2023 DigitalOcean, LLC. Guava has a ready-to-use comparator for doing that: Ordering.explicit(). The solution below is the most efficient in this case: Site design / logo 2023 Stack Exchange Inc; user contributions licensed under CC BY-SA. The best answers are voted up and rise to the top, Not the answer you're looking for? For example, the following code creates a list of Student and in-place . Using Comparator. Oh, ignore, I can do sorted(zip(Index,X,Y,Z)) too. your map should be collected to a LinkedHashMap in order to preserve the order of listB. While we believe that this content benefits our community, we have not yet thoroughly reviewed it. Here is my complete code to achieve this result: But, is there another way to do it? It would be preferable instead to have a method sortCompetitors(), that would sort the list, without leaking it: and remove completely the method getCompetitors(). rev2023.3.3.43278. The basic strategy is to get the values from the HashMap in a list and sort the list. @Debacle What operations are allowed on the backend over listA? No new elements. Thanks for contributing an answer to Code Review Stack Exchange! Returning a negative number indicates that an element is lesser than another. 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(). For bigger arrays / vectors, this solution with numpy is beneficial! Originally posted by David O'Meara: Then when you initialise your Comparator, pass in the list used for ordering. 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. 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. If they are already numpy arrays, then it's simply. Is there a solution to add special characters from software and how to do it. They store items in key, value pairs. Although I am not entirely sure exactly what the OP is asking for, I couldn't help but come to this conclusion as well. Browse other questions tagged, Where developers & technologists share private knowledge with coworkers, Reach developers & technologists worldwide. Found within the Stream interface, the sorted() method has two overloaded variations that we'll be looking into. Browse other questions tagged, Where developers & technologists share private knowledge with coworkers, Reach developers & technologists worldwide. O(n) look up happening roughly O(nlogn) times? Asking for help, clarification, or responding to other answers. 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). All times above are in ranch (not your local) time. Do I need a thermal expansion tank if I already have a pressure tank? We first get the String values in a list. How to Sort a HashMap by Value in Java? | DigitalOcean Premium CPU-Optimized Droplets are now available. Note: the key=operator.itemgetter(1) solves the duplicate issue, zip is not subscriptable you must actually use, If there is more than one matching it gets the first, This does not solve the OPs question. My lists are long enough to make the solutions with time complexity of N^2 unusable. Did you try it with the sample lists. 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. How to match a specific column position till the end of line? The Collections class has two methods for sorting a list: The sort() method sorts the list in ascending order, according to the natural ordering of its elements. How can I randomly select an item from a list? not if you call the sort after merging the list as suggested here. Sign up for Infrastructure as a Newsletter. Why do small African island nations perform better than African continental nations, considering democracy and human development? 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. So for me the requirement was to sort originalList with orderedList. In this tutorial, we'll compare some filtering implementations and discuss their advantages and drawbacks. Sort Elements of a Linked List. @RichieV I recommend using Quicksort or an in-place merge sort implementation. Make the head as the current node and create another node index for later use. Learn more about Stack Overflow the company, and our products. "After the incident", I started to be more careful not to trip over things. Sorting list according to corresponding values from a parallel list [duplicate]. For more information on how to set\use the key parameter as well as the sorted function in general, take a look at this. JavaTpoint offers too many high quality services. 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 don't know if it is only me, but doing : Please add some more context to your post. Does Counterspell prevent from any further spells being cast on a given turn? To avoid having a very inefficient look up, you should index the items in listB and then sort listA based on it. Why are Suriname, Belize, and Guinea-Bissau classified as "Small Island Developing States"? "Sunday" => 0, , "Saturday" => 6. Speed improvement on JB Nizet's answer (from the suggestion he made himself). :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 . Sorting List and Stream on Multiple Fields Java 8 Example Why are physically impossible and logically impossible concepts considered separate in terms of probability? Collections.sort() - Ways to Sort a List in Java - TechVidvan In each iteration, follow the following step . Linked List Operations: Traverse, Insert and Delete What I am doing require to sort collection of factories and loop through all factories and sort collection of their competitors. Surly Straggler vs. other types of steel frames. C:[a,b,c]. You posted your solution two times. Oh, ignore, I can do sorted(zip(Index,X,Y,Z)) too. Site design / logo 2023 Stack Exchange Inc; user contributions licensed under CC BY-SA. In Java how do you sort one list based on another? I like having a list of sorted indices. 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. This method will also work when both lists are not identical: Problem : sorting a list of Pojo on the basis of one of the field's all possible values present in another list. As you can see that we are using Collections.sort() method to sort the list of Strings. You get paid; we donate to tech nonprofits. Using Kolmogorov complexity to measure difficulty of problems? 