Java 检查列表中的元素是否出现在所有事件中

Question

我有一个方法，它接受字符串的 ArrayList，列表中的每个元素等于以下内容的变体:

>AX018718 Equine influenza virus H3N8 // 4 (HA)
CAAAAGCAGGGTGACAAAAACATGATGGATTCCAACACTGTGTCAAGCTTTCAGGTAGACTGTTTTCTTT
GGCATGTCCGCAAACGATTTGCAGACCAAGAACTGGGTGATGCCCCATTCCTTGACCGGCTTCGCCGAGA

此方法分为 Acc(本例中为 AX018718)和 seq(Acc 后面的两行)

然后由另一个名为 pal 的字符串 ArrayList 检查子字符串是否匹配 [AAAATTTT, AAACGTTT, AATATATTT]

我能够获得第一个列表中不同元素的所有匹配，输出为:

AATATATT in organism: AX225014 Was found in position: 15 and at 15
AATATT in organism: AX225014 Was found in position: 1432 and at 1432
AATATT in organism: AX225016 Was found in position: 1404 and at 1404
AATT in organism: AX225016 Was found in position: 169 and at 2205

是否可以检查所有输出的信息是否所有 Acc 都匹配一个 friend ？

在上面的例子中，想要的输出是:

AATATT was found in all of the Acc.

我的工作代码:

public static ArrayList<String> PB2Scan(ArrayList<String> Pal) throws FileNotFoundException, IOException
{
    ArrayList<String> PalindromesSpotted  = new ArrayList<String>();

    File file = new File("IAV_PB2_32640.txt");
    Scanner sc = new Scanner(file);
    sc.useDelimiter(">");
    //initializes the ArrayList
    ArrayList<String> Gene1 = new ArrayList<String>();
    //initializes the writer
    FileWriter fileWriter = new FileWriter("PB2out");
    PrintWriter printwriter = new PrintWriter(fileWriter);
    //Loads the Array List
    while(sc.hasNext()) Gene1.add(sc.next());
    for(int i = 0; i < Gene1.size(); i++) 
    {
    //Acc breaks down the title so the element:
        //>AX225014 Equine influenza virus H3N8 // 1 (PB2)
        //ATGAAGACAACCATTATTTTGATACTACTGACCCATTGGGTCTACAGTCAAAACCCAACCAGTGGCAACA
        //GGCATGTCCGCAAACGATTTGCAGACCAAGAACTGGGTGATGCCCCATTCCTTGACCGGCTTCGCCGAGA
        //comes out as AX225014
    String Acc = Accession(Gene1.get(i));
    //seq takes the same element as above and returns only
    //ATGAAGACAACCATTATTTTGATACTACTGACCCATTGGGTCTACAGTCAAAACCCAACCAGTGGCAACA
    //GGCATGTCCGCAAACGATTTGCAGACCAAGAACTGGGTGATGCCCCATTCCTTGACCGGCTTCGCCGAGA
    String seq = trimHeader(Gene1.get(i));
        for(int x = 0; x<Pal.size(); x++) 
        {
        if(seq.contains(Pal.get(x))){
        String match = (Pal.get(x) + " in organism: " + Acc + " Was found in position: "+ seq.indexOf(Pal.get(x)) + " and at " +seq.lastIndexOf(Pal.get(x)));
        printwriter.println(match);
        PalindromesSpotted.add(match);
        }
        }
    }
    Collections.sort(PalindromesSpotted);
return PalindromesSpotted;
}

Answer 1

首先，您的代码不会写入任何文件来记录结果，因为您没有关闭您的编写器或至少刷新 PrintWriter。事实上，您也不会关闭您的阅读器。您确实应该关闭您的读者和作者以获取免费资源。值得深思。

您可以使您的 PB2Scan() 方法返回一个简单的结果列表，就像现在一样，或者返回一个包含相同 Pal(s) 的 acc 的结果列表，或者可能两者都包含简单的结果列表被记录，并且在该列表的末尾有一个 acc 列表，其中包含也将被记录的相同 Pal(s)。

PB2Scan() 方法的一些附加代码和附加整数参数可以实现此目的。对于附加参数，您可能需要添加如下内容:

public static ArrayList<String> PB2Scan(ArrayList<String> Pal, int resultType) 
                                throws FileNotFoundException, IOException
{ .... }

其中整数 resultType 参数将采用 0 到 2 之间的三个整数值之一:

• 0 - 简单的结果列表，就像代码当前所做的那样；
• 1 - Acc 与 Pal 的相匹配；
• 2 - 简单的结果列表以及与结果列表末尾的 Pal 匹配的 Acc。

您还应该真正将文件读取为 PB2Scan() 方法的参数，因为该文件在下次运行时很容易成为不同的名称。这使得该方法更加通用，而不是硬编码文件名。

public static ArrayList<String> PB2Scan(String filePath, ArrayList<String> Pal, int resultType) 
                                throws FileNotFoundException, IOException { .... }

该方法始终可以写入相同的输出文件，因为它最适合它来自的方法。

使用上述概念而不是写入输出文件(PB2Out.txt)，因为 PalindromesSpotted ArrayList 正在创建，我认为最好在您的ArrayList 或 ArrayList 已完成。为此，另一种方法(writeListToFile())最适合执行该任务。要查明是否有任何相同的 Pal 与其他 Acc 相匹配，最好使用另一个方法 (getPalMatches()) 来完成该任务。

由于任何给定Seq中多个给定Pal的索引位置未正确报告，所以我提供了另一种方法(findSubstringIndexes()) 快速完成该任务。

需要注意的是，下面的代码假设从 trimHeader() 方法获取的 Seq 都是一个字符串，其中没有换行符。

修改后的PB2Scan()方法和上述其他方法如下:

The PB2Scan() Method:

public static ArrayList<String> PB2Scan(String filePath, ArrayList<String> Pal, int resultType) 
                                throws FileNotFoundException, IOException {
    // Make sure the supplied result type is either 
    // 0, 1, or 2. If not then default to 0.
    if (resultType < 0 || resultType > 2) {
        resultType = 0;
    }
    ArrayList<String> PalindromesSpotted = new ArrayList<>();

    File file = new File(filePath);
    Scanner sc = new Scanner(file);
    sc.useDelimiter(">");
    //initializes the ArrayList
    ArrayList<String> Gene1 = new ArrayList<>();
    //Loads the Array List
    while (sc.hasNext()) {
        Gene1.add(sc.next());
    }
    sc.close(); // Close the read in text file.

    for (int i = 0; i < Gene1.size(); i++) {
        //Acc breaks down the title so the element:
        //>AX225014 Equine influenza virus H3N8 // 1 (PB2)
        //ATGAAGACAACCATTATTTTGATACTACTGACCCATTGGGTCTACAGTCAAAACCCAACCAGTGGCAACA
        //GGCATGTCCGCAAACGATTTGCAGACCAAGAACTGGGTGATGCCCCATTCCTTGACCGGCTTCGCCGAGA
        //comes out as AX225014
        String Acc = Accession(Gene1.get(i));

        //seq takes the same element as above and returns only
        //ATGAAGACAACCATTATTTTGATACTACTGACCCATTGGGTCTACAGTCAAAACCCAACCAGTGGCAACA
        //GGCATGTCCGCAAACGATTTGCAGACCAAGAACTGGGTGATGCCCCATTCCTTGACCGGCTTCGCCGAGA
        String seq = trimHeader(Gene1.get(i));
        for (int x = 0; x < Pal.size(); x++) {
            if (seq.contains(Pal.get(x))) {
                String match = Pal.get(x) + " in organism: " + Acc + 
                                " Was found in position(s): " + 
                                findSubstringIndexes(seq, Pal.get(x));
                PalindromesSpotted.add(match);
            }
        }
    }

    // If there is nothing to work with get outta here.
    if (PalindromesSpotted.isEmpty()) {
        return PalindromesSpotted;
    }

    // Sort the ArrayList
    Collections.sort(PalindromesSpotted);
    // Another ArrayList for matching Pal's to Acc's
    ArrayList<String> accMatchingPal = new ArrayList<>();
    switch (resultType) {
        case 0: // if resultType is 0 is supplied
            writeListToFile("PB2Out.txt", PalindromesSpotted);
            return PalindromesSpotted;

        case 1: // if resultType is 1 is supplied
            accMatchingPal = getPalMatches(PalindromesSpotted);
            writeListToFile("PB2Out.txt", accMatchingPal);
            return accMatchingPal;

        default: // if resultType is 2 is supplied
            accMatchingPal = getPalMatches(PalindromesSpotted);
            ArrayList<String> fullList = new ArrayList<>();
            fullList.addAll(PalindromesSpotted);
            // Create a Underline made of = signs in the list.
            fullList.add(String.join("", Collections.nCopies(70, "=")));
            fullList.addAll(accMatchingPal);
            writeListToFile("PB2Out.txt", fullList);
            return fullList;
    }
}   

The findSubstringIndexes() Method:

private static String findSubstringIndexes(String inputString, String stringToFind){
    String indexes = "";
    int index = inputString.indexOf(stringToFind);
    while (index >= 0){
        indexes+= (indexes.equals("")) ? String.valueOf(index) : ", " + String.valueOf(index);
        index = inputString.indexOf(stringToFind, index + stringToFind.length())   ;
    }
    return indexes;
}

The getPalMatches() Method:

private static ArrayList<String> getPalMatches(ArrayList<String> Palindromes) {
    ArrayList<String> accMatching = new ArrayList<>();
    for (int i = 0; i < Palindromes.size(); i++) {
        String matches = "";
        String[] split1 = Palindromes.get(i).split("\\s+");
        String pal1 = split1[0];
        // Make sure the current Pal hasn't already been listed.
        boolean alreadyListed = false;
        for (int there = 0; there < accMatching.size(); there++) {
            String[] th = accMatching.get(there).split("\\s+");
            if (th[0].equals(pal1)) {
                alreadyListed = true;
                break;
            }
        }
        if (alreadyListed) { continue; }
        for (int j = 0; j < Palindromes.size(); j++) {
            String[] split2 = Palindromes.get(j).split("\\s+");
            String pal2 = split2[0];
            if (pal1.equals(pal2)) {
                // Using Ternary Operator to build the matches string
                matches+= (matches.equals("")) ? pal1 + " was found in the following Accessions: "
                        + split2[3] : ", " + split2[3];
            }
        }
        if (!matches.equals("")) {
            accMatching.add(matches);
        }
    }
    return accMatching;
}

The writeListToFile() Method:

private static void writeListToFile(String filePath, ArrayList<String> list, boolean... appendToFile) {
    boolean appendFile = false;
    if (appendToFile.length > 0) { appendFile = appendToFile[0]; }

    try {
        try (BufferedWriter bw = new BufferedWriter(new FileWriter(filePath, appendFile))) {
            for (int i = 0; i < list.size(); i++) {
                bw.append(list.get(i) + System.lineSeparator());
            }
        }
    } catch (IOException ex) {
        ex.printStackTrace();
    }
}