG': 'Serin','UAG': '---','UGG': 'Tryptophan',# UxG

# C
'CUU': 'Leucin','CCU': 'Prolin','CAU': 'Histidin','CGU': 'Arginin',# CxU
'CUC': 'Leucin','CCC': 'Prolin','CAC': 'Histidin','CGC': 'Arginin',# CxC
'CUA': 'Leucin','CCA': 'Prolin','CAA': 'Glutamin','CGA': 'Arginin',# CxA
'CUG': 'Leucin','CCG': 'Prolin','CAG': 'Glutamin','CGG': 'Arginin',# CxG

# A
'AUU': 'Isoleucin','ACU': 'Threonin','AAU': 'Asparagin','AGU': 'Serin',# AxU
'AUC': 'Isoleucin','ACC': 'Threonin','AAC': 'Asparagin','AGC': 'Serin',# AxC
'AUA': 'Isoleucin','ACA': 'Threonin','AAA': 'Lysin','AGA': 'Arginin',# AxA
'AUG': 'Met','ACG': 'Threonin','AAG': 'Lysin','AGG': 'Arginin',# AxG

# G
'GUU': 'Valin','GCU': 'Alanin','GAU': 'Asparaginsäure','GGU': 'Glycin',# GxU
'GUC': 'Valin','GCC': 'Alanin','GAC': 'Asparaginsäure','GGC': 'Glycin',# GxC
'GUA': 'Valin','GCA': 'Alanin','GAA': 'Glutaminsäure','GGA': 'Glycin',# GxA
'GUG': 'Valin','GCG': 'Alanin','GAG': 'Glutaminsäure','GGG': 'Glycin'  # GxG }

def translate_code(seq,init_pos=0):
return {
    DNA_Codons[seq[pos:pos + 3]]
    for pos in range(init_pos,len(seq) - 2,3)
} 

print(translate_code("ACAAUUGACACAUAUCGUCGAGGGUGGCCA"))

我正在寻找的是这样的：

{'Threonin': 2,'Isoleucin': 1,'Asparaginsäure': 1,'Tyrosin': 1,'Arginin': 2,'Glycin': 1,'Tryptophan': 1,'Prolin': 1}

解决方法

import numpy as np    



def count_amino_acids(seq,init_pos=0):

    #First,create an amino acid dictionary from the codon dictionary:
    count_dict = {}    
    
    # go from the initial position,to full length in steps of 3
    for i in np.arange(init_pos,len(seq),3): 
        codon = seq[i:i+3] # get the codon
        aa = DNA_Codons[codon] # look up the amino acid
        
        if aa == '---':       # stop at stop codons
            return count_dict
        
        count_dict[aa] += 1 # increment the counter

    return count_dict
    
count_amino_acids("ACAAUUGACACAUAUCGUCGAGGGUGGCCA")

,

def translate_code(seq,init_pos=0):
    final_codons = {}
    for pos in range(init_pos,len(seq) - 2,3):
        current_codon = DNA_Codons[seq[pos:pos + 3]]
        if current_codon in final_codons:
            final_codons[current_codon] += 1
        else:
            final_codons[current_codon] = 1
    return final_codons

这应该完全按照您指定的方式工作。

,

您可以使用集合模块中的 Counter 类：

要排除您不想要的组合，我建议根本不要将它们放入密码子字典中。我还倒置了你的字典，以减少重复次数并使其更易于维护。

设置：

acids = { 'Alanin':        ['GCA','GCC','GCG','GCU'],'Arginin':       ['AGA','AGG','CGA','CGC','CGG','CGU'],'Asparagin':     ['AAC','AAU'],'Asparaginsäure':['GAC','GAU'],'Cystein':       ['UGC','UGU'],'Glutamin':      ['CAA','CAG'],'Glutaminsäure': ['GAA','GAG'],'Glycin':        ['GGA','GGC','GGG','GGU'],'Histidin':      ['CAC','CAU'],'Isoleucin':     ['AUA','AUC','AUU'],'Leucin':        ['CUA','CUC','CUG','CUU','UUA','UUG'],'Lysin':         ['AAA','AAG'],'Met':           ['AUG'],'Phenylalanin':  ['UUC','UUU'],'Prolin':        ['CCA','CCC','CCG','CCU'],'Serin':         ['AGC','AGU','UCA','UCC','UCG','UCU'],'Threonin':      ['ACA','ACC','ACG','ACU'],'Tryptophan':    ['UGG'],'Tyrosin':       ['UAC','UAU'],'Valin':         ['GUA','GUC','GUG','GUU'] }
codons = { seq:name for name,sequences in acids.items() for seq in sequences }

计数：

from collections import Counter
def translate_code(seq,init_pos=0):
    return Counter( codons[seq[pos:pos + 3]]
                    for pos in range(init_pos,3)
                    if seq[pos:pos + 3] in codons)

输出：

print(translate_code("ACAAUUGACACAUAUCGUCGAGGGUGGCCA"))

Counter({'Threonin': 2,'Arginin': 2,'Isoleucin': 1,'Asparaginsäure': 1,'Tyrosin': 1,'Glycin': 1,'Tryptophan': 1,'Prolin': 1})

注意，Counter 类实际上是一个字典。如果需要，您可以将其转换回普通字典。