Algorithm to find continuous days in a week - algorithm

The user can select any number of week days from a list. An algorithm shall find the longest continuous group of selected days. The start day can be after the end day, if the group spans two weeks. If it makes it simpler, only a group of at least 3 days needs to be detected. With crossing the week border, this makes for a maximum of one group. (There can be no two groups of 3 days within a week that are not connected.)
For example, if the user selects Monday, Tuesday, Wednesday and Saturday from a list, the display should be something like "Monday-Wednesday and Saturday".
Another example is: Wed, Fri, Sat, Sun, Mon -> "Wed, Fri-Mon".
Is there an efficient algorithm for that, preferrably in C# or a similar language? My C# hackwork is now over a page long (incl. few comments) and still not finished.

Use this answer, slightly changed:
Use a modified version of dtb's GroupAdjacentBy which accepts a minCount as a parameter:
public static IEnumerable<IEnumerable<T>> GroupAdjacentBy<T>(
this IEnumerable<T> source, Func<T, T, bool> predicate, int minCount)
{
using (var e = source.GetEnumerator())
{
if (e.MoveNext())
{
var list = new List<T> { e.Current };
var pred = e.Current;
while (e.MoveNext())
{
// if adjacent, add to list
if (predicate(pred, e.Current))
{
list.Add(e.Current);
}
else
{
// otherwise return previous elements:
// if less than minCount elements,
// return each element separately
if (list.Count < minCount)
{
foreach (var i in list)
yield return new List<T> { i };
}
else
{
// otherwise return entire group
yield return list;
}
// create next group
list = new List<T> { e.Current };
}
pred = e.Current;
}
yield return list;
}
}
}
and change the criteria for GroupAdjacentBy to group on week transitions also:
// week starts with Monday, so this should
// represent: Wed, Fri, Sat, Sun, Mon
int[] array = new int[] { 1, 2, 4, 5, 6, 0 };
Func<int, int, bool> adjacentCriteria = (x, y) => (x+1==y) || (x==6 && y==0);
string result = string.Join(", ", array
.GroupAdjacentBy(adjacentCriteria, 3)
.Select(g => new int[] { g.First(), g.Last() }.Distinct())
.Select(g => string.Join("-", g)));
Console.WriteLine(result); // output: 1, 2, 4-0

I've finished my version of it. It's a bit longer than the other one, but then again it also handles the text representation and does exactly this task. How about that?
using System;
using System.Text;
namespace WeekMathTest
{
class Program
{
static void Main(string[] args)
{
string[] weekDayNames = new string[] {
"Mon",
"Tue",
"Wed",
"Thu",
"Fri",
"Sat",
"Sun"
};
WeekDays weekDays = WeekDays.Monday | WeekDays.Tuesday | WeekDays.Thursday | WeekDays.Saturday | WeekDays.Sunday;
Console.WriteLine(WeekDayGroup(weekDays, weekDayNames));
}
static string WeekDayGroup(WeekDays weekDays, string[] weekDayNames)
{
int groupStart = 0, groupEnd = 0, groupLength = 0;
int maxGroupStart = 0, maxGroupEnd = 0, maxGroupLength = 0;
// Iterate all days in a repeated range
// (Sat/Sun doesn't need to be repeated or it would be in the first group)
for (int day = 1; day <= 7 + 5; day++)
{
// Is this day set?
int bitValue = 1 << ((day - 1) % 7);
bool daySet = ((int) weekDays & bitValue) != 0;
if (daySet)
{
if (groupStart == 0)
{
// First day set, remember it as group start
groupStart = day;
groupEnd = day;
groupLength = 1;
}
else
{
// Group has already been started, set new end
groupEnd = day;
groupLength = groupEnd - groupStart + 1;
if (groupLength == 7)
{
// Seen every day of the week, stop here
break;
}
}
}
else
{
if (groupLength >= 3 && groupLength > maxGroupLength)
{
// Group was long enough and longer than the last one, save it
maxGroupStart = groupStart;
maxGroupEnd = groupEnd;
maxGroupLength = groupLength;
}
// Reset operation variables
groupStart = 0;
groupEnd = 0;
groupLength = 0;
}
}
// Final check
if (groupLength >= 3 && groupLength > maxGroupLength)
{
// Group was long enough and longer than the last one, save it
maxGroupStart = groupStart;
maxGroupEnd = groupEnd;
maxGroupLength = groupLength;
}
// Clear all group days from the original value
for (int day = maxGroupStart; day <= maxGroupEnd; day++)
{
int bitValue = 1 << ((day - 1) % 7);
weekDays = (WeekDays) ((int) weekDays & ~bitValue);
}
// Generate output string
StringBuilder sb = new StringBuilder();
for (int day = 1; day <= 7; day++)
{
int bitValue = 1 << ((day - 1) % 7);
bool daySet = ((int) weekDays & bitValue) != 0;
if (daySet)
{
if (sb.Length > 0) sb.Append(", ");
sb.Append(weekDayNames[day - 1]);
}
else if (day == maxGroupStart)
{
if (sb.Length > 0) sb.Append(", ");
sb.Append(weekDayNames[day - 1]);
sb.Append("-");
sb.Append(weekDayNames[(maxGroupEnd - 1) % 7]);
}
}
return sb.ToString();
}
[Flags]
enum WeekDays
{
Monday = 1,
Tuesday = 2,
Wednesday = 4,
Thursday = 8,
Friday = 16,
Saturday = 32,
Sunday = 64
}
}
}

