[英]Remove and mark duplicates in a string array
I'm having this 我有这个
String array[] = {"test","testing again", "test"};
that i want to mark and remove duplicates. 我要标记和删除重复项。 Here is the output that i need: 这是我需要的输出:
2x test
testing again
Can someone help me do this? 有人可以帮我吗? I've tried with Set but it seems it doesnt recogize when a string is already inthere. 我已经尝试过使用Set,但是当字符串已经在其中时,它似乎无法识别。
Here is my code: 这是我的代码:
Set addons = new HashSet<String>();
final String[] arr ={"test","testing again", "test"};
for (int i = 0; i < arr.length; i++) {
Log.d(TAG, "contains adding " + arr[i]);
if (addons.contains(arr[i])) {
//never enters here
Log.d(TAG, "contains " + arr[i]);
addons.remove(arr[i]);
addons.add("2 x " + arr[i]);
} else {
addons.add("1 x " + arr[i]);
}
}
You can do something like this: 您可以执行以下操作:
String[] arr = { "test", "testing again", "test" };
HashMap<String, Integer> counter = new HashMap<>();
for (int i = 0; i < arr.length; i++) {
if (counter.containsKey(arr[i])) {
counter.put(arr[i], counter.get(arr[i]) + 1);
} else {
counter.put(arr[i], 1);
}
}
System.out.println("Occurrences:\n");
for (String key : counter.keySet()) {
System.out.println(key + " x" + counter.get(key));
}
Your example doesn't work because, when you find a new occurrence of a word you remove it and replace it with something like 2x [word]
, when that word comes up again contains(...)
will return false
since it's no longer in the set. 因为,当你发现你删除它,并用类似替换单词的一个新出现的例如不工作2x [word]
,当这个词再次出现contains(...)
将返回false
,因为它不再在集合中。
In java 8: 在Java 8中:
Stream.of("test", "testing again", "test")
.collect(groupingBy(Function.identity(), counting()))
.forEach((str, freq) -> {
System.out.printf("%20s: %d%n", str, freq);
});
try this: 尝试这个:
public static void main(String[] args) {
Set<String> addons = new HashSet<>();
final String[] arr = { "test", "testing again", "test","test","testing again" };
int count = 0;
for (int i = 0; i < arr.length; i++) {
for (int j = 0; j < arr.length; j++) {
if (arr[i].equals(arr[j])) {
count++;
}
}
addons.add(count + " x " + arr[i]);
count = 0;
}
System.out.println(addons);
}
output: 输出:
[2 x testing again, 3 x test]
String[] arr ={"test","testing again", "test"};
Map<String, Integer> results = new HashMap<>();
for (int i = 0; i < arr.length; i++) {
Log.d(TAG, "contains adding " + arr[i]);
if (results.containsKey(arr[i])) {
Log.d(TAG, "contains " + arr[i]);
results.put(arr[i], results.get(arr[i]) + 1);
} else {
results.put(arr[i], 1);
}
}
try following code. 尝试以下代码。
String[] array ={"test","testing again","test"};
Set<String> uniqueWords = new HashSet<String>(Arrays.asList(array));
The problem is that you don't directly add "test" in your Set, but "1 x test" instead. 问题是您没有在集合中直接添加“测试”,而是添加了“ 1 x测试”。
So you better use a Map to save the string, and its number of occurence. 因此,最好使用Map保存字符串及其出现的次数。
String[] array = { "test", "testing again", "test" };
Map<String, Integer> addons = new HashMap<>();
for (String s : array) {
System.out.println("Dealing with [" + s + "]");
if (addons.containsKey(s)) {
System.out.println("\tAlready contains [" + s + "]");
addons.put(s, addons.get(s) + 1); // increment count of s
} else {
System.out.println("\tFirst time seeing [" + s + "]");
addons.put(s, 1); // first time we encounter s
}
}
Use this, where Key to the Map is the String element and Value is the Count of that element. 使用它,其中Map的Key是String元素,Value是该元素的Count。
public static void main(String[] args) {
String array[] = {"test","testing again", "test"};
Map<String, Integer> myMap = new HashMap<>();
for (int i = 0; i < array.length; i++) {
if (myMap.containsKey(array[i])) {
Integer count = myMap.get(array[i]);
myMap.put(array[i], ++count);
} else{
myMap.put(array[i], 1);
}
}
System.out.println(myMap);
}
String array[] = {"test","testing again", "test"};
Map<String, Integer> countMap = new HashMap<>();
for (int i = 0; i<array.length; i++) {
Integer count = countMap.get(array[i]);
if(count == null) {
count = 0;
}
countMap.put(array[i], (count.intValue()+1));
}
System.out.println(countMap.toString());
Output 输出量
{'test'=2, 'testing again'=1}
You can use Multiset from Guava. 您可以从Guava使用Multiset。
String array[] = {"test","testing again", "test"};
Multiset<String> set = HashMultiset.create(Arrays.asList(array));
System.out.println(set);
Output: 输出:
[test x 2, testing again]
Multiset basically counts how many times you tried to add an object. 基本上,Multiset会计算您尝试添加对象的次数。
for (HashMultiset.Entry<String> entry :set.entrySet()) {
System.out.println(entry.getCount() + "x " + entry.getElement());
}
Output: 输出:
2x test
1x testing again
You can use own class that hold duplicates: 您可以使用自己的类来保存重复项:
class SetWithDuplicates extends HashSet<String> {
private final Set<String> duplicates = new HashSet<>();
@Override
public boolean add(String e) {
boolean added = super.add(e);
if(!added) {
duplicates.add(e);
}
return added;
}
public Set<String> duplicates() {
return duplicates;
}
}
And use that similarly like @Ganpat Kaliya: 并像@Ganpat Kaliya一样使用它:
String[] array ={"test","testing again","test"};
SetWithDuplicates <String> uniqueWords = new SetWithDuplicates(Arrays.asList(array));
SetWithDuplicates <String> duplicates = uniqueWords.duplicates();
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