我已将两个文本文件存储到两个单独的数组中。现在，我正在尝试比较两个数组以查找重复值。我的逻辑有问题，无法打印出重复值出现的次数。
文件 1 包含:
1913 2016 1 1913 186
2016 1711 32843 2016 518
3 1913 32843 32001 4
250 5 3500 6 7
8 27 73 9 10
1711 73 11 2 1.4
1.4 12 33.75278 84.38611 1913
19 1930 20 21 1947
22 1955 23 1961 23
1969 27 1995 26 27
1962 28 29 30 1970
31 31
文件 2 包含:
1913 2016 32843 31 27 1.4 4 7 2 23
我试图在 file2 中查找在 file1 中重复的值，以及重复的次数。
我有以下代码:
public static void findDuplicates() {
// array for first file
for (int n = 0; n < nums.size(); n++) {
// matches are false by default
boolean match = false;
int count = 0;
String v = nums.get(n);
// array for second file
for (int k = 0; k < nums1.size(); k++) {
String p = nums1.get(k);
// second file contains values from first file
if (p.contains(v)) {
// there is a match
match = true;
// when there is a match print out matched values and the number of times they appear in second file
if (match) {
count++;
System.out.println( p + " " + "is duped" + " " + count + " " + "times");
}
}
}
}
}
当我编译并运行这段代码时，这是输出:
31 is duped 1 times
有人可以让我知道我在这里做错了什么吗？
编辑
这是我的其余代码:
public static ArrayList<String> nums;
public static ArrayList<String> nums1;
//Create a main method to start the program.
//Add FileNot FoundException in case the file can't be found by computer.
public static void main(String[] args) throws FileNotFoundException{
//The while will help us read the content into our computer piece by piece. It will not stop until the end of assignment.csv.
while(FILE1.hasNext()){
//Create a String variable - TempString. We use TempString to store each piece temporarily.
String TempString = FILE1.next();
String temp1 = TempString.replaceAll("[\\,]", "");
String pattern1 = "[0-9]+\\.{1}[0-9]+";
//Compile the Regular Expression into Pattern and store it in r1 so that the computer can understand the Regular Expression.
Pattern r1 = Pattern.compile(pattern1);
Matcher m1 = r1.matcher(temp1);
String pattern2 = "[0-9]+";
//Compile the Regular Expression into Pattern and store it in r2 so that the computer can understand the Regular Expression.
Pattern r2 = Pattern.compile(pattern2);
Matcher m2 = r2.matcher(temp1);
nums = new ArrayList<String>();
//Recollect, m1 is used to match decimal numbers.
if(!(m1.find())){//if a decimal number CAN'T be found
//We use while statement instead of if statement here.
//If there is only one piece per line, we can use either while statement or if statement.
//However, we have to use while statement if there is more than one piece per line.
while(m2.find()) {//if an integer number CAN be found
//If an Integer is found, we add 1 to Variable count.
count++;
//Even though the number (i.e., m2.group(0)) is an Integer, its data type is String. So we store it to a String variable - number.
String number = m2.group(0);
nums.add(number);
//If the remainder of count by 5 is zero, we display the number and advance to a new line.
if (count % 5 == 0){
System.out.println(number);
}
//Otherwise, we just display the number on the same line and divide numbers by a space.
else
System.out.print(number + " ");
}
}
//If we find a decimal number
else{
//We add 1 to Variable count.
count++;
//Even though the number (i.e., m1.group(0)) is a decimal number, its data type is String. So we store it to a String variable - number.
String number = m1.group(0);
nums.add(number);
//If the remainder of count by 5 is zero, we display the number and advance to a new line.
if (count % 5 == 0) {
System.out.println(number);
}
//Otherwise, we just display the number on the same line and divide numbers by a space.
else
System.out.print(number + " ");
}
}
FILE1.close();//Once we finish the task, we close the file.
while(FILE2.hasNext()){
//Create a String variable - TempString. We use TempString to store each piece temporarily.
