sort list based on another list java sort list based on another list java

This solution is poor when it comes to storage. Here, the sorted() method also follows the natural order, as imposed by the JVM. The signature of the method is: It also returns a stream sorted according to the provided comparator. Most of the solutions above are complicated and I think they will not work if the lists are of different lengths or do not contain the exact same items. We can use the following methods to sort the list: Java Stream interface provides two methods for sorting the list: Stream interface provides a sorted() method to sort a list. Whereas, Integer values are directly sorted using Collection.sort(). The naive implementation that brute force searches listB would not be the best performance-wise, but would be functionally sufficient. There are plenty of ways to achieve this. Guide to Java 8 Collectors: groupingByConcurrent(), Java 8 - Difference Between map() and flatMap(), Java: Finding Duplicate Elements in a Stream, Java - Filter a Stream with Lambda Expressions, Guide to Java 8 Collectors: averagingDouble(), averagingLong() and averagingInt(), Make Clarity from Data - Quickly Learn Data Visualization with Python, // Constructor, getters, setters and toString(), Sorting a List of Integers with Stream.sorted(), Sorting a List of Integers in Descending Order with Stream.sorted(), Sorting a List of Strings with Stream.sorted(), Sorting Custom Objects with Stream.sorted(Comparator operator. There are at least two good idioms for this problem. There is a major issue with this answer: You are inserting a reference to the object originally in listB into listA, which is incorrect behavior if the two objects are equals() but do not refer to the same object - the original object in listA is lost and some references in listA are replaced with references in listB, rather than listA being simply reordered. Other answers didn't bother to import operator and provide more info about this module and its benefits here. 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. @Richard: the keys are computed once before sorting; so the complexity is actually O(N^2). Theoretically Correct vs Practical Notation, Bulk update symbol size units from mm to map units in rule-based symbology. originalList always contains all element from orderedList, but not vice versa. Making statements based on opinion; back them up with references or personal experience. Find the max recommended item from second sublist (3 to end of list) and add it to the newly created list and . Browse other questions tagged, Where developers & technologists share private knowledge with coworkers, Reach developers & technologists worldwide. Now it actually works. Created a default comparator on bookings to sort the list. 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. Excuse any terrible practices I used while writing this code, though. Any suggestions? Sign up for Infrastructure as a Newsletter. Note: Any item not in list1 will be ignored since the algorithm will not know what's the sort order to use. This can create unstable outputs unless you include the original list indices for the lexicographic ordering to keep duplicates in their original order. Just remember Zx and Zy are tuples. The second issue is that if listA and listB do contain references to the same objects (which makes the first issue moot, of course), and they contain the same objects (as the OP implied when he said "reordered"), then this whole thing is the same as, And a third major issue is that by the end of this function you're left with some pretty weird side effects. If so, how close was it? i.e., it defines how two items in the list should be compared. Although I am not entirely sure exactly what the OP is asking for, I couldn't help but come to this conclusion as well. If you preorder a special airline meal (e.g. How do I read / convert an InputStream into a String in Java? To learn more, see our tips on writing great answers. You are using Python 3. Now it produces an iterable object. Why are physically impossible and logically impossible concepts considered separate in terms of probability? then the question should be 'How to sort a dictionary? Linear regulator thermal information missing in datasheet, Short story taking place on a toroidal planet or moon involving flying, Identify those arcade games from a 1983 Brazilian music video, It is also probably wrong to have your class implements. C:[a,b,c]. They reorder the items and want to persist that order (listB), however, due to restrictions I'm unable persist the order on the backend so I have to sort listA after I retrieve it. not if you call the sort after merging the list as suggested here. Sorting values of a dictionary based on a list. more_itertools has a tool for sorting iterables in parallel: I actually came here looking to sort a list by a list where the values matched. @RichieV I recommend using Quicksort or an in-place merge sort implementation. The Comparator.comparing () method accepts a method reference which serves as the basis of the comparison. rev2023.3.3.43278. Sorting in Natural Order and Reverse Order HashMap in java provides quick lookups. It seems what you want would be to use Comparable instead, but even this isn't a good idea in this case. So you could simply have: What I am doing require to sort collection of factories and loop through all factories and sort collection of their competitors. B:[2,1,0], And you want to load them both and then produce: If the elements of the stream are not Comparable, a java.lang.ClassCastException may be thrown upon execution. If you have any suggestions for improvements, please let us know by clicking the report an issue button at the bottom of the tutorial. This is useful when your value is a custom object. I can resort to the use of for constructs but I am curious if there is a shorter way. This will sort all factories according to their price. In our case, we're using the getAge() method as the sorting key. The method signature is: Comparable is also an interface belong to a java.lang package. We're streaming that list, and using the sorted() method with a Comparator. Each factory has an item of its own and a list of other items from competitors. In this tutorial we will sort the HashMap according to value. This work is licensed under a Creative Commons Attribution-NonCommercial- ShareAlike 4.0 International License. Basically, this answer is nonsense. That's O(n^2 logn)! One way of doing this is looping through listB and adding the items to a temporary list if listA contains them: Not completely clear what you want, but if this is the situation: ', not 'How to sorting list based on values from another list?'. Application of Binary Tree. How Intuit democratizes AI development across teams through reusability. As each pair of strings are passed in for comparison, convert them into ints using originalList.indexOf, except that if the index is -1, change the index to originalList.size () Compare the two ints. The returned comparable is serializable. Why is "1000000000000000 in range(1000000000000001)" so fast in Python 3? Site design / logo 2023 Stack Exchange Inc; user contributions licensed under CC BY-SA. How do I sort a list of dictionaries by a value of the dictionary? The source of these elements is usually a Collection or an Array, from which data is provided to the stream. Once you have that, define your own comparison function which compares values based on the indexes of list. 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. Else, run a loop till the last node (i.e. 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. How do you get out of a corner when plotting yourself into a corner, Trying to understand how to get this basic Fourier Series. Oh, ignore, I can do sorted(zip(Index,X,Y,Z)) too. 2. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. If not then just replace SortedMap indexToObj by SortedMap> indexToObjList. Examples: Input: words = {"hello", "geeksforgeeks"}, order = "hlabcdefgijkmnopqrstuvwxyz" Output: "hello", "geeksforgeeks" Explanation: