Rosalind

Rosalind是一个生物信息编程学习网站。抛出了一系列生物学问题，引导使用者通过编程语言解决。网站官方推荐使用python解决，python得益于丰富的开源包，其解决方法在CSDN中也多有涉及。小编使用的主语言是java，因此简单通过CSDN社区进行Java学习笔记记录，不足之处也请大家多多指点~

先来看一下问题描述：

Counting DNA Nucleotides

Problem
A string is simply an ordered collection of symbols selected from some alphabet and formed into a word; the length of a string is the number of symbols that it contains.

An example of a length 21 DNA string (whose alphabet contains the symbols ‘A’, ‘C’, ‘G’, and ‘T’) is “ATGCTTCAGAAAGGTCTTACG.”

Given: A DNA string s of length at most 1000 nt.
Sample input：

GAGTTAGGAATTCGGTCGCGAAACTTGCGATCGTGTTACGGCCTGGGTCATTACGAAATTCCCTAGCTCCGCAGTGTTCCTGGAGTGCCATGTCGTCGCTGCCATGCTCAACCGAGAGCAGCCCGTACTACGTGTCTGGTCCTTACAGGACCTAAGCGAATCAATGTGACTACTTTCATAGGTAGGGTTTGTCGTGTCATGGATACGTCTGACAACAACTCGTGGTTGGGGCTGCGCGCATTGATTGTGAGCGAAATACTCGCAAACCAGACTGTCTGAGATAGTCACATCAGACAACCCTGGTCTCTAGCAAAAATCGTTTTCCTATAAATCACGTAACGCAGTAATCTTCAGGCCTCGCGCCAGTCCGCGATAAGACCATCCCCTGCCCTATCCCGCTAATGGCGAGACCCAAAGGACGAGCCTACGACGATCATCGGACAATCAAGGCGGAAGACGCTAGCACTGATTCTTCGGCTCCTAGACAGCGTAGATTTCAGCTATCACCATAATTTTGGTCGTACGGAGGCCCTTTCCCTCGTCAACTCTGTGTCCACGTACGTAACCAGCCACGACAGTATCTTAAACTCCATGGGTCATATATTCGTACAAGTCCGTCGATTAAGCGTTATGGGCTGCTAATTAGACTCCACCTATGCAGGAGTTGTTGTACCGCATCGGCGATTATCCGCCACTCGAAGAAGTTTAACTGCCTATTATATCTTTGGAGACACTGGTTATGTTTTAATACGCACGCATCTTAGTTCAACGGGACGTGGGCGCACGAGCTATCCTGCTAGGATACTTCACTCGCTTCAGTCACCTATGTCTAGGCCCACTATAAGCCGTGACAT

Return: Four integers (separated by spaces) counting the respective number of times that the symbols ‘A’, ‘C’, ‘G’, and ‘T’ occur in s.

简单来说就是通过输入核酸序列，以ACGT的顺序返回相应的碱基数目。解决方法如下~~

import java.util.Scanner;public class Counting_DNA_Nucleotides {public static void main(String[] args) {Scanner sc = new Scanner(System.in);System.out.println("请输入核酸序列：");//键盘录入核酸序列String line = sc.nextLine();//存储录入的核酸序列为String类型int a=0;int c=0;int g=0;int t=0;for (int i = 0; i <line.length();i++){switch (line.charAt(i)){//括号中的输出类型为char类型，匹配后面的字符时需要将字符加单引号转化为对应的ASCII码方可匹配。case 'A':case 'a':a++;break;case 'C':case 'c':c++;break;case 'G':case 'g':g++;break;case 'T':case 't':t++;break;default:System.out.println("第"+(i+1)+"位不是ACTG");break;}}System.out.println("ACGT分别为："+a+" "+c+" "+g+" "+t);}
}

Sample output：

ACGT分别为：207 229 198 220

但是有些小伙伴不想通过手动输入核酸序列，这里也为大家提供了读取txt文本文件并输出碱基数的方法。两个方法都可行，下述代码只是在上述代码基础上增加了读取文本文件的步骤。

import java.io.BufferedReader;
import java.io.File;
import java.io.FileReader;
import java.io.IOException;public class Counting_DNA_Nucleotides_ReadFile {public static void main(String[] args) {//手动输入Rosalind网站下载的输入文件rosalind_dna.txt的路径地址到filePath中String filePath = "C:/Users/Administrator/Desktop/rosalind_dna.txt";System.out.println("核酸序列读取完毕，结果如下：");System.out.println(readFileContent(filePath));String line = readFileContent(filePath);int a=0;int c=0;int g=0;int t=0;for (int i = 0; i <line.length();i++) {switch (line.charAt(i)) {//括号中的输出类型为char类型，匹配后面的字符时需要将字符加单引号转化为对应的ASCII码方可匹配。case 'A':case 'a':a++;break;case 'C':case 'c':c++;break;case 'G':case 'g':g++;break;case 'T':case 't':t++;break;default:System.out.println("第" + (i + 1) + "位不是ACTG");break;}}System.out.println("ACGT分别为："+a+" "+c+" "+g+" "+t);}//定义方法，输入文本文件路径，以字符串类型返回文本内容public static String readFileContent(String fileName) {File file = new File(fileName);BufferedReader reader = null;StringBuffer sbf = new StringBuffer();try {reader = new BufferedReader(new FileReader(file));String tempStr;while ((tempStr = reader.readLine()) != null) {sbf.append(tempStr);}reader.close();return sbf.toString();} catch (IOException e) {e.printStackTrace();} finally {if (reader != null) {try {reader.close();} catch (IOException e1) {e1.printStackTrace();}}}return sbf.toString();}
}

Sample output：

核酸序列读取完毕，结果如下：
GAGTTAGGAATTCGGTCGCGAAACTTGCGATCGTGTTACGGCCTGGGTCATTACGAAATTCCCTAGCTCCGCAGTGTTCCTGGAGTGCCATGTCGTCGCTGCCATGCTCAACCGAGAGCAGCCCGTACTACGTGTCTGGTCCTTACAGGACCTAAGCGAATCAATGTGACTACTTTCATAGGTAGGGTTTGTCGTGTCATGGATACGTCTGACAACAACTCGTGGTTGGGGCTGCGCGCATTGATTGTGAGCGAAATACTCGCAAACCAGACTGTCTGAGATAGTCACATCAGACAACCCTGGTCTCTAGCAAAAATCGTTTTCCTATAAATCACGTAACGCAGTAATCTTCAGGCCTCGCGCCAGTCCGCGATAAGACCATCCCCTGCCCTATCCCGCTAATGGCGAGACCCAAAGGACGAGCCTACGACGATCATCGGACAATCAAGGCGGAAGACGCTAGCACTGATTCTTCGGCTCCTAGACAGCGTAGATTTCAGCTATCACCATAATTTTGGTCGTACGGAGGCCCTTTCCCTCGTCAACTCTGTGTCCACGTACGTAACCAGCCACGACAGTATCTTAAACTCCATGGGTCATATATTCGTACAAGTCCGTCGATTAAGCGTTATGGGCTGCTAATTAGACTCCACCTATGCAGGAGTTGTTGTACCGCATCGGCGATTATCCGCCACTCGAAGAAGTTTAACTGCCTATTATATCTTTGGAGACACTGGTTATGTTTTAATACGCACGCATCTTAGTTCAACGGGACGTGGGCGCACGAGCTATCCTGCTAGGATACTTCACTCGCTTCAGTCACCTATGTCTAGGCCCACTATAAGCCGTGACAT
ACGT分别为：207 229 198 220

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Rosalind Java| Counting DNA Nucleotides

Rosalind

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