Find centralized, trusted content and collaborate around the technologies you use most. 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. I think that the title of the original question is not accurate. I have created a more general function, that sorts more than two lists based on another one, inspired by @Whatang's answer. 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The second one is easier and faster if you're not using Pandas in your program. Designed by Colorlib. When we try to use sort over a zip object. The basic strategy is to get the values from the HashMap in a list and sort the list. For bigger arrays / vectors, this solution with numpy is beneficial! In Python 2, zip produced a list. Returning a positive number indicates that an element is greater than another. A tree's ordering information is irrelevant. 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. Check out our offerings for compute, storage, networking, and managed databases. This work is licensed under a Creative Commons Attribution-NonCommercial- ShareAlike 4.0 International License. String values require a comparator for sorting. Sometimes, you might want to switch this up and sort in descending order. 12 is less than 21 and no one from L2 is in between. I used java 8 streams to sort lists and put them in ArrayDeques. 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: You get paid; we donate to tech nonprofits. Why is this sentence from The Great Gatsby grammatical? Linear Algebra - Linear transformation question, Acidity of alcohols and basicity of amines, Is there a solution to add special characters from software and how to do it. Surly Straggler vs. other types of steel frames. Else, run a loop till the last node (i.e. This comparator sorts the list of values alphabetically. i.e., it defines how two items in the list should be compared. 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. How can I pair socks from a pile efficiently? MathJax reference. Not the answer you're looking for? Maybe you can delete one of them. How do I align things in the following tabular environment? Did you try it with the sample lists. unit tests. Did any DOS compatibility layers exist for any UNIX-like systems before DOS started to become outmoded? 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. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Is there a solution to add special characters from software and how to do it. All rights reserved. JavaTpoint offers too many high quality services. DigitalOcean makes it simple to launch in the cloud and scale up as you grow whether youre running one virtual machine or ten thousand. - the incident has nothing to do with me; can I use this this way? Did this satellite streak past the Hubble Space Telescope so close that it was out of focus? Sorting in Natural Order and Reverse Order 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. Also easy extendable for similar problems! Why is "1000000000000000 in range(1000000000000001)" so fast in Python 3? T: comparable type of element to be compared. 2. Once you have that, define your own comparison function which compares values based on the indexes of list. For more information on how to set\use the key parameter as well as the sorted function in general, take a look at this. How can this new ban on drag possibly be considered constitutional? 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. 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 . Browse other questions tagged, Where developers & technologists share private knowledge with coworkers, Reach developers & technologists worldwide. 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. Once you have that, define your own comparison function which compares values based on the indexes of list. Though it might not be obvious, this is exactly equivalent to, This is correct, but I'll add the note that if you're trying to sort multiple arrays by the same array, this won't neccessarily work as expected, since the key that is being used to sort is (y,x), not just y. 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. In the case of our integers, this means that they're sorted in ascending order. (This is a very old answer!). zip, sort by the second column, return the first column. I don't know if it is only me, but doing : Please add some more context to your post. Now it produces an iterable object. Another alternative, combining several of the answers. We're streaming that list, and using the sorted() method with a Comparator. In this tutorial, we will learn how to sort a list in the natural order. 1. 2) Does listA and listB contain references to the same objects, or just objects that are equivalent with equals()? So for me the requirement was to sort originalList with orderedList. The common non-linear data structure known as a tree. If you already have a dfwhy converting it to a list, process it, then convert to df again? Theoretically Correct vs Practical Notation. So in a nutshell, we can sort a list by simply calling: java.util.Collections.sort(the list) as shown in the following example: The above class creates a list of four integers and, using the collection sort method, sorts this list (in one line of code) without us having to worry about the sorting algorithm. 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. You get paid; we donate to tech nonprofits. 