String TempString = FILE2.next();
//So I use replaceAll function to eliminate comma (,) and store the new string in temp1.
String temp1 = TempString.replaceAll("[\\,]", "");
String pattern1 = "[0-9]+\\.{1}[0-9]+";
//Compile the Regular Expression into Pattern and store it in r1 so that the computer can understand the Regular Expression.
Pattern r1 = Pattern.compile(pattern1);
//Match the Regular Expression with the piece (temp1) we read from assignment.csv.
Matcher m1 = r1.matcher(temp1);
String pattern2 = "[0-9]+";
//Compile the Regular Expression into Pattern and store it in r2 so that the computer can understand the Regular Expression.
Pattern r2 = Pattern.compile(pattern2);
//Match the Regular Expression with the piece (temp1) we read from assignment.csv.
Matcher m2 = r2.matcher(temp1);
nums1 = new ArrayList<String>();
//We have two types of numbers - Integer and Decimal
//Let's start us Integer.
//Recollect, m1 is used to match decimal numbers.
if(!(m1.find())){//if a decimal number CAN'T be found
//We use while statement instead of if statement here.
//If there is only one piece per line, we can use either while statement or if statement.
//However, we have to use while statement if there is more than one piece per line.
while(m2.find()) {//if an integer number CAN be found
//If an Integer is found, we add 1 to Variable count.
count++;
//Even though the number (i.e., m2.group(0)) is an Integer, its data type is String. So we store it to a String variable - number.
String number = m2.group(0);
nums1.add(number);
//If the remainder of count by 5 is zero, we display the number and advance to a new line.
if (count % 5 == 0){
//System.out.println(number);
}
//Otherwise, we just display the number on the same line and divide numbers by a space.
else
System.out.println(/*number + " "*/);
}
}
//If we find a decimal number
else{
//We add 1 to Variable count.
count++;
//Even though the number (i.e., m1.group(0)) is a decimal number, its data type is String. So we store it to a String variable - number.
String number = m1.group(0);
nums1.add(number);
//If the remainder of count by 5 is zero, we display the number and advance to a new line.
if (count % 5 == 0){
//System.out.println(number);
}
//Otherwise, we just display the number on the same line and divide numbers by a space.
else
System.out.println(/*number + " "*/);
}
findDuplicates();
}
FILE2.close();//Once we finish the task, we close the file.
}
我尽量删除了不必要的代码。
编辑
预期输出应该是:
1913 is duplicated 3 times.
2016 is duplicated 2 times.
32843 is duplicated 1 times.
31 is duplicated 2 times.....
编辑
所以我相信我已经找到了问题所在。出于某种原因，
String p = nums.get(k)
在我的 findDuplicates() 方法中只返回值 31，而不返回其他值。我正在努力解决问题，并会在解决问题时发布答案。
最佳答案
我认为最大的问题是打印行在第二个 for 循环内。
此外，我会删除 boolean 值并只比较 2 个字符串 (p==v)。
所以代码看起来更像这样:
public static void main(String[] args) {
// array for second file
for (int n = 0; n < nums1.size(); n++) {
// matches are false by default
int count = 0;
String v = nums1.get(n);
// array for first file
for (int k = 0; k < nums.size(); k++) {
String p = nums.get(k);
// second file contains values from first file
if (p==v) {
count++;
}
}
System.out.println( v + " " + "is duped" + " " + count + " " + "times");
}
}
}
通过更改，我使代码按预期运行。
您可以查看现场演示 here .
输出:
1913 is duped 4 times
2016 is duped 3 times
32843 is duped 2 times
31 is duped 2 times
27 is duped 3 times
1.4 is duped 2 times
4 is duped 1 times
7 is duped 1 times
2 is duped 1 times
23 is duped 2 times
关于java - 在两个数组中查找重复值，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/39807709/