Related

Google App Script Data Validation inserts too many dropdowns

I have a script that works mostly the way want. It looks at a cell then compares that to a column in another tab, finds the like items and returns that, and creates a dropdown on the cell. This moves down the column until it reaches the end. The problem is that it continues past the last row for about 20 rows. The starting row is row24.
function getInventoryItems() {
var jobSummaryInventoryItems = jobSummary.getRange(24, 8, jobSummary.getLastRow(), 1);
var jobSummaryInventoryItemsValues = jobSummaryInventoryItems.getValues();
var inventoryItems = inventory.getRange(4, 3, inventory.getLastRow(), 1);
var inventoryItemsValues = inventoryItems.getValues();
jobSummary.getRange(24, 8, jobSummary.getLastRow(), 1).setDataValidation(null);
for (z = 0; z < jobSummaryInventoryItemsValues.length; z++) {
if (jobSummaryInventoryItemsValues[z].toString().length > 1) {
var listOfInventory = [];
for (i = 0; i < inventoryItems.getLastRow() - 4; i++) {
if (inventoryItemsValues[i].toString() == jobSummaryInventoryItemsValues[z]) {
break;
}
var w = jobSummaryInventoryItemsValues[z];
if (inventoryItemsValues[i].toString().includes(jobSummaryInventoryItemsValues[z])) {
listOfInventory.push(inventoryItemsValues[i].toString());
}
}
}
if (listOfInventory.length > 0) {
var rangeRule = SpreadsheetApp.newDataValidation().requireValueInList(listOfInventory).build();
jobSummary.getRange(z + 24, 8).setDataValidation(rangeRule);
}
}
Get Inventory Items
function getInventoryItems() {
const ss = SpreadsheetApp.getActive();
const jobSummary = ss.getSheetByName('Job Summary');
const inventory = ss.getSheetByName('Inventory');
const jobSummaryInventoryItems = jobSummary.getRange(24, 8, jobSummary.getLastRow() - 23, 1);
const jobSummaryInventoryItemsValues = jobSummaryInventoryItems.getValues();
const inventoryItems = inventory.getRange(4, 3, inventory.getLastRow() - 3, 1);
const inventoryItemsValues = inventoryItems.getValues();
jobSummary.getRange(24, 8, jobSummary.getLastRow() -23, 1).setDataValidation(null);
for (z = 0; z < jobSummaryInventoryItemsValues.length; z++) {
if (jobSummaryInventoryItemsValues[z].toString().length > 1) {
let listOfInventory = [];
for (i = 0; i < inventoryItems.length; i++) {
if (inventoryItemsValues[i].toString() == jobSummaryInventoryItemsValues[z]) {
break;
}
let w = jobSummaryInventoryItemsValues[z];
if (inventoryItemsValues[i].toString().includes(jobSummaryInventoryItemsValues[z])) {
listOfInventory.push(inventoryItemsValues[i].toString());
}
}
}
if (listOfInventory.length > 0) {
let rangeRule = SpreadsheetApp.newDataValidation().requireValueInList(listOfInventory).build();
jobSummary.getRange(z + 24, 8).setDataValidation(rangeRule);
}
}
}
Sheet.getRange(row,column,number of rows, number of columns)

