sort list based on another list java

We can use Collections.reverseOrder () method, which returns a Comparator, for reverse sorting. The signature of the method is: It also returns a stream sorted according to the provided comparator. Here is an example of how to sort a list and then make the changes in another list according to the changes exactly made to first array list. Sometimes, you might want to switch this up and sort in descending order. If the elements are not comparable, it throws java.lang.ClassCastException. your map should be collected to a LinkedHashMap in order to preserve the order of listB. The naive implementation that brute force searches listB would not be the best performance-wise, but would be functionally sufficient. Follow Up: struct sockaddr storage initialization by network format-string. How do you get out of a corner when plotting yourself into a corner, Trying to understand how to get this basic Fourier Series. We can sort the entries in a HashMap according to keys as well as values. It's a List, and Item has a public String getWeekday() method. http://scienceoss.com/sort-one-list-by-another-list/. That's right but the solutions use completely different methods which could be used for different applications. Do I need a thermal expansion tank if I already have a pressure tank? 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. It would be helpful if you would provide an example of your expected input and output. 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 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. Is it suspicious or odd to stand by the gate of a GA airport watching the planes? It is defined in Stream interface which is present in java.util package. Specifically, we're using the comparingInt() method, and supplying the user's age, via the User::getAge method reference. Check out our offerings for compute, storage, networking, and managed databases. If you notice the above examples, the Value objects implement the Comparator interface. Here is my complete code to achieve this result: But, is there another way to do it? @RichieV I recommend using Quicksort or an in-place merge sort implementation. Sort Elements of a Linked List. For example if. 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. Option 3: List interface sort () [Java 8] Java 8 introduced a sort method in the List interface which can use a comparator. Making statements based on opinion; back them up with references or personal experience. In our case, we're using the getAge() method as the sorting key. Disconnect between goals and daily tasksIs it me, or the industry? Why do many companies reject expired SSL certificates as bugs in bug bounties? 2. It returns a stream sorted according to the natural order. How do you ensure that a red herring doesn't violate Chekhov's gun? You weren't kidding. On the Data tab of the Ribbon, in the Sort & Filter group, click Advanced. Premium CPU-Optimized Droplets are now available. How do I align things in the following tabular environment? Can I tell police to wait and call a lawyer when served with a search warrant? We first get the String values in a list. All rights reserved. Using Kolmogorov complexity to measure difficulty of problems? I can resort to the use of for constructs but I am curious if there is a shorter way. You are using Python 3. you can leverage that solution directly in your existing df. There are a few of these built-in comparators that work with numbers (int, double, and long) - comparingInt(), comparingDouble(), and comparingLong(). Then you can create your custom Comparator that uses the Map to create an order: Then you can sort listA using your custom Comparator. Take a look at this solution, may be this is what you are trying to achieve: O U T P U T All rights reserved. The solution assumes that all the objects in the list to sort have distinct keys. The nature of simulating nature: A Q&A with IBM Quantum researcher Dr. Jamie We've added a "Necessary cookies only" option to the cookie consent popup. Something like this? Connect and share knowledge within a single location that is structured and easy to search. 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. When we try to use sort over a zip object. Are there tables of wastage rates for different fruit and veg? 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. Is there a solution to add special characters from software and how to do it. Can you write oxidation states with negative Roman numerals? I fail to see where the problem is. We can sort a list in natural ordering where the list elements must implement Comparable interface. 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. If changes are possible, you would need to somehow listen for changes to the original list and update the indices inside the custom list. Getting key with maximum value in dictionary? Sorting HashMap by Value Simple Example. Here if the data type of Value is String, then we sort the list using a comparator. MathJax reference. rev2023.3.3.43278. Do roots of these polynomials approach the negative of the Euler-Mascheroni constant? How do I generate random integers within a specific range in Java? It also doesn't care if the List R you want to sort contains Comparable elements so long as the other List L you use to sort them by is uniformly Comparable. You return. If they are already numpy arrays, then it's simply. No spam ever. 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 comparator), Defining a Custom Comparator with Stream.sorted(). HashMap entries are sorted according to String value. Maybe you can delete one of them. DigitalOcean makes it simple to launch in the cloud and scale up as you grow whether youre running one virtual machine or ten thousand. 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. We first get the String values in a list. How do I read / convert an InputStream into a String in Java? Another alternative, combining several of the answers. Working on improving health and education, reducing inequality, and spurring economic growth? This trick will never fails and ensures the mapping between the items in list. I have created a more general function, that sorts more than two lists based on another one, inspired by @Whatang's answer. Oh, ignore, I can do sorted(zip(Index,X,Y,Z)) too. You can use this generic comparator to sort list based on the the other list. The method signature is: Comparable is also an interface belong to a java.lang package. 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. Mail us on [emailprotected], to get more information about given services. zip, sort by the second column, return the first column. The signature of the method is: In the following example, we have used the following methods: The reverseOrder() is a method of Comparator interface which is defined in java.util package. The most obvious solution to me is to use the key keyword arg. With this method: Sorting a 1000 items list 100 times improves speed 10 times on my The second one is easier and faster if you're not using Pandas in your program. What sort of strategies would a medieval military use against a fantasy giant? Sorting values of a dictionary based on a list. good solution! Did you try it with the sample lists. I am also wandering if there is a better way to do that. Use MathJax to format equations. Use MathJax to format equations. Finally, we've used a custom Comparator and defined custom sorting logic. To learn more, see our tips on writing great answers. I think that the title of the original question is not accurate. Create a Map that maps the values of everything in listB to something that can be sorted easily, such as the index, i.e. Stop Googling Git commands and actually learn it! As for won't work..that's right because he posted the wrong question in the title when he talked about lists. How to match a specific column position till the end of line? Acidity of alcohols and basicity of amines. If the list is greater than or equal to 3 split list in two 0 to 2 and 3 to end of 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. I don't know if it is only me, but doing : Please add some more context to your post. 