since you just want code it sounds like from operator import itemgetter sorted(map(itemgetter(1,-1),lists),key=itemgetter(-1)) heres some code that should do what you want … a teacher may want an explanation however… just fair warning solved Python: Sorting and printing lists within lists [closed]
Data = [ ‘<td>1</td>’, ‘<td>2</td>’, ‘<td>3</td>’, ‘<td>4</td>’, ‘<td>A</td>’, ‘<td>B</td>’, ‘<td>C</td>’, ‘<td>D</td>’, ‘<td>I</td>’, ‘<td>II</td>’, ‘<td>III</td>’, ‘<td>IV</td>’, ] lists, result = [], [] for i in range(0, len(Data), 4): lists.append(Data[i:i+4]) for currentList in zip(*lists): result += list(currentList) print result Output [‘<td>1</td>’, ‘<td>A</td>’, ‘<td>I</td>’, ‘<td>2</td>’, ‘<td>B</td>’, ‘<td>II</td>’, ‘<td>3</td>’, ‘<td>C</td>’, ‘<td>III</td>’, ‘<td>4</td>’, ‘<td>D</td>’, ‘<td>IV</td>’] solved Python sort list by … Read more
Assuming I understood your question, these should do it: for person in sorted(cat1): print(person, max(cat1.get(person))) result: ben 9 jeff 6 sam 9 then: for person in sorted(cat1, key=lambda x: max(cat1.get(x)), reverse=True): print(person, max(cat1.get(person))) result: ben 9 sam 9 jeff 6 then: for person in sorted(cat1, key=lambda x: sum(cat1.get(x))/len(cat1.get(x)), reverse=True): print(person, sum(cat1.get(person))/len(cat1.get(person))) result: ben 7.666666666666667 sam … Read more
It would be more readable if you kept the code formatting for the whole code, not just its part. But when I tried to find a main function, I didn’t find any. Therefore there is no reference to main because you haven’t implemented one, at least not in the code you posted here. solved Error … Read more
Besides the Map option, you can consider using some sort of Key value pair and, storing that in your array and sort your array based on the values. See this for potential key value pairs, or make your own. An example: package com.example; import java.util.ArrayList; import java.util.Collections; import java.util.Comparator; import java.util.List; public class KeyValue { … Read more
Most red black tree packages will have an “insert element” method. If you’re not using one already, it might be good to start. If you’re married to a red-black tree implementation that doesn’t have an insert element operation, it’d be a good idea to add such a method, possibly from some good red-black tree doc: … Read more
The question may be a case of “I have X and I need Y” where X is the item which needs attention. If the string really is as you presented it, then Imports System.Text Module Module1 Sub Main() Dim s = “{ “”0″”:{“”variable1″”:””ABC1″”,””variable2″”:””AA””,””variable3″”:””BB””}, “”5″”:{“”variable1″”:””ABC2″”,””variable2″”:””AA””,””variable3″”:””BB””}, “”3″”:{“”variable1″”:””BC3″”,””variable2″”:””AA””,””variable3″”:””BB””}, “”1″”:{“”variable1″”:””DC4″”,””variable2″”:””AA””,””variable3″”:””BB””}, “”4″”:{“”variable1″”:””DD5″”,””variable2″”:””AA””,””variable3″”:””BB””} }” Dim t = s.Split({vbCrLf}, StringSplitOptions.None) Dim u … Read more
The sorting algorithm for Arrays.sort(int[] a) is a tuned quicksort. Ref: https://docs.oracle.com/javase/6/docs/api/java/util/Arrays.html#sort(int[]) Read this article: http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.14.8162&rep=rep1&type=pdf for understanding the Quick-sort used in Array.sort() by JON L. BENTLEY & M. DOUGLAS McILROY In Java 7 pivot dual quick-sort algorithm is used for Arrays.sort(int[] a) i.e. refer this: What’s the difference of dual pivot quick sort and … Read more
Instead of comparing values pointed to by the pointers in statements like this if (n->me < curr->me) you are comparing pointers themselves. You should rewrite the condition like if ( *n->me < *curr->me ) solved insertion sort using linked list and pointers
you can use the Sort overload accepting a Comparison. This should work: public static int MyCompare(double x, double y) { if (x >= 0.0 == y>=0.0) // same sign, compare by absolute value return Math.Abs(x).CompareTo(Math.Abs(y)); if (x < 0.0) return 1; return -1; } usage: var list = new List<double>(); // fill your list // … Read more
In C#, what does a call to Sort with two parameters in brackets mean? You have this line of code: nodes.Sort((x, y) => distances[x] – distances[y]); You are not passing two parameters to the sort method but you are passing one parameter which is a delegate that takes 2 parameters. You are essentially doing the … Read more
var result = string.trim().split(“, “).map(el=>el.split(“|”).map(n=>+n)) this splits the string into an array of these groups ( yet as string ) “1|2, 3|4” => [“1|2″,”3|4”] then maps this array to a new array containing the strings splitted and converted to numbers: => [[1,2],[3,4]] 2 solved Convert string into several integer arrays [duplicate]
From your description, I think you are probably looking for topological sorting. It is based on the assumption that ‘impossible situation’ occurs when one connections suggests that A comes before B but there is some another connection which suggests that B comes before A. Link for topological sort: Topological Sorting solved Looking for a sorting … Read more
dictA = {“f1” : [“m”], “f2” : [“p”,”y”,”o”,”a”,”s”,”d”,”f”,”g”], “f3” : [“w”,”t”], “f5” : [“z”,”x”], “f6” : [“c”,”v”]} result = [] limit_size = 3 values_list = [] for values in dictA.itervalues(): values_list.append(len(values)) for i in range(0,max(values_list)): keys = list(dictA.keys()) count = 0 while count < len(keys): try: result.append(dictA[keys[count]][i]) except IndexError: dictA.pop(keys[count]) count = count + 1 … Read more
Eric Berry wrote a handy class that compares Strings by human values instead of traditional machine values. Below is a modified version of it along with Object comparator (what I think you are looking for) and its testing class. An example on how to use the string comparator: Map<String,String> humanSortedMap = new TreeMap<>(new AlphaNumericStringComparator()); An … Read more