Class is taking more computation time if the same input is given over and over

I am working on an algorithm and it seems to be working fine, apart from one thing.
Let me first show you the code and then I will explain what the code does and what the problem is.
public Triple<List<ROUTE>, Integer, List<Customer>> LocalSearch()
{
int noImprLS = 0;
boolean initialization = false;
List<ROUTE> bestRoutes = startRoutes;
int bestProfit = profit;
List<Customer> bestU = u;
List<ROUTE> tempBestRoutes = startRoutes;
int tempBestProfit = profit;
List<Customer> tempBestU = u;
int tempBestDistance = totalDistance(tempBestRoutes);
ELIMINATOR e = new ELIMINATOR(bestU, bestRoutes, bestProfit, initialization, name, rnd);
while (noImprLS <= noImprUB)
{
System.out.print(noImprLS);
boolean improvement = false;
long starttime = System.nanoTime();
double timeE = 0;
for (int i = 1; i <= N; i++)
{
long starttimeE = System.nanoTime();
e = new ELIMINATOR(bestU, bestRoutes, bestProfit, initialization, name, rnd);
timeE = timeE + (System.nanoTime()-starttimeE)/1000000000.0;
POSTPROCEDURE pp = new POSTPROCEDURE(e.getRoutes(), profitRoutes(e.getRoutes()), e.getU(), name);
for (int p = 0; p < pp.getBestSolution().size(); p++)
{
ROUTE r = pp.getBestSolution().get(p);
addToPOOL(r);
}
int tempprofit = pp.getTP();
int tempdistance = pp.getTD();
if (tempprofit > tempBestProfit)
{
tempBestRoutes = pp.getBestSolution();
tempBestProfit = tempprofit;
tempBestU = pp.getU();
tempBestDistance = tempdistance;
}
else if (tempprofit == tempBestProfit)
{
if (tempdistance < tempBestDistance)
{
tempBestRoutes = pp.getBestSolution();
tempBestProfit = tempprofit;
tempBestU = pp.getU();
tempBestDistance = tempdistance;
}
}
}
if (tempBestProfit > bestProfit)
{
// Move to better neighbor
bestRoutes = tempBestRoutes;
bestProfit = tempBestProfit;
bestU = tempBestU;
noImprLS = 0;
improvement = true;
System.out.print(" total profit: " + bestProfit);
}
else if (tempBestProfit == bestProfit)
{
if (totalDistance(tempBestRoutes) < totalDistance(bestRoutes))
{
// Move to better neighbor
bestRoutes = tempBestRoutes;
bestProfit = tempBestProfit;
bestU = tempBestU;
noImprLS = 0;
improvement = true;
System.out.print(" total profit: " + bestProfit + " total distance: " + totalDistance(bestRoutes));
}
}
if (improvement == false)
{
noImprLS++;
}
long endtime = System.nanoTime();
double duration = (endtime - starttime)/1000000000.0;
System.out.print(" duration: " + duration + " timeE: " + timeE + "\n");
}
Explanation
I know that the code is quite lengthy, but it is all quite important. In this code, I am writing an algorithm for the Team Orienteering Problem with Time Windows (extensive case of the Vehicle Routing Problems). My aim is to find a good set of routes with maximum profit. In the example below, bestRoutes and tempBestRoutes consist of 4 different routes, profit (bestProfit/tempBestProfit) is equal to the total profit of these routes respectively, and (temp)bestU is a list of customers that are not included in my route yet.
The problem now is with ELIMINATOR. This method removes and adds some customers. The output of this class is used for PostProcedure that also changes some facts in the routes.
I hope it is kind of clear now what my code is doing. I am considering N neighbourhoods and I will choose the best one. If the best one is not better than my starting solution, I increase noImprLS with one. I keep on considering new nieghbours until my upperbound on the number of consecutive iterations without improvement is met.
Problem
The problem now is that if I have not found a better solution, and hence I keep on inserting the same routes and profit in ELIMINATOR, my computation time increases.
A few examples where duration indicates how long an iteration within the while loop takes, and timeE indicates what the total time of ELIMINATOR in the for loop is. It is clear that ELIMINATOR causees the duration to increase.
0 total profit: 800 duration: 0.486570471 timeE: 0.16644330999999998
0 total profit: 900 duration: 0.431213528 timeE: 0.11342619799999998
0 total profit: 950 duration: 0.444671005 timeE: 0.12090608200000001
0 total profit: 960 duration: 0.519406695 timeE: 0.16836757300000005
0 duration: 0.460473438 timeE: 0.137813155
1 duration: 0.572109775 timeE: 0.30774360900000003
2 duration: 0.698965292 timeE: 0.471859029
3 duration: 0.918376211 timeE: 0.686916669
4 duration: 1.165481175 timeE: 0.92621492
5 duration: 1.326080436 timeE: 1.0874366910000002
6 duration: 2.006102605 timeE: 1.674879135
7 duration: 2.787172112 timeE: 2.4276636639999993
8 duration: 2.042213493 timeE: 1.7967797849999998
9 duration: 2.652985618 timeE: 2.3503671230000003
