sort list based on another list java

I mean swapItems(), removeItem(), addItem(), setItem() ?? Acidity of alcohols and basicity of amines. How can we prove that the supernatural or paranormal doesn't exist? Then we sort the list. Here is a solution that increases the time complexity by 2n, but accomplishes what you want. Stack Exchange network consists of 181 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. So basically, I have 2 ArrayLists (listA and listB). The answer of riza might be useful when plotting data, since zip(*sorted(zip(X, Y), key=lambda pair: pair[0])) returns both the sorted X and Y sorted with values of X. For example, the following code creates a list of Student and in-place . We can sort the entries in a HashMap according to keys as well as values. You are using Python 3. How To Install Grails on an Ubuntu 12.04 VPS, Simple and reliable cloud website hosting, New! The order of the elements having the same "key" does not matter. In addition, the proposed solution won't work for the initial question as the lists X and Y contain different entries. Connect and share knowledge within a single location that is structured and easy to search. Connect and share knowledge within a single location that is structured and easy to search. This is just an example, but it demonstrates an order that is defined by a list, and not the natural order of the datatype: Now, let's say that listA needs to be sorted according to this ordering. You should instead use [x for (y,x) in sorted(zip(Y,X), key=lambda pair: pair[0])]. People will search this post looking to sort lists not dictionaries. Another solution that may work depending on your setting is not storing instances in listB but instead indices from listA. Sort an array according to the order defined by another array using Sorting and Binary Search: The idea is to sort the A1 [] array and then according to A2 [] store the elements. Thanks for your answer, I learned a lot. T: comparable type of element to be compared. The preferred way to add something to SortedDependingList is by already knowing the index of an element and adding it by calling sortedList.addByIndex(index); If the two lists are guaranteed to contain the same elements, just in a different order, you can use List listA = new ArrayList<>(listB) and this will be O(n) time complexity. rev2023.3.3.43278. If you're using Java 8, you can even get rid of the above FactoryPriceComparator and use the built-in Comparator.comparingDouble(keyExtractor), which creates a comparator comparing the double values returned by the key extractor. unit tests. I can resort to the use of for constructs but I am curious if there is a shorter way. If we sort the Users, and two of them have the same age, they're now sorted by the order of insertion, not their natural order, based on their names. In this quick tutorial, we'll learn how to find items from one list based on values from another list using Java 8 Streams. But it should be: The list is ordered regarding the first element of the pairs, and the comprehension extracts the 'second' element of the pairs. The answer of riza might be useful when plotting data, since zip(*sorted(zip(X, Y), key=lambda pair: pair[0])) returns both the sorted X and Y sorted with values of X. - the incident has nothing to do with me; can I use this this way? Here if the data type of Value is String, then we sort the list using a comparator. Overview. There is a difference between the two: a class is Comparable when it can compare itself to another class of the same type, which is what you are doing here: one Factory is comparing itself to another object. Did you try it with the sample lists. Another alternative, combining several of the answers. :param lists: lists to be sorted :return: a tuple containing the sorted lists """ # Create the initially empty lists to later store the sorted items sorted_lists = tuple([] for _ in range(len(lists))) # Unpack the lists, sort them, zip them and iterate over them for t in sorted(zip(*lists)): # list items are now sorted based on the first list . Why are Suriname, Belize, and Guinea-Bissau classified as "Small Island Developing States"? A tree illustrates a hierarchical structure in contrast to other data structures such an array, stack, queue, and linked list, which are linear in nature. P.S. It puts the capital letter elements first in natural order after that small letters in the natural order, if the list has both small and capital letters. . good solution! How can I check before my flight that the cloud separation requirements in VFR flight rules are met? Why do academics stay as adjuncts for years rather than move around? You posted your solution two times. Does a summoned creature play immediately after being summoned by a ready action? For cases like these, we'll want to write a custom Comparator: And now, when we execute this code, we've got the natural order of names, as well as ages, sorted: Here, we've used a Lambda expression to create a new Comparator implicitly and defined the logic for sorting/comparison. Not the answer you're looking for? Styling contours by colour and by line thickness in QGIS. What is the purpose of this D-shaped ring at the base of the tongue on my hiking boots? For example, when appendFirst is false below will be the output. You can create a pandas Series, using the primary list as data and the other list as index, and then just sort by the index: This is helpful when needing to order a smaller list to values in larger. All rights reserved. Although I am not entirely sure exactly what the OP is asking for, I couldn't help but come to this conclusion as well. Designed by Colorlib. Sorry, that was my typo. Find centralized, trusted content and collaborate around the technologies you use most. 1. This tutorial covered sorting of HashMap according to Value. Once you have a list of sorted indices, a simple list comprehension will do the trick: Note that the sorted index list can also be gotten using numpy.argsort(). Once sorted, we've just printed them out, each in a line: If we wanted save the results of sorting after the program was executed, we would have to collect() the data back in a Collection (a List in this example), since sorted() doesn't modify the source. Surly Straggler vs. other types of steel frames. Sorting values of a dictionary based on a list. The collect() method is used to receive elements from a stream and stored them in a collection. Mark should be before Robert, in a list sorted by name, but in the list we've sorted previously, it's the other way around. That's O(n^2 logn)! Wed like to help. ', not 'How to sorting list based on values from another list?'