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. Overview to Sorting Stream and List on Multiple Fields Using Java 8 We perform sorting on stream and list of objects using the multiple fields using the Comparators and Comparator.thenComparing () method. Making statements based on opinion; back them up with references or personal experience. How to remove an element from a list by index, Sorting an array of objects by property values, String formatting: % vs. .format vs. f-string literal. Key Selector Variant. L1-50 first, L2-50 next, then, L2-45, L2-42, L1-40 and L1-30. MathJax reference. This is actually the proper way of doing it: when you sort a Factory, you cannot sort the inner competitors at the same time, because different objects are being compared. 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? [[name=a, age=age11], [name=a, age=age111], [name=a, age=age1], [name=b, age=age22], [name=b, age=age2], [name=c, age=age33], [name=c, age=age3]]. "Sunday" => 0, , "Saturday" => 6. The naive implementation that brute force searches listB would not be the best performance-wise, but would be functionally sufficient. 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. 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: And then sort your list of people by the order of their id in this mapping: Note: if a person has an ID that is not present in the ids, they will be placed first in the list. From simple plot types to ridge plots, surface plots and spectrograms - understand your data and learn to draw conclusions from it. then the question should be 'How to sort a dictionary? To subscribe to this RSS feed, copy and paste this URL into your RSS reader. 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. The signature of the method is: Let's see another example of Collections.sorts() method. Do you know if there is a way to sort multiple lists at once by one sorted index list? 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? A Comparator can be passed to Collections.sort () or List.sort () method to allow control over the sort order. Sorting Strings in reverse order is as simple as sorting integers in reverse order: In all of the previous examples, we've worked with Comparable types. To learn more, see our tips on writing great answers. Create a Map that maps the values of everything in listB to something that can be sorted easily, such as the index, i.e. Here is Whatangs answer if you want to get both sorted lists (python3). 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. Do roots of these polynomials approach the negative of the Euler-Mascheroni constant? Working on improving health and education, reducing inequality, and spurring economic growth? 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. Did you try it with the sample lists. Linear regulator thermal information missing in datasheet. Site design / logo 2023 Stack Exchange Inc; user contributions licensed under CC BY-SA. This can be elegantly solved with guava's Ordering.explicit: The last version of Guava thas supports Java 6 is Guava 20.0: First create a map, with sortedItem.name to its first index in the list. 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 first get the String values in a list. Java Sorting Java Sorting Learn to use Collections.sort () method to sort a list of objects using some examples. To avoid having a very inefficient look up, you should index the items in listB and then sort listA based on it. Solution based on bubble sort (same length required): If the object references should be the same, you can initialize listA new. In this tutorial we will sort the HashMap according to value. Rather than using a list to get values from the map, well be using LinkedHashMap to create the sorted hashmap directly. Get tutorials, guides, and dev jobs in your inbox. Why is "1000000000000000 in range(1000000000000001)" so fast in Python 3? The toList() return the collector which collects all the input elements into a list, in encounter order. Sorting values of a dictionary based on a list. HashMap in java provides quick lookups. How do you ensure that a red herring doesn't violate Chekhov's gun? Then we sort the list. How to match a specific column position till the end of line? Not the answer you're looking for? will be problematic in the future. Connect and share knowledge within a single location that is structured and easy to search. Here's a simple implementation of that logic. People will search this post looking to sort lists not dictionaries. I am wondering if there is any easier way to do it. See more examples here. That way, I can sort any list in the same order as the source list. Using Kolmogorov complexity to measure difficulty of problems? I mean swapItems(), removeItem(), addItem(), setItem() ?? You posted your solution two times. In Python 2, zip produced a list. Python. you can leverage that solution directly in your existing df. Speed improvement on JB Nizet's answer (from the suggestion he made himself). "After the incident", I started to be more careful not to trip over things. All of them simply return a comparator, with the passed function as the sorting key. All Rights Reserved. The Comparator.comparing () method accepts a method reference which serves as the basis of the comparison. In Java how do you sort one list based on another? What is the purpose of this D-shaped ring at the base of the tongue on my hiking boots? 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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