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. An in-place sort is preferred whenever possible. How can I randomly select an item from a list? 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: Connect and share knowledge within a single location that is structured and easy to search. For Action, select Filter the list, in-place. 2) Does listA and listB contain references to the same objects, or just objects that are equivalent with equals()? 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 Is there a solution to add special characters from software and how to do it, Minimising the environmental effects of my dyson brain, The difference between the phonemes /p/ and /b/ in Japanese. 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. There are plenty of ways to achieve this. Working on improving health and education, reducing inequality, and spurring economic growth? @RichieV I recommend using Quicksort or an in-place merge sort implementation. 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. 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? The method returns a comparator that compares Comparable objects in the natural order. You can setup history as a HashMap or separate class to make this easier. Here is Whatangs answer if you want to get both sorted lists (python3). If you're not used to Lambda expressions, you can create a Comparator beforehand, though, for the sake of code readability, it's advised to shorten it to a Lambda: You can also technically make an anonymous instantiation of the comparator in the sorted() call: And this anonymous call is exactly what gets shortened to the Lambda expression from the first approach. Using this method is fairly simple, so let's take a look at a couple of examples: Here, we make a List instance through the asList() method, providing a few integers and stream() them. How can this new ban on drag possibly be considered constitutional? Oh, ignore, I can do sorted(zip(Index,X,Y,Z)) too. It is from Java 8. Unsubscribe at any time. How do you ensure that a red herring doesn't violate Chekhov's gun? Connect and share knowledge within a single location that is structured and easy to search. :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 . That's O(n^2 logn)! Using Java 8 Streams Let's start with two entity classes - Employee and Department: The . Minimising the environmental effects of my dyson brain. By clicking Post Your Answer, you agree to our terms of service, privacy policy and cookie policy. My solution: The time complexity is O(N * Log(N)). Browse other questions tagged, Where developers & technologists share private knowledge with coworkers, Reach developers & technologists worldwide. For bigger arrays / vectors, this solution with numpy is beneficial! Once you have that, define your own comparison function which compares values based on the indexes of list. Why is "1000000000000000 in range(1000000000000001)" so fast in Python 3? My use case is this: user has a list of items initially (listA). Sorting list based on another list's order. 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. If the age of the users is the same, the first one that was added to the list will be the first in the sorted order. If they are already numpy arrays, then it's simply. We can easily reverse this order as well, simply by chaining the reversed() method after the comparingInt() call: While Comparators produced by methods such as comparing() and comparingInt(), are super-simple to work with and only require a sorting key - sometimes, the automated behavior is not what we're looking for. The order of the elements having the same "key" does not matter. The java.Collections.sort () method is also used to sort the linked list, array, queue, and other data structures. Let's start with two entity classes - Employee and Department: class Employee { Integer employeeId; String employeeName; // getters and setters } class Department { Integer . How is an ETF fee calculated in a trade that ends in less than a year? Is there a solution to add special characters from software and how to do it. To avoid having a very inefficient look up, you should index the items in listB and then sort listA based on it. i.e., it defines how two items in the list should be compared. Using a For-Each Loop Note also, that the SortedDependingList does currently not allow to add an element from listA a second time - in this respect it actually works like a set of elements from listA because this is usually what you want in such a setting. For example, when appendFirst is false below will be the output. Learn more about Stack Overflow the company, and our products. 2. @Debacle What operations are allowed on the backend over listA? 2. 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. 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? The nature of simulating nature: A Q&A with IBM Quantum researcher Dr. Jamie We've added a "Necessary cookies only" option to the cookie consent popup. Why are Suriname, Belize, and Guinea-Bissau classified as "Small Island Developing States"? Code Review Stack Exchange is a question and answer site for peer programmer code reviews. "After the incident", I started to be more careful not to trip over things. We're streaming that list, and using the sorted() method with a Comparator. - Hatefiend The basic strategy is to get the values from the HashMap in a list and sort the list. Wed like to help. If a law is new but its interpretation is vague, can the courts directly ask the drafters the intent and official interpretation of their law? Stream.sorted() by default sorts in natural order. An efficient solution is to first create the mapping from the ID in the ids (your desired IDs order) to the index in that list: val orderById = ids.withIndex ().associate { it.value to it.index } And then sort your list of people by the order of their id in this mapping: val sortedPeople = people . 12 is less than 21 and no one from L2 is in between. We can use this by creating a list of Integers and sort these using the Collections.sort(). 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. The signature of the method is: Let's see another example of Collections.sorts() method. We can now eliminate the anonymous inner class and achieve the same result with simple, functional semantics using lambdas: (Employee e1, Employee e2) -> e1.getName ().compareTo (e2.getName ()); We can test it as below: Once we have the list of values in a sorted manner, we build the HashMap again based on this new list. How can we prove that the supernatural or paranormal doesn't exist? JavaTpoint offers too many high quality services. Has 90% of ice around Antarctica disappeared in less than a decade? To subscribe to this RSS feed, copy and paste this URL into your RSS reader. IMO, you need to persist something else. 1. How to sort one list and re-sort another list keeping same relation python? If you want to do it manually. Can I tell police to wait and call a lawyer when served with a search warrant? 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.

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