. Given an array of strings words [] and the sequential order of alphabets, our task is to sort the array according to the order given. That's easily managed with an index list: Since the decorate-sort-undecorate approach described by Whatang is a little simpler and works in all cases, it's probably better most of the time. Sorting for String values differs from Integer values. It is stable for an ordered stream. By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Surly Straggler vs. other types of steel frames. Browse other questions tagged, Where developers & technologists share private knowledge with coworkers, Reach developers & technologists worldwide. Why did Ukraine abstain from the UNHRC vote on China? Browse other questions tagged, Where developers & technologists share private knowledge with coworkers, Reach developers & technologists worldwide. I have two lists List list1 = new ArrayList(), list2 = new ArrayList(); (Not the same size), of the class Person: I want to create a new list using list1 and list2 sorted by age (descending), but I also another condition that is better explained with an example: He should, because his age is equal to Menard, Alec is from L1 and two Person from L1 can't be one after another is this kind of situation happens. If the data is related then the data should be stored together in a simple class. Assuming that the larger list contains all values in the smaller list, it can be done. It only takes a minute to sign up. Java Sort List Objects - Comparator Summary Collections class sort () method is used to sort a list in Java. That is, the first items (from Y) are compared; and if they are the same then the second items (from X) are compared, and so on. The String class implements Comparable interface. How to sort one list and re-sort another list keeping same relation python? Note: Any item not in list1 will be ignored since the algorithm will not know what's the sort order to use. Asking for help, clarification, or responding to other answers. #kkjavatutorials #JavaAbout this Video:Hello Friends,In this video,we will talk and learn about How to Write a Java program for Sort Map based on Values (Cus. Beware that Integer.compare is only available from java 7. All times above are in ranch (not your local) time. Starting with the example input you provided: This is also known as the Schwartzian_transform after R. Schwartz who popularized this pattern in Perl in the 90s: Note that in this case Y and X are sorted and compared lexicographically. As I understand it, you want to have a combined sorted list but interleave elements from list1 and list2 whenever the age is the same. They store items in key, value pairs. Like Tim Herold wrote, if the object references should be the same, you can just copy listB to listA, either: Or this if you don't want to change the List that listA refers to: If the references are not the same but there is some equivalence relationship between objects in listA and listB, you could sort listA using a custom Comparator that finds the object in listB and uses its index in listB as the sort key. What is the shortest way of sorting X using values from Y to get the following output? The basic strategy is to get the values from the HashMap in a list and sort the list. (This is a very old answer!). Now it produces an iterable object. The method returns a comparator that imposes the reverse of the natural ordering. Java List is similar to arrays except that the length of the list is dynamic and it comes in Java Collection framework. To place them last, you can use a nullsLast comparator: I would just use a map with indexes of each name, to simplify the lookup: Then implement a Comparator that sorts by looking up names in indexOfMap: Note that the order of the first elements in the resulting list is not deterministic (because it's just all elements not present in list2, with no further ordering). Get tutorials, guides, and dev jobs in your inbox. No new elements. Why do small African island nations perform better than African continental nations, considering democracy and human development? In this tutorial we will sort the HashMap according to value. The solution assumes that all the objects in the list to sort have distinct keys. In the case of our integers, this means that they're sorted in ascending order. More general case (sort list Y by any key instead of the default order), http://scienceoss.com/sort-one-list-by-another-list/, How Intuit democratizes AI development across teams through reusability. Sometimes, you might want to switch this up and sort in descending order. It's a List, and Item has a public String getWeekday() method. How can this new ban on drag possibly be considered constitutional? So for me the requirement was to sort originalList with orderedList. The most obvious solution to me is to use the key keyword arg. Why is this sentence from The Great Gatsby grammatical? More general case (sort list Y by any key instead of the default order), http://scienceoss.com/sort-one-list-by-another-list/, How Intuit democratizes AI development across teams through reusability. DigitalOcean makes it simple to launch in the cloud and scale up as you grow whether youre running one virtual machine or ten thousand. All of them simply return a comparator, with the passed function as the sorting key. It only takes a minute to sign up. The solution below is simple and should fix those issues: Location of index in list2 is tracked using cur_loclist. Once you have a list of sorted indices, a simple list comprehension will do the trick: Note that the sorted index list can also be gotten using numpy.argsort(). Zip the two lists together, sort it, then take the parts you want: Also, if you don't mind using numpy arrays (or in fact already are dealing with numpy arrays), here is another nice solution: I found it here: I suspect the easiest way to do this will be by writing a custom implementation of java.util.Comparator which can be used in a call to Collections.sort(). - the incident has nothing to do with me; can I use this this way? We can use Collections.sort() method to sort a list in the natural ascending order. There are a few of these built-in comparators that work