DNA链序列的碱基计数

Question

如何用python编写一段代码，读取DNA序列链并返回它的重复碱基列表，描述这三件事：哪个是碱基（AGTC），它在链中的位置以及重复的次数。例如：

ACTTTTGTCTAAACCCCGTCCTATAACT

这个函数的输出是：list_base=[（'T'，3,4），（'A'，11,3），（'C'，14,6）]

Answer 1

我做了以下工作：

import re
from collections import defaultdict

seq = "ACTTTTGTCTAAACCCCCCGTCCTATATATAACT"
bases = ['A','G','C','T']

indexes = defaultdict(list)
counts = dict()

for base in bases:
    comSeq = re.compile(base)
    matches = comSeq.findall(seq)
    count = len(matches)
    counts[base] = count

    start = 0

    for match in matches:
        index = seq.find(base, start)
        indexes[base].append(index)
        start = index +1

print(indexes)
print(counts)

dict索引为您提供链中基的每个位置：

{'A': [0, 10, 11, 12, 24, 26, 28, 30, 31], 'G': [6, 19], 'C': [1, 8, 13, 14,
 15, 16, 17, 18, 21, 22, 32], 'T': [2, 3, 4, 5, 7, 9, 20, 23, 25, 27, 29, 33]}

dict counts为您提供了基在链中出现的次数：

{'A': 9, 'G': 2, 'C': 11, 'T': 12}

这可能不是最好、最有效的代码，我也不确定你想要什么，希望这能有所帮助

Answer 2

这就是你要找的吗

DNA_seq = 'ACTTTTGTCTAAACCCCCCGTCCTATATATAACT'
count_dic = {'A': [0,0], "G": [0,0], "C": [0,0], "T": [0,0]}
for i in range(len(DNA_seq)-1):
    j=i
    seq_count = 1
    while DNA_seq[j] == DNA_seq[j+1]:
        seq_count +=1 
        j +=1
        if seq_count > count_dic[DNA_seq[i]][1]:
            count_dic[DNA_seq[i]][1] = seq_count
            count_dic[DNA_seq[i]][0] = i + 1

count_dic的内容是

{'A': [11, 3], 'G': [0, 0], 'C': [14, 6], 'T': [3, 4]}