米脂
我有一个带有p值列的dataframe，我想对这些p值进行选择。> pvalues_anova[1] 9.693919e-01 9.781728e-01 9.918415e-01 9.716883e-01 1.667183e-02[6] 9.952762e-02 5.386854e-01 9.997699e-01 8.714044e-01 7.211856e-01[11] 9.536330e-01 9.239667e-01 9.645590e-01 9.478572e-01 6.243775e-01[16] 5.608563e-01 1.371190e-04 9.601970e-01 9.988648e-01 9.698365e-01[21] 2.795891e-06 1.290176e-01 7.125751e-01 5.193604e-01 4.835312e-04选择方式：anovatest<- results[ - which(results$pvalues_anova < 0.8) ,]如果我在R中使用它，这个函数可以很好地工作，但是如果我在另一个应用程序(银河)中运行它，没有e-01G.4.835312e-04不会被扔出去。还有其他方法来标注p值吗，比如0.0004835312而不是4.835312e-04?你可以：使用以列表作为值的映射。Map<KeyType, List<ValueType>>.创建一个新的包装类，并将此包装器的实例放在映射中。Map<KeyType, WrapperType>.使用元组类(节省创建大量包装器)。Map<KeyType, Tuple<Value1Type, Value2Type>>.并排使用多重地图。实例1.以列表为值的映射// create our mapMap<String, List<Person>> peopleByForename = new HashMap<>(); // populate itList<Person> people = new ArrayList<>();people.add(new Person("Bob Smith"));people.add(new Person("Bob Jones"));peopleByForename.put("Bob", people);// read from itList<Person> bobs = peopleByForename["Bob"];Person bob1 = bobs[0];Person bob2 = bobs[1];这种方法的缺点是列表没有绑定到两个值。2.使用包装类// define our wrapperclass Wrapper { public Wrapper(Person person1, Person person2) { this.person1 = person1; this.person2 = person2; } public Person getPerson1 { return this.person1; } public Person getPerson2 { return this.person2; } private Person person1; private Person person2;}// create our mapMap<String, Wrapper> peopleByForename = new HashMap<>();// populate itWrapper people = new Wrapper();peopleByForename.put("Bob", new Wrapper(new Person("Bob Smith"), new Person("Bob Jones"));// read from itWrapper bobs = peopleByForename.get("Bob");Person bob1 = bobs.getPerson1;Person bob2 = bobs.getPerson2;这种方法的缺点是，您必须为所有这些非常简单的容器类编写大量的锅炉板代码。3.使用元组// you'll have to write or download a Tuple class in Java, (.NET ships with one)// create our mapMap<String, Tuple2<Person, Person> peopleByForename = new HashMap<>();// populate itpeopleByForename.put("Bob", new Tuple2(new Person("Bob Smith", new Person("Bob Jones"));// read from itTuple<Person, Person> bobs = peopleByForename["Bob"];Person bob1 = bobs.Item1;Person bob2 = bobs.Item2;在我看来，这是最好的解决办法。4.多重地图// create our mapsMap<String, Person> firstPersonByForename = new HashMap<>();Map<String, Person> secondPersonByForename = new HashMap<>();// populate themfirstPersonByForename.put("Bob", new Person("Bob Smith"));secondPersonByForename.put("Bob", new Person("Bob Jones"));// read from themPerson bob1 = firstPersonByForename["Bob"];Person bob2 = secondPersonByForename["Bob"];此解决方案的缺点是，这两个映射之间的关系并不明显，编程错误可能导致两个映射不同步。