python生信01

 

001、生成

nN

nnNN

nnnNNN

....

a、

[root@pc1 test1]# ls
test.py
[root@pc1 test1]# cat test.py        ## 测试程序
#!/usr/bin/env python3
# -*- coding: utf-8 -*-

for i in range(1,11):
        for j in range(1,i):
                print("n", end = "")
        for k in range(1,i):
                print("N", end = "")
        print("")
[root@pc1 test1]# python3 test.py     ## 执行程序

nN
nnNN
nnnNNN
nnnnNNNN
nnnnnNNNNN
nnnnnnNNNNNN
nnnnnnnNNNNNNN
nnnnnnnnNNNNNNNN
nnnnnnnnnNNNNNNNNN

 

b、

[root@pc1 test1]# ls
test.py
[root@pc1 test1]# cat test.py            ### 程序
#!/usr/bin/env python3
# -*- coding:utf-8 -*-
for i in range(1,10):
        k = 1
        while k <= i:
                print("n", end = "")
                k += 1
        k= 1
        while k <=i:
                print("N", end = "")
                k += 1
        print("")
[root@pc1 test1]# python3 test.py         ## 运行程序
nN
nnNN
nnnNNN
nnnnNNNN
nnnnnNNNNN
nnnnnnNNNNNN
nnnnnnnNNNNNNN
nnnnnnnnNNNNNNNN
nnnnnnnnnNNNNNNNNN

 

002、计算3bp组合的频率

a、

[root@pc1 test1]# ls
test.py
[root@pc1 test1]# cat test.py        ## 测试程序
#!/usr/bin/env python3
# -*- coding: utf-8 -*-

dna="CTCCTGAACAACCCCACTGTACTTCCC"    ## 测试序列
dict1 = dict()
while len(dna) >= 3:
        codon = dna[:3]
        if codon in dict1:
                dict1[codon] += 1
        else:
                dict1[codon] = 1
        dna = dna[1:]
for i,j in dict1.items():
        print(i, j)
[root@pc1 test1]# python3 test.py | head -n 5     ## 运算结果
CTC 1
TCC 2
CCT 1
CTG 2
TGA 1

 

b、

 

 

 

 

 

 

 

参考：https://blog.csdn.net/m0_57099761/article/details/123464340