10 duration: 2.422183993 timeE: 2.1859969810000006
The ELIMINATOR CODE:
import java.util.ArrayList;
import java.util.Iterator;
import java.util.List;
import java.util.Random;
public class ELIMINATOR extends BASIS
{
private List<Customer> u;
private List<ROUTE> routes;
private int totalprofit;
private Random prob;
public ELIMINATOR(List<Customer> u, List<ROUTE> routes, int profit, boolean initialization, String name, Random rnd)
{
super(name);
this.u = u;
this.routes = routes;
this.totalprofit = profit;
this.prob = rnd;
if (initialization == true)
{
addCustomers();
for (ROUTE route : routes)
{
checkCorrectNess(route, "after adding procedure in eliminator");
}
}
else
{
removeCustomers();
for (ROUTE route : routes)
{
checkCorrectNess(route, "after removing procedure in eliminator");
}
addCustomers();
for (ROUTE route : routes)
{
checkCorrectNess(route, "after removing and adding procedure in eliminator");
}
}
}
public void removeCustomers()
{
double Ph = 0.1;
double Pl = 0.3;
double total_profit = totalprofit;
int num_customers = 0;
// Calculate the total profit and total number of customers in the routes
for(ROUTE route : routes)
{
num_customers = num_customers + (route.getLocations().size()-2);
}
// Calculate average profit
double average_profit = total_profit/num_customers;
// For each customer on each route, determine whether he/she will be removed
for(ROUTE r : routes)
{
List<RouteNode> route = r.getLocations();
int routesize = route.size();
int j = 1;
while (j < routesize-1)
{
boolean removed = false;
RouteNode node = route.get(j);
if (node.customer.getProfit() >= average_profit)
{
if (prob.nextDouble() < Ph)
{
removed = true;
RouteNode node_toberemoved = node;
int index_node = route.indexOf(node);
route.remove(index_node);
u.add(node.customer);
route = removal(route, node_toberemoved, index_node);
r.setLocations(route);
r.setDistance(distanceOneRoute(route));
r.setProfit(profitOneRoute(route));
checkCorrectNess(r, "remove customers eliminator");
}
}
else
{
if (prob.nextDouble() < Pl)
{
removed = true;
RouteNode node_toberemoved = node;
int index_node = route.indexOf(node);
route.remove(index_node);
u.add(node.customer);
route = removal(route, node_toberemoved, index_node);
r.setLocations(route);
r.setDistance(distanceOneRoute(route));
r.setProfit(profitOneRoute(route));
checkCorrectNess(r, "remove customers eliminator");
}
}
if (removed == false)
{
j++;
}
else
{
routesize = route.size();
total_profit = total_profit-node.customer.getProfit();
average_profit = total_profit/num_customers;
}
}
}
totalprofit = profitRoutes(routes);
}
public void addCustomers()
{
List<Customer> u_copy = new ArrayList<Customer>(u);
List<Customer> u_temp = new ArrayList<Customer>(u);
for (Customer c : u_temp)
{
boolean added = false;
for (ROUTE r : routes)
{
checkCorrectNess(r, "add customers eliminator");
if (added == true)
{
break;
}
Customer customer = c;
u_copy.remove(c);
List<RouteNode> route = r.getLocations();
for (int i = 0; i < route.size()-1; i++)
{
RouteNode possibleNode = new RouteNode();
possibleNode.customer = customer;
List<Integer> distances = calculateDistances(route.get(i), possibleNode, route.get(i+1));
// Calculate shift for customer under consideration
int arrivalTime = route.get(i).timeStartService+ route.get(i).customer.getService() + distances.get(0);
int wait = Math.max(0, customer.getOpeningTW()-arrivalTime);
int serviceDuration = customer.getService();
int shift = distances.get(0) + wait + serviceDuration + distances.get(2) - distances.get(1);
// Determine Start Service
int startServiceTime = Math.max(customer.getOpeningTW(), arrivalTime);
// Obtain waiting time of next customer
int waiting_next = route.get(i+1).wait;
// Obtain MaxShift of next customer
int maxShift = route.get(i+1).maxShift;
if (shift <= (waiting_next + maxShift) & startServiceTime <= customer.getClosingTW() )
{
// Customer can be inserted
added = true;
RouteNode newNode = new RouteNode();
newNode.customer = customer;
newNode.arrivalTime = arrivalTime;
newNode.timeStartService = startServiceTime;
newNode.shift = shift;
newNode.wait = wait;
int pos_insertion = i + 1;
route = ADD(route, newNode, pos_insertion);
r.setLocations(route);
r.setDistance(distanceOneRoute(route));
r.setProfit(profitOneRoute(route));
checkCorrectNess(r, "add customers eliminator");
// exit the last for loop
break;
}
}
}
if (added == false)
{
u_copy.add(c);
}
}
u = u_copy;
totalprofit = profitRoutes(routes);
}
/**
* Returns list of unvisited customers
* #return
*/
public List<Customer> getU()
{
return u;
}
/**
* Returns list of routes
* #return
*/
public List<ROUTE> getRoutes()
{
return routes;
}
}

