英文:
Java List String sort Based on Integer with multiple underscore's
问题
你好，你可以使用Java 8中的lambda表达式和Comparator来实现这个排序逻辑。以下是你的代码片段的排序实现：
keyValues.sort(Comparator.comparing(s -> Arrays.asList(s.split("_"))
.stream()
.map(Integer::parseInt)
.collect(Collectors.toList())));
这将按照你描述的逻辑对keyValues进行排序。
英文:
Hi All I have arrayList of String datatype but needs to sort like integer with multiple underscore's using Java 8 only ,For your reference I have added sample data, the original data came Liferay 7.1 DDL
List<String> keyValues = new ArrayList<String>();
keyValues.add("100_10_20_1");
keyValues.add("100001");
keyValues.add("100002");
keyValues.add("100002_1");
keyValues.add("100003");
keyValues.add("100");
keyValues.add("100_1");
keyValues.add("100_2");
keyValues.add("100_1_1");
keyValues.add("100_10_20");
keyValues.add("10000001");
keyValues.add("100_10_20_2");
i tried to replace the underscore with 0 but what happens sometimes 100_1 replace _ with 0 then it becomes 10001 but what if i already have 10001 value present in list ? even same thing for empty if i replace _ with "" then it becomes 1001 but what if i already have 1001 then again i am facing issue
Exact logic based on underscore was
100
100_1
100_1_2
/after that only/
100_2
100_2_2
/after that/
103
104
105
10000
/similarly the pattern repeats/
10000_1
10000_2
10000_2_1
100002
i need the exact output as
100
100_1
100_1_1
100_2
100_10_20
100_10_20_1
100_10_20_2
100001
100002
100002_1
100003
10000001
i tried with replacing underscore with 0 or empty space but there is possiblity of if it had already that number present in list,if i replace with dot and storing as Double then multiple pointer exception will occur
答案1
得分: 4
我创建了一个比较器，然后根据每个字符串具有的最小标记数量进行分割和处理。
static Comparator<String> stringComparator = new Comparator<String>() {
@Override
public int compare(String o1, String o2) {
String[] s1 = o1.split("_");
String[] s2 = o2.split("_");
int minLen = Math.min(s1.length, s2.length);
for (int i=0; i < minLen; i++){
if (!s1[i].equals(s2[i])){
return Integer.parseInt(s1[i]) - Integer.parseInt(s2[i]);
}
}
return o1.compareTo(o2);
}
};
调用方式如下：
List<String> keyValues = new ArrayList<String>();
keyValues.add("100001");
keyValues.add("100002");
keyValues.add("100002_1");
keyValues.add("100003");
keyValues.add("100");
keyValues.add("100_1");
keyValues.add("100_2");
keyValues.add("100_1_1");
keyValues.add("100_10_20");
keyValues.add("10000001");
keyValues.stream().sorted(stringComparator).forEach(System.out::println);
输出结果：
100
100_1
100_1_1
100_2
100_10_20
100001
100002
100002_1
100003
10000001
英文:
I made a comparator then split the string and process based on the min amount of tokens each string has.
static Comparator<String> stringComparator = new Comparator<String>() {
@Override
public int compare(String o1, String o2) {
String[] s1 = o1.split("_");
String[] s2 = o2.split("_");
int minLen = Math.min(s1.length, s2.length);
for (int i=0; i < minLen; i++){
if (!s1[i].equals(s2[i])){
return Integer.parseInt(s1[i]) - Integer.parseInt(s2[i]);
// return s1[i].compareTo(s2[i]); //not natural order/sort
}
}
return o1.compareTo(o2);
}
};
Calling it like so:
List<String> keyValues = new ArrayList<String>();
keyValues.add("100001");
keyValues.add("100002");
keyValues.add("100002_1");
keyValues.add("100003");
keyValues.add("100");
keyValues.add("100_1");
keyValues.add("100_2");
keyValues.add("100_1_1");
keyValues.add("100_10_20");
keyValues.add("10000001");
keyValues.stream().sorted(stringComparator).forEach(System.out::println);
Gives output:
100
100_1
100_1_1
100_2
100_10_20
100001
100002
100002_1
100003
10000001
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