[英]How can I sort a list of clothing sizes (e.g. 4XL, S, 2XL)?
我需要您的幫助來跟蹤查詢。 鑒於以下列表:
["2XL", "3XL", "4XL", "5XL", "6XL", "L", "M", "S", "XL"]
我如何對其進行排序以使其按此順序排列?
[“S”、“M”、“L”、“XL”、“2XL”、“3XL”、“4XL”、“5XL”、“6XL”]
請注意,並非所有尺寸都存在。
構建一個比較器,它會根據您想要的順序進行查找:
Comparator<String> sizeOrder = Comparator.comparingInt(desiredOrder::indexOf);
哪里
desiredOrder = Arrays.asList("S", "M", "L", "XL", "2XL", "3XL", "4XL", "5XL", "6XL");
然后:
yourList.sort(sizeOrder);
如果需要,您可以為查找構建一個Map<String, Integer>
:
Map<String, Integer> lookup =
IntStream.range(0, desiredOrder.length())
.boxed()
.collect(Collectors.toMap(desiredOrder::get, i -> i));
然后做:
Comparator<String> sizeOrder = Comparator.comparing(lookup::get);
我不相信這會比使用List.indexOf
更List.indexOf
,因為desiredOrder
列表太小了。
與所有與性能相關的內容一樣:使用您認為最易讀的; profile 如果您認為這是一個性能瓶頸,那么只有嘗試替代方案。
一般方法將關注大小字符串背后的模式,而不僅僅是容納樣本輸入。 您有一個由S
、 M
或L
表示的基本方向和可選的修飾符(除非M
)改變幅度。
static Pattern SIZE_PATTERN=Pattern.compile("((\\d+)?X)?[LS]|M", Pattern.CASE_INSENSITIVE);
static int numerical(String size) {
Matcher m = SIZE_PATTERN.matcher(size);
if(!m.matches()) throw new IllegalArgumentException(size);
char c = size.charAt(m.end()-1);
int n = c == 'S'? -1: c == 'L'? 1: 0;
if(m.start(1)>=0) n *= 2;
if(m.start(2)>=0) n *= Integer.parseInt(m.group(2));
return n;
}
然后,您可以對尺寸列表進行排序,例如
List<String> sizes = Arrays.asList("2XL", "5XL", "M", "S", "6XL", "XS", "3XS", "L", "XL");
sizes.sort(Comparator.comparing(Q48298432::numerical));
System.out.print(sizes.toString());
其中Q48298432
應替換為包含numerical
方法的類的名稱。
使用可能更有效且當然更清晰的enum
路線的替代方法。
// Sizes in sort order.
enum Size {
SMALL("S"),
MEDIUM("M"),
LARGE("L"),
EXTRA_LARGE("XL"),
EXTRA2_LARGE("2XL"),
EXTRA3_LARGE("3XL"),
EXTRA4_LARGE("4XL"),
EXTRA5_LARGE("5XL"),
EXTRA6_LARGE("6XL");
private final String indicator;
Size(String indicator) {
this.indicator = indicator;
}
static final Map<String,Size> lookup = Arrays.asList(values()).stream()
.collect(Collectors.toMap(
// Key is the indicator.
s -> s.indicator,
// Value is the size.
s-> s));
public static Size lookup(String s) {
return lookup.get(s.toUpperCase());
}
// Could be improved to handle failed lookups.
public static final Comparator<String> sizeOrder = (o1, o2) -> lookup(o1).ordinal() - lookup(o2).ordinal();
}
public void test(String[] args) {
List<String> test = Arrays.asList("S","6XL", "L");
Collections.sort(test, Size.sizeOrder);
System.out.println(test);
}
另一種方式可能是這樣的:
List<String> desiredOrder = Arrays.asList("S", "M", "L", "XL", "2XL", "3XL", "4XL", "5XL", "6XL");
String [] array = new String[desiredOrder.size()];
list.forEach(s->{
if (desiredOrder.indexOf(s) != -1)
array[desiredOrder.indexOf(s)] = s;
});
List<String> set= Stream.of(array).filter(Objects::nonNull).collect(Collectors.toList());
或者
Set<Integer> integers = new HashSet<>();
list.forEach(s -> {
if (desiredOrder.indexOf(s) != -1) {
integers.add(desiredOrder.indexOf(s));
}
});
List<String> sortedItems = integers.stream()
.sorted()
.map(i->desiredOrder.get(i))
.collect(Collectors.toList());
這是完成大小排序的基本算法。 該算法的基本思想是將大小的每個部分( number
、 modifier
、 unit
)分解為相應的區域( [0, 1000)
、 [1000, 1000_000)
、 [1000_000, reset]
),然后使用dir
來決定modifier
方向。 例如:
class Size {
public static int sizeToInt(String size) {
int n = size.length(), unit = size.charAt(n - 1), dir = unit == 'S' ? 1 : -1;
//use a separate method to parse the number & extra modifier if needed
int i = 0;
int number = n > 1&&Character.isDigit(size.charAt(i)) ? size.charAt(i++) : '1';
int modifier = i + 1 < n ? size.charAt(i) : 0;
return -1 * (unit * 1000_000 + dir * (modifier * 1000 + number));
}
}
上面的解決方案僅適用於您的簡單情況,對於復雜情況,您需要解析number
& modifier
並決定如何對region
進行分區。 然后您可以按以下代碼按Comparator
對大小進行排序:
Comparator<String> sizeComparator = Comparator.comparingInt(Size::sizeToInt);
List<String> sizes = Arrays.asList(
"3XL", "2XL", "XL",
"5L", "2L", "L",
"5M", "2M", "M",
"S", "2S", "3S",
"XS", "2XS", "3XS"
);
sizes.sort(sizeComparator);
System.out.println(sizes);
// ^--- [
// "3XS", "2XS", "XS",
// "3S", "2S", "S",
// "M", "2M", "5M",
// "L", "2L", "5L",
// "XL", "2XL", "3XL"
// ]
創建一個包含所有可能大小的 Enum 類,並向其中添加getEnum
方法以支持以數字開頭的大小:
public enum SizeEnum
{
S("S"), M("M"), L("L"), XL("XL"), TWO_XL("2XL"), THREE_XL("3XL"), FOUR_XL("4XL"), FIVE_XL("5XL"), SIX_XL("6XL");
private String value;
SizeEnum(String value){
this.value = value;
}
public String getValue() {
return value;
}
public static SizeEnum getEnum(String value) {
return Arrays.stream(values())
.filter(v -> v.getValue().equalsIgnoreCase(value))
.findAny()
.orElseThrow(() -> new IllegalArgumentException());
}
}
在比較器中使用SizeEnum
類:
List<String> sizes = Arrays.asList("2XL", "3XL", "4XL", "5XL", "6XL", "L", "M", "S", "XL");
sizes.sort(Comparator.comparing(v -> SizeEnum.getEnum(v.toUpperCase())));
sizes.stream().forEach(l -> System.out.println(l));
public static void SortSizes(String [] sizes, String [] sortedRadixArr) throws Exception{
//creating an int array with same length as sortedRadixArr
//Here each index of intRadix array is mapped to the index of sortedRadixArr
//Eg: intRadix = [0,0,0,0,0] ----> ["S","M","L","XL","2XL"]
int [] intRadix = new int[sortedRadixArr.length];
for(int i = 0; i < sizes.length; i++){
boolean matchFlag = false;
for(int j=0; j< sortedRadixArr.length; j++){
if(sizes[i] == sortedRadixArr[j]){
//Whenever the String is matched with the String of sortedRadixArr array,
//increment the value of intRadix array at the equivalent index of sortedRadixArr.
//This helps us to count number same strings(sizes). like two or more "S".
//We can't do that with sortedRadixArr as it is a String array.
intRadix[j]++;
matchFlag = true;
break;
}
}
if(!matchFlag){
throw new Exception("Invalid size found!");
}
}
int count = 0;
for(int k = 0; k < intRadix.length; k++){
//Loop thru intRadix array. It contains count of respective sizes(Strings) at each index.
//We will re-write sizes array in sorted order with help of intRadix as below.
while(intRadix[k] > 0){
sizes[count++] = sortedRadixArr[k];
intRadix[k]--;
}
}
}
}
這里, sortedRadixArr
是基數數組,即具有所有可能值的值:
["S", "M", "L", "XL", "2XL", "3XL", "4XL", "5XL", "6XL"]
sizes
是您要排序的數組:
["2XL", "3XL", "4XL", "5XL", "6XL", "L", "M", "S", "XL"]
它接受重復值。
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