Sorting a DataTable using LINQ

I have a DataTable which has the following structure.
StartDate (Type DateTime)
EndDate (Type DateTime)
Description (Type Text)
What I need to check is if the dates overlap or not, i.e., if the Start Date of 1 item is less than the EndDate of the previous item.
Here is the code I wrote.
NB: I already have checks to ensure that for each record, the Start Date has to be lesser than the End Date
dtDates.DefaultView.Sort = "StartDate ASC"; //Sort the DataTable based on the Start Dates
if(dtDates.Rows.Count > 1) //Check only if more than 1 Date is specified
{
for (int i = 1; i < dtDates.Rows.Count; i++)
{
if(TypeConvert.ToDate(dtDates.Rows[i]["StartDate"] < TypeConvert.ToDate(dtDates.Rows[i-1]["EndDate"])
{
retVal = true;
currentRow = i;
break;
}
}
}
The above code works fine.
But now, I want to implement it using LINQ
So, here is the code I am trying to attempt.
How can I do the whole comparison using LINQ?
public static bool CheckIfDatesOverlap(DataTable dataTable)
{
bool retVal = false;
int currentRow = 0;
List<DataRow> listDates = (from DataRow dgv in dataTable.Rows
select dgv).OrderBy(x => x[StartDate]).ToList();
if (listDates.Count > 1) //Perform Comparision only if more than 1 row is present
{
for (int i = 1; i < listDates.Count; i++)
{
if (TypeConvert.ToDateTime(listDates[i][StartDate]) < TypeConvert.ToDateTime(listDates[i][EndDate]))
{
retVal = true; //There are duplicates, hence return true
currentRow = i;
break;
}
else
{
//Do nothing as dates do not overlap
}
}
}
else
{
retVal = false; //As there is only 1 row, return false
}
if (retVal)
{
string message = "Dates Overlap";
//Display message
}
return retVal;
}
If you use SelectWithPrevious described here then you can write the following:
bool datesOverlap = table.Rows
.Cast<DataRow>()
.SelectWithPrevious((current, previous) => new
{
PreviousEndDate = (DateTime)previous["EndDate"],
StartDate = (DateTime)current["StartDate"]
})
.Any(x => x.StartDate < x.PreviousEndDate);

