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. 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. A example will show this. This can create unstable outputs unless you include the original list indices for the lexicographic ordering to keep duplicates in their original order. For more information on how to set\use the key parameter as well as the sorted function in general, take a look at this. Assume that the dictionary and the words only contain lowercase alphabets. Given parallel lists, how can I sort one while permuting (rearranging) the other in the same way? By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. 1. Here is Whatangs answer if you want to get both sorted lists (python3). Sorting Strings is a tiny bit different, since it's a bit less intuitive on how to compare them. Ultimately, you can also just use the comparing() method, which accepts a sorting key function, just like the other ones. MathJax reference. Disconnect between goals and daily tasksIs it me, or the industry? The most obvious solution to me is to use the key keyword arg. That way, I can sort any list in the same order as the source list. The solution below is simple and does not require any imports. My use case is this: user has a list of items initially (listA). Theoretically Correct vs Practical Notation. If so, how close was it? As you can see from the output, the linked list elements are sorted in ascending order by the sort method. Finally, we've used a custom Comparator and defined custom sorting logic. All of the values at the end of the list will be in their order dictated by the list2. Another alternative, combining several of the answers. 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, The most efficient way to merge two lists in Java, Java merge sort implementation efficiency. We can sort a list in natural ordering where the list elements must implement Comparable interface. By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. They're functional in nature, and it's worth noting that operations on a stream produce a result, but do not modify its source. Linear regulator thermal information missing in datasheet. Another alternative, combining several of the answers. All rights reserved. How do you ensure that a red herring doesn't violate Chekhov's gun? May be just the indexes of the items that the user changed. Each factory has an item of its own and a list of other items from competitors. If the list is less than 3 do nothing. 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. I can resort to the use of for constructs but I am curious if there is a shorter way. This is quite inefficient, though, and you should probably create a Map from listA to lookup the positions of the items faster. Did you try it with the sample lists. Styling contours by colour and by line thickness in QGIS. If they are already numpy arrays, then it's simply. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. 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. 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. Specifically, we're using the comparingInt() method, and supplying the user's age, via the User::getAge method reference. What do you mean when you say that you're unable to persist the order "on the backend"? This is an old question but some of the answers I see posted don't actually work because zip is not scriptable. 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. 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.) Why did Ukraine abstain from the UNHRC vote on China? Sort Elements of a Linked List. Since Comparator is a functional interface, we can use lambda expressions to write its implementation in a single line. Application of Binary Tree. 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. Check out our offerings for compute, storage, networking, and managed databases. The method returns a comparator that compares Comparable objects in the natural order. A:[c,b,a] What is the purpose of this D-shaped ring at the base of the tongue on my hiking boots? It is from Java 8. Do roots of these polynomials approach the negative of the Euler-Mascheroni constant? Can I tell police to wait and call a lawyer when served with a search warrant? No new elements. :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 . 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: Did any DOS compatibility layers exist for any UNIX-like systems before DOS started to become outmoded? We can also pass a Comparator implementation to define the sorting rules. The solution below is the most efficient in this case: Site design / logo 2023 Stack Exchange Inc; user contributions licensed under CC BY-SA. 2023 ITCodar.com. Sorting HashMap by Value Simple Example. 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. Follow Up: struct sockaddr storage initialization by network format-string. Once, we have sorted the list, we build the HashMap based on this sorted list. Something like this? We've used the respective comparison approaches for the names and ages - comparing names lexicographically using compareTo(), if the age values are the same, and comparing ages regularly via the > operator. Copyright 2011-2021 www.javatpoint.com. Sort a List of Integers 5 1 List<Integer> numbers = Arrays.asList(6, 2, 1, 4, 9); 2 System.out.println(numbers); 3 4 numbers.sort(Comparator.naturalOrder()); 5 System.out.println(numbers);. B:[2,1,0], And you want to load them both and then produce: My lists are long enough to make the solutions with time complexity of N^2 unusable. An in-place sort is preferred whenever possible. Sorting values of a dictionary based on a list. Why is this sentence from The Great Gatsby grammatical? We are sorting the names according to firstName, we can also use lastName to sort. Given an array of strings words [] and the sequential order of alphabets, our task is to sort the array according to the order given. 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. The basic strategy is to get the values from the HashMap in a list and sort the list. The String class implements Comparable interface. This method returns a lexicographic-order comparator with another comparator. This gives you more direct control over how to sort the input, so you can get sorting stability by simply stating the specific key to sort by. Can I tell police to wait and call a lawyer when served with a search warrant? The size of both list must be same to use this trick. We first get the String values in a list. Are there tables of wastage rates for different fruit and veg? All rights reserved. Connect and share knowledge within a single location that is structured and easy to search. This could be done by wrapping listA inside a custom sorted list like so: Then you can use this custom list as follows: Of course, this custom list will only be valid as long as the elements in the original list do not change. 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. 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. 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. Here is my complete code to achieve this result: But, is there another way to do it? Connect and share knowledge within a single location that is structured and easy to search. We've sorted Comparable integers and Strings, in ascending and descending order, as well as used a built-in Comparator for custom objects. 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(). @Hatefiend interesting, could you point to a reference on how to achieve that? This method will also work when both lists are not identical: /** * Sorts list objectsToOrder based on the order of orderedObjects. Why is this sentence from The Great Gatsby grammatical? Option 3: List interface sort () [Java 8] Java 8 introduced a sort method in the List interface which can use a comparator. Else, run a loop till the last node (i.e. Just encountered the same problem. Premium CPU-Optimized Droplets are now available. Learn more about Stack Overflow the company, and our products. All the elements in the list must implement Comparable interface, otherwise IllegalArgumentException is thrown. I like this because I can do multiple lists with one index. vegan) just to try it, does this inconvenience the caterers and staff? This is an old question but some of the answers I see posted don't actually work because zip is not scriptable. Site design / logo 2023 Stack Exchange Inc; user contributions licensed under CC BY-SA. Other answers didn't bother to import operator and provide more info about this module and its benefits here. You weren't kidding. This trick will never fails and ensures the mapping between the items in list. Browse other questions tagged, Where developers & technologists share private knowledge with coworkers, Reach developers & technologists worldwide. QED. 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? However, if we're working with some custom objects, which might not be Comparable by design, and would still like to sort them using this method - we'll need to supply a Comparator to the sorted() call. Here if the data type of Value is String, then we sort the list using a comparator. Can you write oxidation states with negative Roman numerals? Let's save this result into a sortedList: Here we see that the original list stayed unmodified, but we did save the results of the sorting in a new list, allowing us to use both if we need so later on. Once you have that, define your own comparison function which compares values based on the indexes of list.