with numbers (int, double, and long) - comparingInt(), comparingDouble(), and comparingLong(). Sorting list according to corresponding values from a parallel list [duplicate]. Assume that the dictionary and the words only contain lowercase alphabets. All rights reserved. If you have 2 lists of identical number of items and where every item in list 1 is related to list 2 in the same order (e.g a = 0 , b = 1, etc.) When we try to use sort over a zip object. How to handle a hobby that makes income in US. This is an old question but some of the answers I see posted don't actually work because zip is not scriptable. More elegant code or using some built in Java class? Short story taking place on a toroidal planet or moon involving flying. Note: Any item not in list1 will be ignored since the algorithm will not know what's the sort order to use. Whereas, Integer values are directly sorted using Collection.sort(). To get a value from the HashMap, we use the key corresponding to that entry. Learn more about Stack Overflow the company, and our products. How do I call one constructor from another in Java? Try this. Does this require that the values in X are unqiue? Let's start with two entity classes - Employee and Department: class Employee { Integer employeeId; String employeeName; // getters and setters } class Department { Integer . This method will also work when both lists are not identical: /** * Sorts list objectsToOrder based on the order of orderedObjects. Speed improvement on JB Nizet's answer (from the suggestion he made himself). If you have any suggestions for improvements, please let us know by clicking the report an issue button at the bottom of the tutorial. Basically, this answer is nonsense. The solution below is simple and does not require any imports. But because you also like to be able to sort history based on frequency, I would recommend a History class: Then create a HashMap to quickly fill history, and convert it into a TreeSet to sort: Java List.Add() Unsupportedoperationexception, Keyword for the Outer Class from an Anonymous Inner Class, Org.Hibernate.Hibernateexception: Access to Dialectresolutioninfo Cannot Be Null When 'Hibernate.Dialect' Not Set, Convert Timestamp in Milliseconds to String Formatted Time in Java, How to Query Xml Using Namespaces in Java with Xpath, Convenient Way to Parse Incoming Multipart/Form-Data Parameters in a Servlet, How to Convert the Date from One Format to Another Date Object in Another Format Without Using Any Deprecated Classes, Eclipse 2021-09 Code Completion Not Showing All Methods and Classes, Rotating Coordinate Plane for Data and Text in Java, Java Socket Why Server Can Not Reply Client, How to Fix the "Java.Security.Cert.Certificateexception: No Subject Alternative Names Present" Error, Remove All Occurrences of Char from String, How to Use 3Des Encryption/Decryption in Java, Creating Multiple Log Files of Different Content with Log4J, Very Confused by Java 8 Comparator Type Inference, Copy a Stream to Avoid "Stream Has Already Been Operated Upon or Closed", Overload with Different Return Type in Java, Eclipse: How to Build an Executable Jar with External Jar, Stale Element Reference: Element Is Not Attached to the Page Document, Method for Evaluating Math Expressions in Java, How to Use a Tablename Variable for a Java Prepared Statement Insert, Why am I Getting Java.Lang.Illegalstateexception "Not on Fx Application Thread" on Javafx, What Is a Question Mark "" and Colon ":" Operator Used For, How to Validate Two or More Fields in Combination, About Us | Contact Us | Privacy Policy | Free Tutorials. Premium CPU-Optimized Droplets are now available. @Hatefiend interesting, could you point to a reference on how to achieve that? Sort Elements of a Linked List. We can use the following methods to sort the list: Using stream.sorted () method Using Comparator.reverseOrder () method Using Comparator.naturalOrder () method Using Collections.reverseOrder () method Using Collections.sort () method Java Stream interface Java Stream interface provides two methods for sorting the list: sorted () method "After the incident", I started to be more careful not to trip over things. To avoid having a very inefficient look up, you should index the items in listB and then sort listA based on it. test bed for array based list implementation, Reading rows based on column value in POI. Once we have the list of values in a sorted manner, we build the HashMap again based on this new list. I think most of the solutions above will not work if the 2 lists are of different sizes or contain different items. How do I read / convert an InputStream into a String in Java? If we talk about the working of this method, then the method works on ASCII values. Is it suspicious or odd to stand by the gate of a GA airport watching the planes? All Rights Reserved. I have created a more general function, that sorts more than two lists based on another one, inspired by @Whatang's answer. We can sort a list in natural ordering where the list elements must implement Comparable interface. Did this satellite streak past the Hubble Space Telescope so close that it was out of focus? Is it possible to create a concave light? Working on improving health and education, reducing inequality, and spurring economic growth? Best answer! Any suggestions? This class has two parameters, firstName and lastName. I like this because I can do multiple lists with one index. This is an old question but some of the answers I see posted don't actually work because zip is not scriptable. Once, we have sorted the list, we build the HashMap based on this sorted list. This is useful when your value is a custom object. See more examples here. All of the values at the end of the list will be in their order dictated by the list2. My code is GPL licensed, can I issue a license to have my code be distributed in a specific MIT licensed project? Has 90% of ice around Antarctica disappeared in less than a decade? Do roots of these polynomials approach the negative of the Euler-Mascheroni constant? Not the answer you're looking for? HashMap entries are sorted according to String value. How can I pair socks from a pile efficiently? Otherwise, I see a lot of answers here using Collections.sort(), however there is an alternative method which is guaranteed O(2n) runtime, which should theoretically be faster than sort's worst time complexity of O(nlog(n)), at the cost of 2n storage.