Question

要通过随机散布两个有序数组的项来创建一个数组，而在新数组中，A仍在B之前，p在q之前，等等。

该怎么做？我可以使用underscorejs或jquery。

[A,B,C,D]
[p,q,r,s]
==>
[p,q,A,r,B,C,D,s], or [A,B,p,C,q,r,s,D] or [A,p,q,B,r,C,s,D] or...

The following is not OK.
[B,p,A,...] or [p,A,r,...]

Answer 1

您的问题包含答案。

采用随机数组。
将第一个元素从其移至结果数组。
重复直到所有元素都移动。

可能的解决方案之一：

function crossJoin(tabl1, tabl2, header) {
 var output = [];
  var days = tabl1;
  var customer = tabl2;
  if(header) {output.push([header[0][1], header[0][0]])};
  for(var i = 1, iLen = customer.length; i < iLen; i++) {
//     output.push(days[i]);
    for(var j = 1, jLen = days.length; j < jLen; j++) { 
      output.push(days[j],customer[i]);

    }
  }
  return output;  
}

Answer 2

最后是香草js。

   public class Food {
     public String id, name, price;
   }


  private void parseXML(String urlResponse) {
        XmlPullParserFactory parserFactory;
        try {
            parserFactory = XmlPullParserFactory.newInstance();
            XmlPullParser parser = parserFactory.newPullParser();
            InputStream is = new ByteArrayInputStream(urlResponse.getBytes());
            parser.setFeature(XmlPullParser.FEATURE_PROCESS_NAMESPACES, false);
            parser.setInput(is, null);

            processParsing(parser);

        } catch (XmlPullParserException e) {

        } catch (IOException e) {
        }
    }


    private void processParsing(XmlPullParser parser) throws IOException, XmlPullParserException{
        ArrayList<Food> foods = new ArrayList<>();
        int eventType = parser.getEventType();
        Food currentFood = null;

        while (eventType != XmlPullParser.END_DOCUMENT) {
            String eltName = null;

            switch (eventType) {
                case XmlPullParser.START_TAG:
                    eltName = parser.getName();

                    if ("food".equals(eltName)) {
                        currentFood = new Food();
                        foods.add(currentFood);
                    } else if (currentFood != null) {
                        if ("name".equals(eltName)) {
                            currentFood.name = parser.nextText();
                        } else if ("id".equals(eltName)) {
                            currentFood.id = parser.nextText();
                        } else if ("price".equals(eltName)) {
                            currentFood.price = parser.nextText();
                        }
                    }
                    break;
            }

            eventType = parser.next();
        }

        printFoods(foods);
    }

Answer 3

使用下划线，您可以在一般情况下（即任意数量的数组）使用sample：

function mix(...arrays) {
  var res = [];
  while (_.some(arrays, a => a.length)) {
    res.push(_.sample(_.filter(arrays, a => a.length)).shift());
  }
  return res;
}

console.log(mix(['A', 'B', 'C', 'D'], ['p', 'q', 'r', 's'], [1, 2, 3, 4, 5]));

<script src="https://cdnjs.cloudflare.com/ajax/libs/underscore.js/1.9.1/underscore-min.js"></script>

连接两个有序数组，随机混合项目，但保持原始顺序

3 个答案: