我有一个字符串CCAATA CCGT
,我正在尝试获取连续子序列的固定长度 n。然后,我想得到这样的东西:
该字符串中每个子序列的索引。 0-3、1-4、2-5 等
0 thru 3 : CCAA
1 thru 4 : CAAT
2 thru 5 : AATA
3 thru 6 : ATAC
4 thru 7 : TACC
5 thru 8 : ACCG
6 thru 9 : CCGT
列表大小为 7。在这里,我正在循环浏览列表并获取索引和lastIndexOf。之后,3 thru 6 : ATAC
,我得到
线程"main"中的异常 java.lang.IndexOutOfBoundsException: Index: 7, Size: 7
for (int i = 0; i < list.size(); i++) {
System.out.println(ss.indexOf(list.get(i))
+ " thru " + ss.lastIndexOf(list.get(i + n - 1)) + " : "
+ list.get(i));
演示:
import java.util.ArrayList;
public class Subsequences {
public static void main(String[] args) {
String s = "CCAATA CCGT";
ArrayList<String> list = new ArrayList<String>(); // list of subsequence
int n = 4; // subsequences of length
String ss = s.replaceAll("\s+", "");
String substr = null;
for (int i = 0; i <= ss.length() - n; i++) {
substr = ss.substring(i, i + n);
list.add(substr);
}
for (int i = 0; i < list.size(); i++) {
System.out.println(ss.indexOf(list.get(i))
+ " thru " + ss.lastIndexOf(list.get(i + n - 1)) + " : "
+ list.get(i));
}
}
}
有什么提示吗?
n
添加到lastIndexOf
中,因为您将substring
隔开了 4。List
中的每个条目由 4 个字符组成。将索引检查更改为此
(ss.lastIndexOf(list.get(i)) + n - 1)
最后看起来像这样
for (int i = 0; i < list.size(); i++) {
System.out.println(ss.indexOf(list.get(i))
+ " thru " + (ss.lastIndexOf(list.get(i)) + n - 1) + " : "
+ list.get(i));
}
输出:
0 thru 3 : CCAA
1 thru 4 : CAAT
2 thru 5 : AATA
3 thru 6 : ATAC
4 thru 7 : TACC
5 thru 8 : ACCG
6 thru 9 : CCGT
你的问题在list.get(i + n - 1)
.您当前正在迭代,以便每个子序列的开头范围从 0
到 list.size() - 1
。最后一个有意义的子序列是位置list.size() - n
到list.size() - 1
的n
字符。
for (int i = 0; i < list.size() - n; i++) {
System.out.println(ss.indexOf(list.get(i))
+ " thru " + ss.lastIndexOf(list.get(i + n - 1)) + " : "
+ list.get(i));
}
<</div>
div class="one_answers"> 删除所有空格,循环:
String data = "CCAATA CCGT";
String replaced = data.replaceAll("\s", "");
for (int i = 0; i < replaced.length() - 4 + 1; i++) {
System.out.println(replaced.subSequence(i, i + 4));
}
输出:
CCAA
CAAT
AATA
ATAC
TACC
ACCG
CCGT
在你的循环中
for (int i = 0; i < list.size(); i++) {
System.out.println(ss.indexOf(list.get(i))
+ " thru " + ss.lastIndexOf(list.get(i + n - 1))
+ " : " + list.get(i));
}
当你做list.get(i + n - 1)
并且你的i
是4时,成瘾的结果将是4 + 4 - 1 = 7,并且你无法获得具有相同或更大索引的列表的成员你的list.size()
,所以系统抛出异常
要获得预期的结果,您可以执行以下操作:
import java.util.ArrayList;
public class Subsequences {
public static void main(String[] args) {
String s = "CCAATA CCGT";
ArrayList<String> list = new ArrayList<String>(); // list of subsequence
int n = 4; // subsequences of length
String ss = s.replaceAll("\s+", "");
String substr = null;
for (int i = 0; i <= ss.length() - n; i++) {
substr = ss.substring(i, i + n);
list.add(substr);
}
// --------Here the edits-------
for (int i = 0; i < list.size(); i++)
System.println(i + " thru " + (i+n-1) + " : " + list.get(i))
// -----------------------------
}
}
您也可以使用简单的正则表达式来执行此操作。删除空格并运行此正则表达式:
(?=(.{4}))
样本:
package com.see;
import java.util.ArrayList;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class RegexTest {
private static final String TEST_STR = "CCAATA CCGT";
public ArrayList<String> getMatchedStrings(String input) {
ArrayList<String> matches = new ArrayList<String>();
input = input.replaceAll("\s", "");
Pattern pattern = Pattern.compile("(?=(.{4}))");
Matcher matcher = pattern.matcher(input);
while (matcher.find())
matches.add(matcher.group(1));
return matches;
}
public static void main(String[] args) {
RegexTest rt = new RegexTest();
for (String string : rt.getMatchedStrings(TEST_STR)) {
System.out.println(string);
}
}
}