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. Why do academics stay as adjuncts for years rather than move around? Has 90% of ice around Antarctica disappeared in less than a decade? The naive implementation that brute force searches listB would not be the best performance-wise, but would be functionally sufficient. My question is how to call compare method of factoryPriceComparator to sort factories? In Python 2, zip produced a list. How can I pair socks from a pile efficiently? How do I make a flat list out of a list of lists? Try this. Sort a List of Objects by Field in Java - Hire Amir In this tutorial, we will learn how to sort a list in the natural order. Beware that Integer.compare is only available from java 7. However, some may lead to under-performing solutions if not done properly. It seems what you want would be to use Comparable instead, but even this isn't a good idea in this case. Asking for help, clarification, or responding to other answers. In Java there are set of classes which can be useful to sort lists or arrays. The size of both list must be same to use this trick. I was in a rush. Your problem statement is not very clear. If you have any suggestions for improvements, please let us know by clicking the report an issue button at the bottom of the tutorial. Something like this? Excuse any terrible practices I used while writing this code, though. QED. Staging Ground Beta 1 Recap, and Reviewers needed for Beta 2, Sorting a list in Python using the result from sorting another list, How to rearrange one list based on a second list of indices, How to sort a list according to another list? They're functional in nature, and it's worth noting that operations on a stream produce a result, but do not modify its source. Since Comparator is a functional interface, we can use lambda expressions to write its implementation in a single line. you can leverage that solution directly in your existing df. How do I call one constructor from another in Java? Linear regulator thermal information missing in datasheet. P.S. I am wondering if there is any easier way to do it. Follow Up: struct sockaddr storage initialization by network format-string. unit tests. Surly Straggler vs. other types of steel frames. It is defined in Stream interface which is present in java.util package. In the case of our integers, this means that they're sorted in ascending order. Unsubscribe at any time. It is from Java 8. I think that the title of the original question is not accurate. Acidity of alcohols and basicity of amines. I see where you are going with it, but you need to rethink what you were going for and edit this answer. If values in the HashMap are of type Integer, the code will be as follows : Here HashMap values are sorted according to Integer values. Java 8 - How to sort ArrayList using Stream API - BenchResources.Net The java.Collections.sort () method is also used to sort the linked list, array, queue, and other data structures. 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. MathJax reference. Why do academics stay as adjuncts for years rather than move around? Once you have that, define your own comparison function which compares values based on the indexes of list Y. MathJax reference. 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. Browse other questions tagged, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site. - the incident has nothing to do with me; can I use this this way? super T> comparator), Defining a Custom Comparator with Stream.sorted(). We can sort the entries in a HashMap according to keys as well as values. Note that you can shorten this to a one-liner if you care to: As Wenmin Mu and Jack Peng have pointed out, this assumes that the values in X are all distinct. We can use Collections.sort() method to sort a list in the natural ascending order. All rights reserved. then the question should be 'How to sort a dictionary? 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. This solution is poor when it comes to storage. An in-place sort is preferred whenever possible. I did a static include of. This is a very nice way to sort the list, and to clarify, calling with appendFirst=true will sort the list as [d, c, e, a, b], @boxed__l: It will sort the elements contained in both lists in the same order and add at the end the elements only contained in A. I think most of the solutions above will not work if the 2 lists are of different sizes or contain different items. It is the method of Java Collections class which belong to a java.lang package. 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. In case of Strings, they're sorted lexicographically: If we wanted the newly sorted list saved, the same procedure as with the integers applies here: Check out our hands-on, practical guide to learning Git, with best-practices, industry-accepted standards, and included cheat sheet. Note that the class must implement Comparable interface. In addition, the proposed solution won't work for the initial question as the lists X and Y contain different entries. Is there a solution to add special characters from software and how to do it. Warning: If you run it with empty lists it crashes. Sorting values of a dictionary based on a list. 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. Just encountered the same problem. For example, when appendFirst is false below will be the output. You should instead use [x for (y,x) in sorted(zip(Y,X), key=lambda pair: pair[0])]. Sorting list based on another list's order. People will search this post looking to sort lists not dictionaries. Both of these variations are instance methods, which require an object of its class to be created before it can be used: This methods returns a stream consisting of the elements of the stream, sorted according to natural order - the ordering provided by the JVM. HashMap in java provides quick lookups. 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