JTable + sorting column, but row is not sorted

I have created a JTable, the table contains 4 rows, and for a specific row, I have overridden the sorting and its working fine only on that column.
Status Scheduled Date Scheduled time Status
false 30/01/2012 02:00:00 Scheduled
false 29/01/2012 14:58:00 Scheduled
false 29/01/2012 15:50:00 Scheduled
For Scheduled Date, which I try to sort, it would sort, but the respecitve rows are not being updated.
Here is my code for sorting
public static void sortColumn(DefaultTableModel model, int colIndex,
boolean sortingOrder) {
Vector<?> data = model.getDataVector();
Object[] colData = new Object[model.getRowCount()];
SortedSet<Object> dataCollected = null;
List<Date> dateCollected;
boolean dateFlag = false;
dateCollected = new ArrayList<Date>();
// Copy the column data in an array
for (int i = 0; i < colData.length; i++) {
Object tempData = ((Vector<?>) data.get(i)).get(colIndex);
if ((colIndex == 1 || colIndex == 4)
&& tempData.toString().contains("/")) {
String[] _scheduledDate1 = ((String) tempData).split("/");
Calendar _cal1 = Calendar.getInstance();
_cal1.set(Integer.parseInt(_scheduledDate1[2]),
Integer.parseInt(_scheduledDate1[1]) - 1,
Integer.parseInt(_scheduledDate1[0]));
dateCollected.add(_cal1.getTime());
dateFlag = true;
} else {
colData[i] = ((Vector<?>) data.get(i)).get(colIndex);
}
}
// DateCompare compare = new DateCompare();
if (!dateFlag) {
dataCollected = new TreeSet<Object>();
dataCollected.add(colData);
dateFlag = false;
}
// Copy the sorted values back into the table model
if ((colIndex == 1 || colIndex == 4) && dateFlag) {
SimpleDateFormat sdf = new SimpleDateFormat("dd/MM/yyyy");
sortOrder = !sortOrder;
if (sortOrder) {
Collections.sort(dateCollected);
} else {
Collections.sort(dateCollected, Collections.reverseOrder());
}
colData = dateCollected.toArray();
for (int i = 0; i < colData.length; i++) {
((Vector<Object>) data.get(i)).set(colIndex,
sdf.format(((Date) colData[i]).getTime()));
}
} else {
for (int i = 0; i < colData.length; i++) {
((Vector<Object>) data.get(i)).set(colIndex, colData[i]);
}
}
model.fireTableStructureChanged();
}
How to I get the entire row update accordingly?
I found the problem, my object against I was comparing was wrong, I've change the code for the same it all works fine.
I implemented QuickSort algorithm to sort the vector on the specific column I need.

