如何在序列中找到三个字母？

Question

我有一个序列如下：

my_file_m= "TCCATTCTCTACCCAGCCCCCACTCTGACCCCTTTACTCTGACCCCTTTATTGTCTACTCCTCAGAGCCCCCAGTCTGTA
TCCTTCTAACTTAGAAAGGGGATTATGGCTCAGGGTCCAACTCTGTGCTCAGAGCTTTCAACAACTACTCAGAAACACAA
GATGCTGGGACAGTGACCTGGACTGTGGGCCTCTCATGCACCACCATCAAGGACTCAAATGGGCTTTCCGAATTCACTGG
AGCCTCGAATGTCCATTCCTGAGTTCTGCAAAGGGAGAGTGGTCAGGTTGCCTCTGTCTCAGAATGAGGCTGGATAAGAT"

我想找出具体的三个字母 TAA、TGA 和 TAG 的位置和数量。如果有的话，我想给它们上色。

我从加载字母开始

my_file = open(my_file_m)
mine = my_file.read()
print(mine)

我不能使用 .count 也不能​​使用 find，因为我有三个输入。有什么想法如何找到它们并突出显示它们吗？

Answer 1

这是我对您问题的解决方案：

注意：此代码还可以查找重叠序列。根据您是否要允许重叠，您必须删除 '?='

import re 

class bcolors:
    HEADER = '\033[95m'
    OKBLUE = '\033[94m'
    OKGREEN = '\033[92m'
    WARNING = '\033[93m'
    FAIL = '\033[91m'
    ENDC = '\033[0m'
    BOLD = '\033[1m'
    UNDERLINE = '\033[4m'

my_file_m= '''TTCCATTCTCTACCCAGCCCCCACTCTGACCCCTTTACTCTGACCCCTTTATTGTCTACTCCTCAGAGCCCCCAGTCTGTATCCTTCTAACTTAGAAAGGGGATTATGGCTCAGGGTCCAACTCTGTGCTCAGAGCTTTCAACAACTACTCAGAAACACAAGATGCTGGGACAGTGACCTGGACTGTGGGCCTCTCATGCACCACCATCAAGGACTCAAATGGGCTTTCCGAATTCACTGGAGCCTCGAATGTCCATTCCTGAGTTCTGCAAAGGGAGAGTGGTCAGGTTGCCTCTGTCTCAGAATGAGGCTGGATAAGAT'''


pat = re.compile(r'(?=(TAA|AAT|TGA|TAG))') # Very important, if you do not need overlaps then remove '?='
matches = re.finditer(pat,my_file_m)
result1 = [int(match.start(1)) for match in matches] # find all the starting positions of the string
result2 = [range(x,x+3) for x in result1 ] # find all the positions of the characters (given that we search for patterns of length 3, can be modified for other lengths too )
result3 = set().union(*result2) # generate a union

for chari in range(len(my_file_m)): # colorize based on if it is in a sequence or not
    if(chari in result3):
        print bcolors.OKGREEN + my_file_m[chari]  + bcolors.ENDC,
    else:
        print my_file_m[chari],

---

清洁工：

import re 
import sys

my_file_m= '''TAATTCCATTCTCTACCCAGCCCCCACTCTGACCCCTTTACTCTGACCCCTTTATTGTCTACTCCTCAGAGCCCCCAGTCTGTATCCTTCTAACTTAGAAAGGGGATTATGGCTCAGGGTCCAACTCTGTGCTCAGAGCTTTCAACAACTACTCAGAAACACAAGATGCTGGGACAGTGACCTGGACTGTGGGCCTCTCATGCACCACCATCAAGGACTCAAATGGGCTTTCCGAATTCACTGGAGCCTCGAATGTCCATTCCTGAGTTCTGCAAAGGGAGAGTGGTCAGGTTGCCTCTGTCTCAGAATGAGGCTGGATAAGAT'''

pat = re.compile(r'(?=(TAA|TGA|TAG))') # Very important, if you do not need overlaps then remove '?='
lettersToColor = set().union(*[range(m.start(1),m.start(1)+3) for m in re.finditer(pat, my_file_m)])

for chari in range(len(my_file_m)): # colorize based on if it is in a sequence or not
    if(chari in lettersToColor):
        sys.stdout.write('\033[92m' + my_file_m[chari]  +'\033[0m')
    else:
        sys.stdout.write(my_file_m[chari])

归功于：here 和 here

输出：

Answer 2

使用标准库中的re.findall 函数和collection.Counter

import re
from collections import Counter

pat = re.compile(r"(TAA|TGA|TAG)")
c = re.findall(pat,my_file_m)

print(c)
print(Counter(c))

输出

['TGA', 'TGA', 'TAA', 'TAG', 'TGA', 'TGA', 'TGA', 'TAA']
Counter({'TGA': 5, 'TAA': 2, 'TAG': 1})

Answer 3

您是否需要将 DNA 序列每三个字母分开来绘制遗传密码？

如果有，请看下面的代码。

my_file_m= '''TCCATTCTCTACCCAGCCCCCACTCTGACCCCTTTACTCTGACCCCTTTATTGTCTACTCCTCAGAGCCCCCAGTCTGTA
TCCTTCTAACTTAGAAAGGGGATTATGGCTCAGGGTCCAACTCTGTGCTCAGAGCTTTCAACAACTACTCAGAAACACAA
GATGCTGGGACAGTGACCTGGACTGTGGGCCTCTCATGCACCACCATCAAGGACTCAAATGGGCTTTCCGAATTCACTGG
AGCCTCGAATGTCCATTCCTGAGTTCTGCAAAGGGAGAGTGGTCAGGTTGCCTCTGTCTCAGAATGAGGCTGGATAAGAT'''

mm = "".join(my_file_m.split())                 # delete the new line characters

messenger = map(''.join, zip(*[iter(mm)]*3))    # split every three letters

print messenger.count('TAA')
print messenger.count('TGA')
print messenger.count('TAG')

输出

0
1
0