How to get the Max() of a Count() with LINQ

I'm new to LINQ and I have this situation. I have this table:
ID Date Range
1 10/10/10 9-10
2 10/10/10 9-10
3 10/10/10 9-10
4 10/10/10 8-9
5 10/11/10 1-2
6 10/11/10 1-2
7 10/12/10 5-6
I just want to list the Maximun of rows per date by range, like this:
Date Range Total
10/10/10 9-10 3
10/11/10 1-2 2
10/12/10 5-6 1
I want to do this by using LINQ, do you have any ideas of how to do this?
I think something along these lines should work:
List<MyTable> items = GetItems();
var orderedByMax = from i in items
group i by i.Date into g
let q = g.GroupBy(i => i.Range)
.Select(g2 => new {Range = g2.Key, Count = g2.Count()})
.OrderByDescending(i => i.Count)
let max = q.FirstOrDefault()
select new {
Date = g.Key,
Range = max.Range,
Total = max.Count
};
Using extension methods:
List<MyTable> items = GetItems();
var rangeTotals = items.GroupBy(x => new { x.Date, x.Range }) // Group by Date + Range
.Select(g => new {
Date = g.Key.Date,
Range = g.Key.Range,
Total = g.Count() // Count total of identical ranges per date
});
var rangeMaxTotals = rangeTotals.Where(rt => !rangeTotals.Any(z => z.Date == rt.Date && z.Total > rt.Total)); // Get maximum totals for each date
unfortunately I can't test this at the moment but give this a try:
List<MyTable> items = GetItems();
items.Max(t=>t.Range.Distinct().Count());
This approach:
1) Groups by Date
2) For each Date, groups by Range and calculates the Total
3) For each Date, selects the item with the greatest Total
4) You end up with your result
public sealed class Program
{
public static void Main(string[] args)
{
var items = new[]
{
new { ID = 1, Date = new DateTime(10, 10, 10), Range = "9-10" },
new { ID = 2, Date = new DateTime(10, 10, 10), Range = "9-10" },
new { ID = 3, Date = new DateTime(10, 10, 10), Range = "9-10" },
new { ID = 4, Date = new DateTime(10, 10, 10), Range = "8-9" },
new { ID = 5, Date = new DateTime(10, 10, 11), Range = "1-2" },
new { ID = 6, Date = new DateTime(10, 10, 11), Range = "1-2" },
new { ID = 7, Date = new DateTime(10, 10, 12), Range = "5-6" },
};
var itemsWithTotals = items
.GroupBy(item => item.Date) // Group by Date.
.Select(groupByDate => groupByDate
.GroupBy(item => item.Range) // Group by Range.
.Select(groupByRange => new
{
Date = groupByDate.Key,
Range = groupByRange.Key,
Total = groupByRange.Count()
}) // Got the totals for each grouping.
.MaxElement(item => item.Total)); // For each Date, grab the item (grouped by Range) with the greatest Total.
foreach (var item in itemsWithTotals)
Console.WriteLine("{0} {1} {2}", item.Date.ToShortDateString(), item.Range, item.Total);
Console.Read();
}
}
/// <summary>
/// From the book LINQ in Action, Listing 5.35.
/// </summary>
static class ExtensionMethods
{
public static TElement MaxElement<TElement, TData>(this IEnumerable<TElement> source, Func<TElement, TData> selector) where TData : IComparable<TData>
{
if (source == null)
throw new ArgumentNullException("source");
if (selector == null)
throw new ArgumentNullException("selector");
bool firstElement = true;
TElement result = default(TElement);
TData maxValue = default(TData);
foreach (TElement element in source)
{
var candidate = selector(element);
if (firstElement || (candidate.CompareTo(maxValue) > 0))
{
firstElement = false;
maxValue = candidate;
result = element;
}
}
return result;
}
}
According to LINQ in Action (Chapter 5.3.3 - Will LINQ to Objects hurt the performance of my code?), using the MaxElement extension method is one of the most effecient approaches. I think the performance would be O(4n); one for the first GroupBy, two for the second GroupBy, three for the Count(), and four for loop within MaxElement.
DrDro's approach is going to be more like O(n^2) since it loops the entire list for each item in the list.
StriplingWarrior's approach is going to be closer to O(n log n) because it sorts the items. Though I'll admit, there may be some crazy magic in there that I don't understand.

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